maths chapter wise important questions
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1
QUESTION BANK II PUC SCIENCE
I. Very Short answer questions. (1x19=19)
1. Define Symmetric relation.
Ans: A relation „R‟ on the set „A‟ is said to be symmetric if for all a, b, A, aRb Implies
bRa. i.e. (a, b) R (b, a) R
2. Let A = {4, 6, 8, 20} R = {(a, b); a + b = 25, a, b A} Show that the relation „R‟ Is
empty.
Ans: This is an empty set, as no pair (a, b) satisfies the condition a + b = 25.
3. Give an example of a relation defined on a suitable set which is
i. reflexive, symmetric and transitive.
ii. reflexive, symmetric but not transitive.
iii. reflexive, transitive but not symmetric.
iv. symmetric, transitive but not reflexive.
vi. symmetric, but not reflexive and not transitive.
vii. not reflexive, not symmetric, and not transitive.
Solution: Consider a Set A = {a, b, c}
i. Define a relation R, on „A‟ as.
R1 = {(a, a), (b, b), (c, c)}. Clearly R1 is reflexive, symmetric and transitive
ii. Consider the relation R2 on A as.
R2 = {(a, a), (b, b), (c, c), (a, b), (b, a), (a, c), (c, a)}.
R2 is symmetric, reflexive
But R2 is not transitive – for (b, a) R2 and (a, c) R2 but (b, c) R2.
iii. Consider the relation R3 on A as
R3 = {(a, a), (b, b), (c, c), (a, b)}
R3 is reflexive and transitive but not symmetric for (a, b) R3 but (b, a) R3
iv. Consider the relation R4 on A as
R4 = {(a, a), (b, b), (c, c), (b, a)} is symmetric and transitive but not reflexive.
Since (c, c) R4.
v. Consider the relation R5 on A as
R5 = {(a, a), (b, b), (c, c), (a, b), (c, a)}
R5 is reflexive. R5 is not symmetric because (a, b) R5 but (b, a) R5 Also R5 is
not transitive because (c, a) R5, (a, b) R5 but (c, b) R5.
vi. Consider the relation R6 on A by
R6 = {(a, a), (c, c), (a, b), (b, a)}
R6 is Symmetric but R6 is not reflexive because (b, b) R6 . Also R6 is not transitive
for (b, a) R6 , (a, b) R6 but (b, b) R6.
vii. The relation R7 defined on A as
R7 = {(a, b), (b, c) } is not reflexive, not symmetric and not transitive.
* FUNCTION * (One mark question and answers)
4. Let A = {1, 2, 3} B = {4, 5, 6, 7,} and let f = {(1, 4), (2, 5), (3, 6)} be a function from
„A‟ to „B‟. show that „f‟ is not onto.
Ans: 7 B has no pre image in A. so „f‟ is not onto.
5. If, f : R R is defined by f(x) = 4x-1 x R prove that „f‟ is one-one.
Solution: For any two elements x1, x2 R such that f(x1) = f (x2)
we have 4x1–1 = 4x2–1 4x1 = 4x2
x1 = x2 Hence „f‟ is one – one.
6. Define transitive relation.
2
Ans: A relation R on the set “S” is said to be transitive relation if aRb and bRc aRc.
i.e. if (a, b) R and (b, c) R (a, c) R.
7. Let f : R R be the function defined by f (x) = 2x-3 x R . write 1
f
Ans; 1
= f(x) = 2x–3 x = (y + 3)2
y
–1 –11 1
( ) = (y+3) i.e. ( ) = (x+3)2 2
f y f y
8. Define a binary operation on a set.
Ans: Let S be a non-empty set. The function *: S X S S which associates each ordered
pair (a, b) of the elements of S to a unique element of S denotes by a*b is called a
binary operation or a binary composition on S.
9. Determine whether or not each of the definition defined below is a binary operation
justify.
a) on Z+ defined by a*b = | a-b |
b) on Z+ defined by a*b = a
Solution: a) We have for a, b, Z+ a*b = | a – b |. We know the
| a – b | is always positive. + a, b Z , a*b = | a–b | is a positive integer.
Hence * is a binary operation on Z+
b) Cleary a*b = a Z for all ,a b Z
Thus * is a binary operation on Z+
* SHORT ANSINER QUESTIONS* (Answers to two marks questions)
10. Define an equivalence relation. Give on example of a relation which is transitive but
not reflexive.
Solution: A relation “R” on a Set S is called on equivalence relations. if „R‟ is reflexive,
symmetric, and transitive.
Ex. The relation “<” (less than) defined on the set R of all real numbers is transitive,
but not reflexive.
11. Show that the signum function f : R R
defined by
1 x > 0
( ) 0 x = 0
–1 x < 0
if
f x if
if
is neither one-one nor onto.
Solution: From the graph of the function we have
f(2) = 1 and f(3) = 1, i.e. f(2) = f(3) = 1 but 2 3
„f‟ is not one-one
Again for 4 R (co domain) there exists no x R (domain)
Such that f(x) = 4 because f(x) = 1 or –1 for x 0.
„f‟ is not onto.
3
12. State whether the function f : R defined by f(x) = 1 + x2 x R is one-one, onto
or objective justify your answer.
Solution: We have f(x) = 1 + x2 for all x R .
clearly we observe that f(x) = 1 + 22 = 5 and f(–2) = 1 + (–2)
2 = 5 i.e, f(2) = f(–2)
but 2 –2 „f‟ is not one-one.
Again for y –2 there exist no real number x, such that f(x) = –2 because if f(x) = –2
1 + x2 = –2 x
2 = –3
x = –3 R
„f‟ is not onto.
13. Show that the greatest integer function f : R R defined by f(x) = [x] is neither one-
one nor onto.
Solution: We have f(x) = [x] = greatest integer less than or equal to x.
i.e. [2.3] = 2, [2] = 2 but 2.3 2.
Now we have [x] = 2 for all x [2, 3]
„f‟ is not one-one.
Again 3
2R (codomain), but there existing no x R (domain) such that f(x) = 3
2
because [x] is always an integer x R
„f‟ is not onto.
14. Cheek the injectivity and surjectivity of the function f : Z Z defined by f(x) = x3
x Z
Solution: We have f(x1) = f(x2) 3 3
1 2 1 2x x x x
„f‟ is injective.
Now 7 Z (codomain) and it has no pre-image in Z because if f(x) = 7 then 3 37 7x x Z
„f‟ is not surjective.
15. If f(x) = | x | and g(x) = | 5x – 2 | then find (i) got and (ii) fog.
Solution: We have f(x) = | x | and g(x) = | 5x–2 |
consider go f(x) = g( | x | ) = 5 | x | – 2
| 5 – 2 | 0
= | –5 – 2 | 0
x x
x x
Now fog(x) = f [g(x)] = f( |5x–2 | )
= | 5x–2 |
16. If f ; R R be given by 3( ) 3 –f x x1
3
find fo f(x).
Solution: We have 1
33( ) 3 –f x x
Now to 1
33( ) ( ) 3 – = f(y)fof x f f x f x
where 1
33 = (3 –y x
1
33( ) ( ) 3 –fof x f y y
1
3 31 13 33 3= 3 – 3 – = (x ) .x x
Hence fo f(x) = x.
4
17. Consider f : {1, 2, 3} {a, b, c} given by f(1) = a, f(2) = b f(3) = C find –1
f and
show that
–1–1
f f
Solution: Consider f : {1, 2, 3} {a, b, c} given by
f(1) = a, f(2) = b and f(3) = C
(1) , (2) and f(3)=4f a f b
1 1
( ) 1, ( ) 2f a f b
and 1
( ) 3f c
–1
,1 , ,2 , ,3 ( )f a b c g say
we observe that ‟g‟ is also objective.
–1
1, , 2, , 3,g a b c f
–1–1 –1
g f f f
18. If f(x) = 4 3 2,
36 – 4
xx
x
show that
(fof) (x) = x for all 23
x . what is the inverse f ?
Solution:
4 34 3
4 ( ) 3 6 – 4( ) = 4 36 ( ) – 4
6. – 46 – 4
x
f x xfof x f f xxf x
x
4 (4 3) 3(6 – 4) 34
6 (4 3) – 4( – 4) 34
x x xx
x x
–1
fof I f f
19. Let f : R R be defined by f(x) = 10x+7, Find the function g : R R such that
gof = fog = Ig
Solution: We have f(x) = 10x+7
By data, gof(x) = Ig(x) g [f(x)] = x.
g [10x+7] = x.
Let y = 10x+7 x – 7
10
y
Then g(10x+7) = – 7
10 710
yg y
20. Consider a function f : : 0,2
f R
given by f(x) = Sinx. and
g : : 0,2
f R
given by g(x) = cosx. Show that „f‟ and „g‟ are one-one but
(f + g) is not one-one.
Solution: Since for any two distinct elements x1 and x2 in 0,2
, sin x1 sin x2
and cos x1 cos x2
Both „f‟ and g are one-one
But (f + g) (o) = f(o) + g(o) = 1 and
5
= f sin cos 12 2 2 2 2
f g g
Therefore (f + g) is not one – one
21. Examine whether the binary operation * defined below are commutative, associative
a. On Q defined loy a * b = ab + 1
b. On Z+ defined loy a * b = a
b
c. On Q defined loy a * b = ab/2
Solution: a) For every a, b, Q we have
i) a * b = ab + 1 Q
Now a * b = ab + 1 = ba + 1 (Usual multiplication is commutative)
= b * a
* is commutative in Q.
ii) consider 4, 5, 6 Q.
4 * ( 5 * 6) = 4 * (30 + 1) = 4 * 31)
= 4 (31) + 1 = 124 + 1 = 125
and (4*5) * 6 = 21 * 6 = (21 x 6) + 1 = 157
a * (b * c) = (a * b) * c is not true for all a, b, c, Q. Hence * is not associative.
b) For all a, b, * Z+, clearly a * b = a
b Z
+
i.e. * is a binary operation in Z+
i) If a then ab b
a
i.e. a * b b * a for a b.
Thus * is not commutative in Z+
ii) Consider 2, 3, 4, Z+
(2 * 3) * 4 = 23 * 4 = 8 * 4 = 8
4 = 2
12
and 2 * (3 * 4) = 2 * 34 = (2)
34 = 2
81
( 2 * 3) * 4 2 * ( 3 * 4)
Thus (a*b)*c = a*(b*c) is not true , , ,a b c Z
Hence * is not associative in Z+.
C. We have a * b = a,b Q2
ab
i) Now a * b = 2
ab = 2
ba = b * a
(Usual multiplication is commutative)
* is commutative in Q.
ii. Now consider a * (b * c) = a * 2
bc
2
42
bca
abc
( * )* *2 4
ab abca b c c
(a * b) * c = a * (b * c) for all a, b, c, Q.
Hence * is associative in Q.
22. Write the multiplicative modulo 12 table for the set A = {1, 5, 7, 11}. Find the
identity element w.r .t X12 X12 1 5 7 11
Solution: The elements in the row corresponding 1 1 5 7 11
to the element 1, coincides with the elements 5 5 1 11 7
of a above the horizontal line in the same order. 7 7 11 1 5
Thus 1 acts as an identify element. 11 11 7 5 1
6
* ANSIEWERS TO THREE MARKS QUESTIONS*
23. The relation R defined on the Set A = {1, 2, 3, 4, 5} by R = {(a, b) ; | a–b | is even}.
Show that the relation „R‟ is on equivalence relation. (consider „O‟ as an even num
ber)
Proof: For any a A we have a–a = o considered as even i.e. (a, a,) R for all a A
Thus R is reflexive relation.
Let (a, b) R |a–b | is even
|b–a | is also even (b, a ) R.
Thus the relation R is symmetric.
Let (a, b) R and (b, c) R.
| a–b | is even and | b–c | even 1 1 2a b k and | b – c | =2l for k, l Z.
| a–b | + | b–c | = 2k + 2 l = 2 (k + l)
a–b + b–c = 2 (n) where n = (k+l)
a–c = 2n
| a–c | = even (a, c) R
„R‟ Is transitive.
Hence „R‟ Is an equivalence relation.
24. Let A = R–{3} and B = R–{1}. Consider the function f : A B defined by f(x) =
23
xx
Is „f‟ is one-one and onto ? Justify your answer.
Solution: We have f(x) = 2
3
x
x
for all x R
Now let f(x1) = f(x2) 1 2
2 2
2 2
3 3
x x
x x
1 2 2 1 1 2 1 22 3 6 2 3 6x x x x x x x x
1 2x x
Thus 1 2 1 2( ) ( )f x f x x x i.e. „f‟ is one-one.
Let y B thus .y l
Now 2
( ) 2 ( 3)3
xf x y y x y x
x
3 23 2 ( 1) ( 1).
1
yy x y x y
y
Also 2 3 2 3
3 for if 31 1
y y
y y
2 3 3 3 2 3y y which is not true.
Thus for every y B there exists
2 3
1
yx A
y
such that ( )f x y
i.e. „f‟ is onto.
Hence „f‟ is bijective function.
25. If „f‟ and g are two functions defined by f(x) = 2x + 5 and g(x) = x2 + 1 find 1 gof(2),
ii) fog(2) and iii) gog(2).
Solution: We have 2( ) 2 5, ( ) 1.f x x g x x
Now ( ) [ ( )] (2 5)gof x g f x g x
7
2( ) (2 5) 1gof x x
24 25 20 1x x
24 20 26x x
2(2) 4(2) 20(2) 26gof
16 40 26 82
ii) 2 2( ) ( 1) 2( 1) 5fog x f x x
2( ) 2 7fog x x
2(2) 2(2) 7 8 7 15fog
iii) 2( ) ( 1)gog x g x
2 2 4 2( 1) 1 2 2x x x
(2) 16 8 2 26gog
26. Let * be the binary operation on N given by a * b = l.c.m of „a‟ and „b‟ Is *
commutative? Is * associative? Find the identify in N w.r.t *.
Solution: We have a * b = l.c.m. of „a‟ and „b‟ ,a b N clearly l.c.m of two positive
integers is a positive integer. Thus N is closed under the operation *.
i) Clearly l.c.m of „a‟ and b = l.c.m. of „b‟ and „a‟
a*b = b*a for all a, b, N
Thus * is commutative.
ii) Let a, b, c, N be arbitrary.
Now a*(b * c) = a*(l.c.m. of b and c)
= l.c.m. of [„a‟ and (l.c.m of b and c)]
= l.c.m of (a,b,c) ................ (1)
Again, (a * b) * c
= (l.c.m of „a‟ and b) * c
= l.c.m of [(l.c.m. of „a‟ and b) and c]
= l.c.m of (a,b,c)................(2)
Thus (a * b) *c = a* (b * c) , ,a b c N
Hence * is associative.
iii) Let e be the identify element in N. Then for all a N, we have
a * e = e * a = a
* = l.c.m of a and e = a
* = l.c.m. of e and a = a
a e
e a
l.c.m of a and e = l.c.m. of e and a = a
for all a N. implies e = 1.
Thus 1 acts as identify element in N.
* FIVE MARKS QUESTIONS AND ANSWERS*
27. Consider : [ 5, )f R defined by 2( ) 9 6 5f x x x show that „f‟ is invertible
with –1 6 1
( )3
yf y
Solution: We have : [ 5, )f R by 2( ) 9 6 5f x x x
Let 1,a a R Such that f(a1) = f(a2)
2 2
1 1 2 29 6 5 9 6 5a a a a
8
2 2
1 2 1 29 9 6 6 0a a a a
1 2 1 2 1 29 6 0a a a a a a
1 2
2 2
1 2 9 9 6 0a a a a
1 2 1 2 1 20 9 9 6 as ,a a a a o a a R
1 2a a Thus „f‟ is one-one.
Let 2( ) 9 6 5y f x x x
29 6 5 0x x y
6 36 36( 5)
18
yx
1 1 5 1 6
3 3
y yx
1 1 6( )
3
yf y
For every element ,y B there exists a pre-Image x in [ 5, . So „f‟ is onto Thus „f‟
is one-one and onto and therefore invertible Hence inverse function of „f‟ is given by
1 1 6( )
3
yf y
28. If :f A A defined by 4 3
( )6 4
xf x
x
where 2 .
3A R show that „f‟ is
invertible and 1
f f
Solution: We have 4 3
( )6 4
xf x
x
with 2
3x
Now 1 21 2
1 2
4 3 4 3( ) ( )
6 4 6 4
x xf x f x
x x
1 2 1 2 1 2 1 224 16 18 12 24 18 16 12x x x x x x x x
2 1 1 234 34x x x x
Thus „f‟ is one-one.
Now let y A and f(x) ( )f x y
y A and 4 3
6 4
xy
x
y A and 4 3
6 3
yx
x
Thus for every y A , there exists
4 3 236 3
yx A y
y
such that
( ) .f x y That is „f‟ is onto.
Hence „f‟ is bijective.
Now 1
( ) ( )f x y f y x
9
4 3 4 3
6 4 6 4
y xx y
x x
Thus 1
:f A A
is defined by 1 4 3( )
6 4
xf x
x
Clearly 1
.f f
29. Let f; w :f w w be defined by
1 n is odd
( )1 if 'n' iseven
nf n
n
Show that „f‟ is invertible and find
The inverse of „f‟. Hence w is the set of all whole numbers.
Solution: Let x1, x2 w be arbitrary
Let x1 and x2 are even.
1 2 1 2 1 2( ) ( ) 1 1f x f x x x x x
Let 1x and 2x are odd
1 2 1 2 1 2( ) ( ) 1 1f x f x x x x x
Let x1 and x2 are odd 1 2 1 2( ) ( ) 1 1f x f x x x
1 2x x
In both the cases, 1 2 1 2( ) ( )f x f x x x
Supposing 1x is odd and 2x is even
then 1 2x x . Now 1 1( ) 1f x x and 2 2( ) 1f x x
Also 1 21 1x x
i.e. 1 2 1 2 ( ) ( )x x f x f x
Hence „f‟ is one-one
Let 0m w (co domain) and f(n) = m
Now 1 if 'n' is odd
( ) ( )1 if 'n' is even.
n mf n m f n
n m
1 if 'n' is odd
( )1 if 'n' is even.
n mf n
n m
Thus for every m w (co domain)
there exists n w (= m+1 or m–1)
Such that f(n) = m.
Also (considering O as even)
( ) 0 1 1f o i.e. 1
(1) 0f
Thus „f‟ is onto
Hence „f‟ is a bijection.
Let 1
( ) ( ) .f n m f m n
1 is odd
( )1 is even
m n mf m
m n m
1 is even
( )1 is odd
m n nf m
m n n
10
Thus 1 1 is even( )
1 is odd
n nf n
n n
Is inverse function.
INVERSE TRIGONOMETRIC FUNCTIONS
1 MARK QUESTIONS WITH SOLUTIONS
1. Find the principal value of (
√ )
Solution let (
√ )
√
(
√ )
2. Find the principal value of √
( √ )
√
(
)
( √ ) (
)
3. Find the principal value of (
)
Solution (
)
it does not lie b/w –
–
(
) * ( —
)+ * (
)+
(
)
solution
( (
))
, it does not lie b/w 0 and
( (
)) ( (
))
= (
)
=
(
)
5. ( √ )
√
√
( √ )
6. Evaluate ( ) [– ]
[– ]
[ ( )]
7. Find the value of
Solution we have
8. Find the value of
(
)
2 MARKS QUESTIONS WITH SOLUTIONS
(
) (
)
(
)
(
)
(
)
(
)
*
+
–
(
)
(
)
(
)
(
) (
)
*
–
+ (
)
*
+
(
)
*
+
(
)
* (
)+ (
)
(
)
(
)
(
)
(
)
(
)
(
)
√
√
√ –
(√
)
(√
) (
√
)
(
)
(
) (
) (
)
(
)
.
(
) (
)
(
) (
)
(
) *
(
)
(
)+
[ (
)]
(
)
(
)
(
)
(
)
3 marks Question with solutions
(
) (
)
(
) (
)
*
(
) (
)+
*
(
) (
)+
(
+
–
√
√ √
√
.
√ √
√
(
)
(
)
Topic: Matrices
Question bank with solutions
One mark question ( V S A)
1. Define matrix
2. Define a diagonal matrix
3. Define scalar matrix
4. Define symmetric matrix
5. Define skew-symmetric matrix
6.In a matrix [
2 5 19 −17
35 −25
2 12
√3 1 −5 17
]
find 1) order of the matrix
2) Write the elements of 𝑎13 , 𝑎21 , 𝑎33 , 𝑎24 , 𝑎23
7. If a matrix 8 elements what is the possible order it can have ?
8. If a matrix 18 elements what is the possible order it can have?
9. construct 2 × 2 matrix [𝑎𝑖𝑗] whose elements are given by
1) 𝑎𝑖𝑗 = (𝑖 + 𝑗) 2 2) 𝑎𝑖𝑗 = (𝑖+𝑗) 2
2
10. construct the 2 × 3 matrix whose elements are given by 𝑎𝑖𝑗= |𝑖 − 𝑗|
11. Construct the 3× 2 matrix whose elements are given by 𝑎𝑖𝑗= 𝑖
𝑗
12. Find x, y, z if [4 3𝑥 5
]= [𝑦 𝑧1 5
]
13. Find x, y, z if [𝑥 + 𝑦 25 + 𝑧 𝑥𝑦
]= [6 25 8
]
14. Find the matrix x such that 2A + B + X =0 where A = [−1 23 4
] and B = [3 −21 5
]
15. If A = [1 2 32 3 1
] B = [3 −1 3
−1 0 2] Find 2A – B
16. Find X if Y =[3 21 4
] and 2X+Y = [1 0
−3 2]
17. Find X If X+Y = [7 02 5
] and X-Y = [3 00 3
]
18. Simplify cos 𝜃 [cos 𝜃 sin 𝜃−sin 𝜃 cos 𝜃
] + sin 𝜃 [sin 𝜃 −cos 𝜃cos 𝜃 sin 𝜃
]
19. Find X If 2[1 30 𝑥
] + [𝑦 01 2
] = [5 61 8
]
20. If A = [1 23 4
] Find A + 𝐴1
21. A = [1 −2 30 1 4
] and B = [0 2 56 −3 1
] Find 3A + 2B
22. if A = [sin 𝜃 cos 𝜃
−cos 𝜃 sin 𝜃] Verify A A1 = I
23. if B = [cos 𝜃 sin 𝜃−sin 𝜃 cos 𝜃
] verify B B1= I
24. If A = [012] B = [1 5 7] Find AB
25. Compute 1) [1 −22 3
] [1 2 32 3 1
]
2) [3 −1 3
−1 0 2] [
2 −31 03 1
]
26. Find X and Y [2𝑥 + 𝑦 3𝑦
6 4] = [
6 06 4
]
27. What is the number of possible square matrix order 3 with each entries 0 or 1
28. Find X and Y if [5 − 𝑥 2𝑦 − 8
0 3] is a scalar matrix
29. Find X [4 𝑥 + 2
2𝑥 − 3 𝑥 + 1] is a symmetric matrix
II. Two mark and Three marks questions (SA)
1.Radha , fauzia, simran are the student of 12th class Radha has 15 note book and 6 pens ,
Fauzia has 10 books 2 pens and Simran has 13 books and 5 pens express this in to matrix
forms.
2. Construct 3× 2 matrix whose elements are given by 𝑎𝑖𝑗 =1
2 |𝑖 − 3𝑗|
3. Find X,Y,Z from the equation [
𝑥 + 𝑦 + 𝑧𝑥 + 𝑧𝑦 + 𝑧
] = [957]
4. Find a,b,c, d From the equation [𝑎 − 𝑏 2𝑎 + 𝑐2𝑎 − 𝑏 3𝑐 + 𝑑
] = = [−1 50 13
]
5. If A = [8 04 −23 6
] B = [2 −24 2
−5 1] Find X such that 2A + 3X = 5B
6. Find X and Y 2[𝑥 57 𝑦 − 3
] + [3 −41 2
] = [7 615 14
]
7. Find X and Y if x [23] + 𝑦 [
−11
]= [105
]
8. Given 3 [𝑥 𝑦𝑧 𝑤
] = [𝑥 6
−1 2𝑤] + [
4 𝑥 + 𝑦𝑧 + 𝑤 3
] Fine the values of X,Y,Z and W
9. If 𝐴𝑋 = [cos 𝑥 sin 𝑥−sin 𝑥 cos 𝑥
] and 𝐴𝑌 = [cos 𝑦 sin 𝑦−sin 𝑦 cos 𝑦
] Show that 𝐴𝑋𝐴𝑌 = 𝐴𝑋+𝑌
10. If A = [3 −24 −2
] and I = [1 00 1
] Find K If A2 = KA – 2I
11. If A = [3 √3 24 2 0
] and B =[2 −1 21 2 4
] verify (A + B )1 = A1 +B1
12. For any matrix A with real number entries , A+ A1 is symmetric matrix and A – A1
Skew-symmetric matrix
13. For any matrix A = [1 56 7
] verify that A+ A1 is symmetric matrix
14. For any matrix A = [1 56 7
] verify that A – A1 Skew-symmetric matrix
15. If A and B be the invertible matrices of same order then (AB)-1 = B-1A-1
16. By using elementary operation Find the inverse of the matrix [1 22 −1
]
17. By using elementary operation Find the inverse of the matrix [1 32 7
]
18. By using elementary operation Find the inverse of the matrix [1 −22 1
]
19. Find P-1 if it exists and P = [10 −2−5 1
]
20. If A = [3 1
−1 2] Show that A2 -5A +7I = 0
21. If A = [2 30 −4
] and B = [1 52 0
] Show that (AB)1 = B1A1
III. Five mark questions ( LA)
1.If A = [1 1 −12 0 33 −1 2
] B = [1 30 2
−1 4] and C =[
1 2 3 −42 0 −2 1
]
Find A B , BC and show that (AB )C = A(BC)
2. If A = [0 6 7
−6 0 87 −8 0
] B = [0 1 11 0 21 2 0
] C = [2
−23
] calculate AC, BC and (A+B) C
Deduce that (A+B) C = AC + BC
3. If A = [1 2 33 −2 14 2 1
] Show that A3 – 23A - 40 I = 0
4. If A = [1 2 −35 0 21 −1 1
] B = [3 −1 24 2 52 0 3
] and C = [4 1 20 3 21 −2 3
]
verify A+ (B-C) = (A+B ) –C
5. If A =
[ 2
31
5
31
3
2
3
4
37
32
2
3]
and B =
[ 2
5
3
51
1
5
2
5
4
57
5
6
5
2
5]
find 3A – 5B
6. If A = [2 0 12 1 31 −1 0
] find A2 – 5 A + 6 I ?
7. If A = [1 0 20 2 12 0 3
] prove that A3 – 6A2 + 7A + 2I = 0
8. Express the matrix B = [2 −2 −4
−1 3 41 −2 −3
] Find the sum of symmetric and skew-
symmetric matrix
9. Express the matrix B = [6 −2 2
−2 3 −12 −1 3
] Find the sum of symmetric and skew-
symmetric matrix
10. If A = [1 22 1
] B = [2 01 3
] C = [1 12 3
] calculate AB , BC, A(B+C)
Verify that AB + AC = A(B+C)
11. If F(x) = [cos 𝑥 − sin 𝑥 0sin 𝑥 cos 𝑥 0
0 0 1] show that F(x) F(y) = F(x+y)
12. If A =[−245
] and B = [1 3 −6] verify (AB)1 = B1A1
13. If A = [cos 𝜃 sin 𝜃−sin 𝜃 cos 𝜃
] Prove that An = [cos 𝑛𝜃 sin 𝑛𝜃−sin 𝑛𝜃 cos 𝑛𝜃
]
********
Solutions
One mark questions (VSA)
1. The numbers arranged in rectangular array of rows and columns by the brockets is
called matrix
2. A square matrix is said to be diagonal matrix if all non diagonals elements are zeros
3. A diagonal matrix is said to be scalar marics if it’s diagonal elements are equal
4. If a square matrix A = [𝑎𝑖𝑗]𝑚×𝑚is said to be symmetric if and only if A1 = A
5. If a square matrix A = [𝑎𝑖𝑗]𝑚×𝑚is said to be skew-symmetric if and only if A1 = -A
6. 1) order of the matrix is 3 × 4 2) 19, -2, -5 , 12, 5
2
7. Possible orders are (1,8) (8,1) (2,4) (4,2) is 1X8 , 8X 1 , 2X 4 , 4X 2
8. Possible orders are (1,18) (18,1) (3,6) (6,3) (2,9) (9,2) is 1 X18 , 18X1 , 3X6 , 6X3 ,
2X9, 9X2 ,
9. 1) [𝑎𝑖𝑗] = [4 99 16
] 2) [𝑎𝑖𝑗] = [2
9
29
28]
10. [0 1 21 0 2
]
11. [
11
2
2 1
33
2
]
12. X= 1 Y = 4 Z= 3
13. X = 2 Y = 4 Z = 0
14. X = -2A - B = [2 −4
−6 −8] - [
3 −21 5
] = [−1 −2−7 −13
]
15. 2A –B = [−1 5 35 6 0
]
16. By solving X = [−1 −1−2 −1
]
17. By solving above matrix X = [5 01 4
] and Y = [2 01 1
]
18. By multiplying we get the answer [1 00 1
] = I
19. 2+Y = 5 implies Y = 3 and 2x+2 = 8 implies x =3
20. A + A1 = [2 55 8
]
21. 3 A +2 B = [3 −2 1912 −3 14
]
22. A A1 = [sin 𝜃 cos 𝜃
−cos 𝜃 sin 𝜃] [
sin 𝜃 −cos 𝜃cos 𝜃 sin 𝜃
] = [1 00 1
] = I after multiplying
23. BB 1 = [cos𝜃 sin 𝜃−sin 𝜃 cos 𝜃
] [cos 𝜃 −sin 𝜃sin 𝜃 cos𝜃
] = [1 00 1
]= I after multiplying
24. (AB)1 = [0 1 20 5 100 7 14
]
25. 1) [−3 −4 18 13 3
] 2) [14 −64 5
] after multiplying
26. 3 |𝐴| = K |𝐴| implies K = 3
27. Y = 0, X = 3 by solving
28. The square matrix of order 3X3 = 9 and 2 entries
Then possible entries is 29 = 512
29. [5 − 𝑥 2𝑦 − 8
0 3]= [
3 00 3
] then X = 2 Y = 4
30. [4 𝑥 + 2
2𝑥 − 3 𝑥 + 1] = [
4 2𝑥 − 3𝑥 + 2 𝑥 + 1
] implies X = 5
Solutions : Two mark and Three marks questions (SA)
1. books pens
Radha : 15 6 this can be expressed as [15 610 213 5
] or [15 10 136 2 5
]
Fauzia : 10 2
Simran: 13 5
2. 𝑎𝑖𝑗 =
[ 1
5
21
22
03
2]
3. X+ Y + Z = 9 X +Z = 5 Y + Z = 7
7 + Z = 9 X + 2 = 5 Y + 2 = 7
Z= 2 X = 3 Y = 5
4. By solving equality a =1 , b= 2, c =3 and d = 4
5. X =
[ −2 −
10
3
414
3
−31
3
−7
3 ]
6. compare two matrices X = 2, Y = 9
7. by solving we get X = 3 , Y = -4
8. by solving and compare we get X = 2 , Y = 4, Z = 1, w = 3
9. 𝐴𝑋 𝐴𝑌 = [cos 𝑥 sin 𝑥−sin 𝑥 cos 𝑥
] [cos 𝑦 sin 𝑦−sin 𝑦 cos 𝑦
] = [cos(𝑥 + 𝑦) sin(x +𝑦)−sin(𝑥 + 𝑦) cos(𝑥 + 𝑦)
] = 𝐴𝑋+𝑌
10. A2 = KA – 2I
[1 −24 −4
] = [3𝑘 − 2 −2𝑘
4𝑘 −2𝑘 − 2]
Then 4K = 4
K = 1
11. ( A+B)1 = [5 5
√3 − 1 44 4
] and A1 + B 1 = [5 5
√3 − 1 44 4
]
Hence ( A+B)1 = A1 + B 1
12. B = A + A1 , B1 = (A + A1 )1 = A1+A = B ∴ B = A +A1 is symmetric
C = A –A 1 , C 1 = (A -A1 )1 = A1-A = -(A- A1) = - C ∴ C = A – A1 is skew- symmetric
13. Z = A + A1 = [1 56 7
] + [1 65 7
] = [2 1111 14
] = Z 1 ∴ Z = Z1 = A + A1 is symmetric
14. Z1 = (A - A1)1 = ([1 56 7
] − [1 65 7
])1
= ([0 −11 0
] )1
= -Z ∴ Z1 = - Z, A –A1
skew- symmetric
15. (AB) (AB)-1 = I
A-1(AB) (AB)-1 = A-1I I A = A
B(AB)-1 = A-1 IA-1 = A-1
B-1B(AB)-1 = B-1A-1 AA-1 = I
(AB)-1 = B-1A-1 BB-1 I
16. A = [1 22 −1
]
A = IA
[1 22 −1
] = [1 00 1
] A 𝑅2 = 𝑅2 − 2𝑅1
[1 20 −5
] = [1 0
−2 1]A 𝑅2 = −
1
5𝑅2
[1 20 1
] = [1 02
5−
1
5
] A 𝑅1 = 𝑅1 − 2𝑅2
[1 00 1
] = [
1
5
2
52
5−
1
5
] A
∴ A -1 = [
1
5
2
52
5−
1
5
]
17. By above process ∴A -1 = [7 −3
−2 1]
18. By above prose’s ∴A -1 = [
1
5
2
5−2
5
1
5
]
19. P = [10 −2−5 1
]
P = IP
[10 −2−5 1
] = [1 00 1
] P
By elementary operation
[1 −
1
5
0 0] = [
1
100
1
21] p
p-1 does not exists
20. A2 -5A +7I = [8 5
−5 3] - [
15 5−5 10
] + [7 00 7
] = [0 00 0
] = 0
21. By mathematical induction we get the solution
22. If A=A1 , B = B1 , (AB)1 = AB
(AB)1 = B1 A1 = BA ∴ AB = BA AB Is symmetric
23. A2 = A A By product of two matrix get the solution
24. (AB)1 =[8 −810 0
]
B1A1 = [8 −810 0
]
∴ (AB)1 = B1A1
25. By solving x = 2 , y = 4 , z = 3
Solutions : Five mark questions (LA)
1.AB =[2 1
−1 181 5
] (AB) C = [4 4 4 −735 −2 −39 2231 2 −27 11
]
BC = [7 2 −3 −14 0 −4 27 −2 −11 8
] A(BC) = [4 4 4 −735 −2 −39 2231 2 −27 11
]
Hence (AB) C = A(BC)
2. (A +B) C = [102028
] AC = [91230
] BC = [18
−2] AC + BC = [
102028
]
Hence (A +B) C = AC + BC
3. A = [1 2 33 −2 14 2 1
] A2 = [19 4 81 12 814 6 15
] A3 = [63 46 6960 −6 2392 46 63
]
LHS = A3 – 23A – 40 I = 0 By simplification
4. A + (B – C) = [0 0 −39 −1 52 1 1
] and (A+B) –C = [0 0 −39 −1 52 1 1
]
Hence A + (B – C) = (A+B) –C
5. 3A -5B = [2 3 51 2 47 6 2
] - [2 3 51 2 47 6 2
] = [0 0 00 0 00 0 0
] = 0
6. A2 – 5A + 6I = [1 −1 −3
−1 −1 −10−5 4 4
] by simplification
7. If A = [1 0 20 2 12 0 3
] by calculating A2 , A3 take LHS = RHS
8. B = [2 −2 −4
−1 3 41 −2 −3
] by theorem number 2
B = 1
2 (B +B 1) +
1
2 (B -B 1) hence they are equal
9. B = [6 −2 2
−2 3 −12 −1 3
] by theorem number 2
B = 1
2 (B +B 1) +
1
2 (B -B 1) hence they are equal
11. If AB = [4 65 3
] AC = [5 74 5
] A(B+C) = [9 139 8
] = AB + AC
12. F(x).F(y) = [cos 𝑥 − sin 𝑥 0sin 𝑥 cos 𝑥 0
0 0 1] [
cos 𝑦 − sin 𝑦 0sin 𝑦 cos 𝑦 0
0 0 1]
= [cos(𝑥 + 𝑦) − sin(𝑥 + 𝑦) 0sin(𝑥 + 𝑦) cos(𝑥 + 𝑦) 0
0 0 1
] = F(x+y)
13. LHS = (AB)1 = [−2 4 5−6 12 1512 −24 −30
] = B1A1 = RHS
14. By mathematical induction we get the solution
1
Question Bank
5. CONTINUITY AND DIFFERRENTIABILITY
ONE MARK QUESTIONS
1. Find the derivative of 2cos( )x with respect to x.
Sol: let 2cos( )y x
2 2 2sin( ) ( ) 2 sin( )dy d
x x x xdx dx
2. Find the derivative of 2 3 4 5x x x x xe e e e e with respect to x.
Sol : let 2 3 4 5x x x x xy e e e e e
2 3 4 52 3 42 3 4 5x x x x xdy
e xe x e x e x edx
3. Find the derivative of log(logx) with respect to x.
Sol : log(log )y x
1 1
(log )log log
dy dx
dx x dx x x
4. Find the derivative of cos(sinx) with respect to x.
Sol : cos(sin )y x
sin(sin ) (sin ) cos sin(sin )dy d
x x x xdx dx
5. Find the derivative of sec(tan )x with respect to x
Sol : sec(tan )y x
2sec tan tan tan sec ( )
sec tan tan tan (tan )2
x x xdy dx x x
dx dx x
6. Find the derivative of the function cos( )x with respect to x.
Sol : cosy x
sin
sin2
xdy dx x
dx dx x
2
7. If 2 33 2x xy e e find dy
dx
Sol : 2 33 2x xy e e
2 3 2 3 2 33 2 6 6 6( )x x x x x xdy d de e e e e e
dx dx dx
8. Find the derivative of 5cosx – 3sinx with respect to x.
Sol : 5cos 3siny x x
5sin 3cosdy
x xdx
9. The function 1
( )5
f xx
is not continuous at x = 5. Justify the statement.
Sol : 1
( )5
f xx
is a quotient function. The function f(x) is not defined at x = 5 because
1 1(5)
5 5 0f
is not defined. Therefore f(x) is continuous for all values of x except x = 5.
10. Find dy
dx if x y
Sol : x y
d d
x ydx dx
( ) ( ) 0d d
x ydx dx
1dy
dx
11. If tan(2 3)y x find dy
dx
Sol : tan(2 3)y x
2 2sec (2 3) 2 3 2sec (2 3)dy d
x x xdx dx
12. Find the derivative of f given by 1( ) tanf x x assuming it exists.
Sol : 1tany x 2
1
1
dy
dx x
3
13. Prove that the function ( ) nf x x is continuous at x = n, where n is a positive integer.
Sol : ( ) nf x x , n N . Here, f(x) is a polynomial function and fD R
lim ( ) lim ( ).n n
x n x nf x x n f n
Therefore f(x) is continuous at n N .
14. Find the derivative of 1sin xe
with respect to x.
Sol : 1sin xy e
1
1sin
sin 1
2sin
1
xxdy d e
e xdx dx x
15. Find dy
dx, if 2x at , 2y at .
Sol : 2x at , 2y at
2dx
atdt
, 2dy
adt
2 1
2
dy
dy adtdxdx at t
dt
TWO MARK QUESTIONS:
1. Examine whether the function f given by 2( )f x x is continuous at x = 0.
Sol : 2( )f x x at x = 0; (0) 0f .
Then 2
0 0lim ( ) lim 0x x
f x x
0
lim ( ) 0 (0)x
f x f
.
f is continuous at x = 0.
2. Discuss the continuity of the function f defined by 1
( )f xx
, 0x .
Sol : Fix any non zero real number c, we have 1 1
lim ( ) limx c x c
f xx c
4
Also, since for 0c , 1
( )f cc
, we have lim ( ) ( )x c
f x f c
and hence, f is continuous at every
point in the domain of f. Thus f is continuous function.
3. Find the derivative of the function sin
xey
x with respect to x.
Sol : sin
xey
x
2
sin sin
sin
x xd dx e e x
dy dx dx
dx x
2 2
sin cossin cos
sin sin
xx x e x xdy e x e x
dx x x
4. Discuss the continuity of the function f given by 3 2( ) 1f x x x
Sol : Clearly f is defined at every real number c and its value at c is 3 2 1c c . We also know that
3 2 3 2lim ( ) lim 1 1x c x c
f x x x c c
Thus lim ( ) ( )x c
f x f c
, and hence f is continuous at every real number. This means f is a
continuous function.
5. Verify Rolle’s theorem for the function 2 2y x , a = -2 and b = 2
Sol : The function 2 2y x is continuous in [- 2 , 2] and differentiable in (-2, 2). Also
( 2) (2) 6f f and hence the value of ( )f x at -2 and 2 coincide. Rolle’s theorem states that
there is a point ( 2,2)c , where |( ) 0f c . Since |( ) 2f x x , we get c = 0. Thus at c = 0, we
have |( ) 0f c and 0 ( 2,2)c .
6. If f and g be two real functions continuous at real number c. Then show that f + g is continuous at
x = c.
Sol : The continuity of f + g at x = c, clearly it is defined at x = c we have
lim lim ( ) ( )x c x c
f g x f x g x
lim ( ) lim ( ) ( ) ( )x c x c
f x g x f c g c
( )f g c
Hence f + g is continuous at x = c.
5
7. Find dy
dx if, cosx a , siny a .
Sol : cosx a , siny a
sindx
ad
cos
dya
d
cos
cotsin
dy
dy addxdx a
d
8. Discuss the continuity of the function f given by ( ) | |f x x at x = 0.
Sol : By definition , if x<0
( ), if x 0
xf x
x
Clearly the function is defined at x = 0 and f(0) = 0.
Let hand limit of f at x = 0 is 0 0
lim ( ) lim( ) 0x x
f x x
Right hand limit of f at x = 0 is 0 0
lim ( ) lim 0x x
f x x
.
Thus the left hand limit, right hand limit and the value of the function coincide at x = 0.Hence, f is
continuous at x = 0.
9. Find the derivative of the function sin( )
cos( )
ax b
cx d
with respect to x.
Sol : sin( )
cos( )
ax by
cx d
2
cos( ) sin( ) sin( ) cos( )
cos ( )
d dcx d ax b ax b cx d
dy dx dx
dx cx d
2
cos( )cos( ) sin( )sin( )
cos ( )
dy a cx d ax b c ax b cx d
dx cx d
10. Discuss the continuity of sine function.
Sol : ( ) sinf x x is defined for every real number. Let c be a real number. Put x = c + h.
If x c we know that 0h . Therefore
lim ( ) limsinx c x c
f x x
6
0 0
limsin( ) lim sin .cosh cos .sinhh h
c h c c
0 0
limsin .cosh limcos .sinh sin 0 sin ( )h h
c c c c f c
Thus lim ( ) ( )x c
f x f c
Therefore f is a continuous function.
11. Differentiate sin xx 0x with respect to x.
Sol : sin xy x
Taking log on both sides
sinlog log xy x
log sin logy x x
1
sin (log ) (sin ) logdy d d
x x x xy dx dx dx
1 sin
cos logdy x
x xy dx x
sin
cos logdy x
y x xdx x
sin sincos logxdy x
x x xdx x
12. Differentiate the function 1sin tan xe with respect to x.
Sol : 1sin tan xy e
1 1cos tan tanx xdy de e
dx dx
1
2
cos tan
1
x x
x
e edy
dx e
13. If 22x at , 4y at find dy
dx
Sol : 22x at , 4y at
4dx
atdt
, 34dy
atdt
7
3
24
4
dy
dy atdt tdxdx at
dt
14. If x yxy e , prove that ( 1)
( 1)
dy y x
dx x y
Sol : x yxy e
Differentiating both sides with respect to x.
( ) x yd dxy e
dx dx
1x ydy dyx y e
dx dx
x y x ydy dyx y e e
dx dx
x y x ydy dyx e e y
dx dx
x y x ydyx e e y
dx
( 1)
( 1)
x y
x y
dy e y xy y y x
dx x e x xy x y
15. If cos cos2 cos3y x x x find dy
dx
Sol : cos cos2 cos3y x x x
Taking log on both sides, we get
log log(cos cos2 cos3 )y x x x
log logcos logcos2 logcos3y x x x
1 sin 2sin 2 3sin3
cos cos 2 cos3
dy x x x
y dx x x x
cos cos 2 cos3 tan 2 tan 2 3tan3dy
x x x x x xdx
16. If x y a prove that dy y
dx x
8
Sol : x y a
Differentiate w.r.t x we get
1 1
02 2
dy
dxx y
1 1
2 2
dy
dxy x
dy y
dx x
17. Find the derivative of sin cos
sin cosx x
x x
with respect to x.
Sol : (sin cos )
sin cosx x
y x x
Taking log on both sides
(sin cos )log log(sin cos ) x xy x x
log (sin cos )log(sin cos )y x x x x
Differentiate w.r.t x
1
sin cos log(sin cos ) (sin cos ) log(sin cos )dy d d
x x x x x x x xy dx dx dx
1 cos sin
(sin cos ) (cos sin ) log(sin cos )sin cos
dy x xx x x x x x
y dx x x
(sin cos )
sin cos cos sin 1 log(sin cos )x xdy
x x x x x xdx
18. If 1sinx
y x find dy
dx
Sol : 1sinx
y x
Taking logarithm on both sides
1log log sinx
y x
1log log siny x x
1 11log sin ( ) log sin
dy d dx x x x
y dx dx dx
1
1 2
1log sin
sin 1
dy xx
y dx x x
9
1
1 2log sin
sin 1
dy xy x
dx x x
= 1 1
1 2sin log sin
sin 1
x xx x
x x
19. If sin logey x prove that 21 ydy
dx x
Sol : sin logey x
cos log
cos log loge
e e
xdy dx x
dx dx x
221 sin(log 1e x ydy
dx x x
THREE MARK QUESTIONS:
1. If 2
1
2
1cos
1
xy
x
0 < x < 1 Find
dy
dx
Sol : 2
1
2
1cos
1
xy
x
Put tanx
2
1 1
2
1 tancos cos cos 2
1 tany
12 2tany x
Differentiate w.r. t x
2
1
1
dy
dx x
2. If 3 2 2 3 81x x y xy y . Find dy
dx
Sol : 3 2 2 3 81x x y xy y
Differentiate w.r. t x
2 2 2 23 2 2 3 0dy dy dy
x x xy x y y ydx dx dx
2 2 2 22 3 3 2dy
x xy y x xy ydx
10
2 2
2 2
3 2
2 3
x xy ydy
dx x xy y
3. Diffentiate 2
2
3 4
3 4 5
x xy
x x
w.r. t x.
Sol : 2
2
3 4
3 4 5
x xy
x x
Taking logarithm on both sides, we have
2 21log log 3 log 4 log 3 4 5
2y x x x x
Differentiating on both sides w.r.t x, we get
2 2
1 1 1 2 6 4
2 3 4 3 4 5
dy x x
y dx x x x x
2 2
1 2 6 4
2 3 4 3 4 5
dy y x x
dx x x x x
2
2 2 2
3 41 1 2 6 4
2 3 4 5 3 4 3 4 5
x xdy x x
dx x x x x x x
4. Find dy
dx if 1
2
2sin
1
xy
x
Sol : 1
2
2sin
1
xy
x
Put tanx
1 1
2
2 tansin sin sin 2
1 tany
12 2tany x
Differentiating w.r.t x, we get
2
2
1
dy
dx x
5. Differentiate the function cos
logx
x with respect to x.
11
Sol : cos
logx
y x
Taking logarithm on both sides
cos
log log logx
y x
log cos log logy x x
Differentiating w.r. t. x on both sides, we get
1
cos log log cos log logdy d d
x x x xy dx dx dx
1 cos
sin log loglog
dy xx x
y dx x x
coscos cos
sin log log log sin log loglog log
xdy x xy x x x x x
dx x x x x
6. Find dy
dx if x yy x
Sol : x yy x
Taking logarithm on both sides
log logx yy x
log logx y y x
Differentiating with respect to x, on both sides, we get
(log ) log ( ) (log ) log ( )d d d d
x y y x y x x ydx dx dx dx
log logx dy y dy
y xy dx x dx
log logx y x dy y x y
y dx x
log
log
y y x ydy
dx x x y x
7. Differentiate 2sin x with respect to cos xe
Sol : let 2sinu x and cos xv e
Differentiate w.r.t x
12
2sin cosdu
x xdx
and cossin xdvxe
dx
cos cos
2sin cos 2cos
sin x x
du
du x x xdxdvdv xe e
dx
8. Differentiate 1 sintan
1 cos
x
x
with respect to x.
Sol : 1 sintan
1 cos
xy
x
1 1 1
2
2sin cos sin2 2 2tan tan tan tan
22cos cos
2 2
x x xx
yx x
2
xy
Differentiating w.r.t x on both sides
1
2
dy
dx
9. Verify mean value theorem if 3 2( ) 5 3f x x x x in the interval [a, b] where a = 1 and b = 3. Find
(1,3)c for which |( ) 0f c .
Sol: Given 3 2( ) 5 3f x x x x [1,3]x which is a polynomial function.
Since a polynomial function is continuous and derivable at all x R
(1) f(x) is continuous on [1,3] (2) f(x) is derivable on (1, 3)
Therefore condition of mean value theorem satisfied on [1,3]. Hence, at least one real (1,3)c
3 2
|3 5(3) 3(3) 1 5(1) 3(1)(3) (1)
( )3 1 2
f ff c
| 20( ) 10
2f c
| 2( ) 3 10 3f x x x ; | 2( ) 3 10 3 10f c c c
23 10 7 0c c
23 7 3 7 0c c c ;
13
(3 7) (3 7) 0c c c c = 1 (1,3) 7
(1,3)3
c .
Hence the mean theorem satisfied for given function in the given interval.
10. If 1cosy x find 2
2
d y
dx in terms of y alone.
Sol : 1cosy x
cosx y
Differentiating w.r.t y, we get
sindx
ydy
cosdy
ecydx
Again differentiating w.r. t x , we get
2
2cos cot
d y dyecy y
dx dx
22
2cos cot
d yec y y
dx
11. Find the derivative of log
logx
x with respect to x.
Sol : log
logx
y x
Taking logarithm on both sides
log
log log logx
y x
log log log logy x x
Differentiating w.r. t x on both sides
1
log log log log log logdy d d
x x x xy dx dx dx
log log1 log
log
xdy x
y dx x x x
loglog log log log1 1log
xx xdyy x
dx x x x x
14
12. Find dy
dx if 3 2 5cos .siny x x
Sol : 3 2 5cos siny x x
Differentiating w.r.t x on both sides
3 2 5 2 5 3cos sin sin cosdy d d
x x x xdx dx dx
3 5 5 4 5 3 2cos 2sin cos 5 sin sin 3dy
x x x x x x xdx
4 5 5 3 2 5 310 sin cos cos 3 sin sindy
x x x x x x xdx
13. Verify Rolle’s theorem for the function 2( ) 2 8f x x x , [ 4,2]x .
Sol : Given 2( ) 2 8f x x x , [ 4,2]x . Since a polynomial function is continuous and derivable on
R. (1) f(x) is continuous on [-4,2] (2) f(x) is derivable on [-4,2].
Also 2( 4) ( 4) 2( 4) 8 0f and 2(2) 2 2 2 8 0f ( 4) (2)f f .
This means that all the conditions of Rolle’s theorem are satisfied by f(x) in [-4,2].
Therefore there exist at least one real number ( 4,2)c such that |( ) 0f c .
2( ) 2 8f x x x |( ) 2 2f x x
|( ) 0f c 2 2 0c 1 ( 4,2)c
Rolle’s theorem is verified with c = - 1
14. If 1sin tx a
and 1cos ty a
then prove that dy y
dx x
Sol : 1sin tx a
1cos ty a
11
sin2
t
x a
11
cos2
t
y a
Differentiating w.r.t “t” we get
11
sin12
1log sin
2
tdx da a t
dt dx
11cos
121
log cos2
tdy da a t
dt dx
11sin
2
2
log
2 1
t
dx a a
dt t
11cos
2
2
log
2 1
t
dy a a
dt t
15
1
1
11
1cos
2
sin2
1cossin
2
2
log
2 1
log
2 1
t
t
tt
ady ady atdt
dxdx a aadt t
dy y
dx x
15. Find the derivative of sin2x xx with respect to x.
Sol : let sin2x xy x = u – v
Where xu x and sin2 xv
Taking log on both sides
log log xu x and sinlog log 2 xv
log logu x x and log sin log2v x
Differentiate with respect to x we get
1
log logdu d d
x x x xu dx dx dx
and 1
cos log 2dv
xv dx
1.log 1 logxdu xu x x x
dx x
and sincos log 2 2 cos log 2xdv
v x xdx
y u v
sin1 log 2 cos log 2x xdy du dvx x x
dx dx dx
16. If ( sin )x a and (1 cos )y a prove that tan2
dy
dx
Sol : ( sin )x a (1 cos )y a
Differentiating w.r.t x on both sides
(1 cos )dx
ad
(0 sin ) sin
dya a
d
sin
(1 cos )
dy
dy addxdx a
d
2
2 sin cos sin2 2 2 tan
22 cos cos
2 2
ady
dxa
17. If a function f(x) is differentiable at x = c , prove that it is continuous at x = c.
16
Sol : Since f is differentiable at c, we have |( ) ( )lim ( )x c
f x f cf c
x c
But for x c , we have ( ) ( )
( ) ( ) .f x f c
f x f c x cx c
Therefore ( ) ( )
lim[ ( ) ( )] lim .x c x c
f x f cf x f c x c
x c
Or ( ) ( )
lim ( ) lim ( ) lim limx c x c x c x c
f x f cf x f c x c
x c
|( ) . 0f c = 0
lim ( ) ( )x c
f x f c
. Hence f is continuous at x = c.
FIVE MARK QUESTIONS
1. If 3cos log 4sin logy x x prove that 2
2 1 0x y xy y .
Sol : 3cos log 4sin logy x x
Differentiating w.r.t x on both sides
1
3sin log 4cos logx xy
x x
1 3sin log 4cos logxy x x
Again differentiating on both sides we get
2 1
3cos log 4sin log1
x xxy y
x x
2
2 1 3cos log 4sin logx y xy x x
2
2 1x y xy y 2
2 1 0x y xy y
2. If 2 33 2x xy e e prove that 2
25 6 0
d y dyy
dx dx
Sol : 2 33 2x xy e e
2 3 2 36 6 6x x x xdye e e e
dx
2
2 3 2 3
212 18 6 2 3x x x xd y
e e e edx
17
Hence 2
3 3 2 3 2 3
25 6 6(2 3 ) 30 6 3 2x x x x xd y dy
y e e x e e e edx dx
2
3 3 2 3 2 3
25 6 12 18 30 30 18 12 0x x x x xd y dy
y e e x e e e edx dx
3. If 2
1tany x prove that 2
2 2
2 11 2 1 2x y x x y .
Sol : 2
1tany x
Differentiating w.r.t on both sides
1
1 2
2 tan
1
xy
x
2 1
11 2tanx y x
Again differentiating w.r.t x on both sides
2
2 1 2
21 2
1x y xy
x
On cross multiplication, we get
2
2 2
2 11 2 1 2x y x x y
4. If mx nxy Ae Be , Show that 2
20
d y dym n mny
dx dx .
Sol : mx nxy Ae Be
Differentiating w.r.t x on both sides
mx nxdyAme Bne
dx Again differentiate w.r.t x on both sides
2
2 2
2
mx nxd yAm e Bn e
dx
Hence 2
2 2
2
mx nx mx nxd y dym n mny m Ae n Be m n Ame Bne mny
dx dx
2 2 2 2mx nx mx nx mx nxm Ae n Be Am e Bmne Amne n Be mny
nx mx mx nxBmne Amne mny mn Ae Be mny
0mny mny mx nxy Ae Be
18
5. If 1siny x prove that 2
2
21 0
d y dyx x
dx dx
Sol : 1siny x
Differentiate w.r.t x, we get
2
1
1
dy
dx x
On cross multiplication
21 1dy
xdx
Again Differentiate w.r.t x, we get
2
2
2 2
21 0
2 1
d y x dyx
dx dxx
Taking Lcm and simplifying, we get
2
2
21 0
d y dyx x
dx dx
6. If 1cosy x prove that 2
2 11 0x y xy
Sol : 1siny x
Differentiate w.r.t x, we get
2
1
1
dy
dx x
On cross multiplication
2
11 1x y
Again Differentiate w.r.t x, we get
2
2 12
21 0
2 1
xx y y
x
Taking Lcm and simplifying, we get 2
2 11 0x y xy
7. If 5cos 3siny x x , prove that 2
20
d yy
dx
Sol : 5cos 3siny x x
Differentiating w.r.t x , on both sides
19
5sin 3cosdy
x xdx
Again differentiating w.r.t x we get
2
25cos 3sin 5cos 3sin
d yx x x x
dx
2
2
d yy
dx
2
20
d yy
dx
8. If 1 1ye x , prove that
22
2
d y dy
dx dx
Sol : 1 1ye x
Differentiate w.r.t x on both sides
1 1 0y yd de x x e
dx dx
1 0y y dye x e
dx
1
1
dy
dx x
Again differentiate w.r.t x , we get
2
22
1
1
d y
dx x
22
2
d y dy
dx dx
9. If 7 7500 600x xy e e then show that 2
249
d yy
dx
Sol : 7 7500 600x xy e e
Differentiate w.r.t x
7 7500 7 600 7x xdye e
dx
Again differentiate w.r.t x
2
7 7
2500 49 600 49x xd y
e edx
2
7 7
249 500 600x xd y
e edx
2
249
d yy
dx
APPLICATION OF DERIVATIVES
TWO MARK QUESTIONS:
1) Find the rate of change of the area of a circle w.r.t to its radius ‘r’
when r = 4 cm?
Ans: Area of circle A = r2, dA/dr = ? when r = 4 cm
Differentiate w.r.t. ‘r’
dA/dr = (2r)
= (2)(4)
= 8 sq. cms
Therefore area of the circle is increasing at the rate of 8 sq. cms.
2) An edge of a variable cube is increasing at the rate of 3cm/s. How fast is
the volume of the cube increasing when the edge is 10cm long?
Ans: Volume of a cube V = x3., Given: dx/dt = 3cm/s. dV/dt = ?
when x = 10cm
Didifferentiate w.r.t ‘t’
dV/dt = 3x2(dx/dt)
= 3(10)2 . (3)
= 900 c.c/s
Therefore volume of the cube increasing at the rate of 900 c.c/s.
3) Show that the function f(x) = x3 -3x2 + 4x, xR is strictly increasing on R.
Ans: f(x) = x3 – 3x2 + 4x
Differentiate w.r. t x
f(x) = 3x2 – 6x + 4
= 3(x2 -2x+1) + 1
= 3(x-1) 2 + 1 0, xR
Therefore f is strictly increasing on R.
4) Show that the function f(x) = e2x is strictly increasing on R.
Ans: f(x) = e2x,
Differentiate w.r.t x
f(x) = 2. e2x
clearly f(x) 0 xR (since exponential function is always
positive)
Therefore f is strictly increasing on R.
5) Find the intervals in which f(x) = x2 + 2x – 5 is strictly increasing or
decreasing.
Ans: f(x) = x2 + 2x – 5
Differentiate w.r.t x
f(x) = 2x + 2
= 2(x+1)
Now f(x) = 0, 2(x+1) = 0
x = -1.
Now x = -1 divides the real line into 2 disjoint intervals namely
(-, -1) and (-1, ).
In (-, -1), f(x) 0
In (-1,), f(x) > 0.
f is strictly decreasing in (-, -1) and f is strictly increasing in (-1,).
6) Find the slope of the tangent to the curve y = (x-1)/(x-2), x≠2 at x = 10.
Ans: y =
Differentiate w.r.t x
dy/dx = (x-1) [(-1)/(x-2)2] + [1/(x-2)](1)
slope of tangent = dy/dx x = 10
= (10-1)[(-1)/(10-2)2] + [1/(10-2)](1)
= -9/64 + 1/8 = -1/64
7) Find the points at which the tangent to the curve y = x3 – 3x2 – 9x + 7
is parallel to x axis.
Ans: y = x3 – 3x2 – 9x + 7.
Differentiate w. r. t. x
dy/dx = 3x2 – 6x -9 = slope of the tangent.
Given tangent is parallel to x axis.
Slope of the tangent = slope of x axis.
3x2 – 6x – 9 = 0
X2 -2x -3 = 0
(x-3)(x+1) = 0
x = 3, x = -1
When x = 3, y = (3)3 – 3(3)2 – 9(3) + 7 = -20
When x = -1, y = (-1)3 -3(-1)2 -9(-1) + 7 = 12
Therefore the points are (3,-20) , ( -1,12) .
8) Using differentials, find approximate value of 25.3 up to 3 decimal
places.
Ans: y = x, Let x = 25 and x = 0.3
Then y = x + x - x
= 25.3 -25
25.3 = y + 5
Now dy = (dy/dx) x = (1/2x) (0.3)
= (1/225) (0.3) = 0.3/10 = 0.03.
Therefore approximate value of 25.3 is 5 + 0.03 = 5.03
9) If the radius of a sphere is measured as 7 m with an error of 0.02m,
then find the approximate error in calculating its volume.
Ans: Given radius of the sphere r = 7m and r = 0.02 m.
Volume of sphere V = (4/3) r3.
Differentiate w.r.t ‘r’
dV/dr = (4/3) (3r2)
Therefore dV = (dV/dr) r
= (4r2)(r)
= (4) (49) (0.02) = 3.92 m3.
Therefore the approximate error in calculating the volume is 3.92 m3.
10) If the radius of sphere is measured as 9m with an error of 0.03m,
then find the approximate error in calculating its surface area.
Ans: Radius of the sphere r = 9m, r = 0.03m.
Surface area of sphere S = 4r2.
Differentiate w.r.t. ‘r’
dS/dr = 4(2r)
Now dS = (dS/dr) (r)
= (4)(2)(r)r
=(8) (9) (0.03)
=2.16m3
THREE MARK QUESTIONS:
1) Find the local maxima and local minima if any, of the function f(x) = x2
and also find the local maximum and local minimum values.
Ans: f(x) = x2 Differentiate w.r.t. x
f’(x) = 2x
f (x) = 0 2x = 0 x=0
f(x) =2x
f(x) = 2 > 0
By second derivative test x= 0 is a point of local minima .
local minimum m value = f(0) = 02 = 0
2) Find the local maxima and local minima if any, of the function
f(x) =x3 – 6x2 + 9x + 15 and also find the local maximum and local
minimum values.
Ans: f(x) = x3- 6x2+9x+15
Differentiate w.r.t. x
f(x) = 3x2-12x+9
f(x) = 6x-12
Now f(x) = 0 3x2-12x+9=0
X2-4x+3=0
(x-3)(x-1) = 0
x=3, x=1
Now f(3) = 6(3)- 12 = 6>0
f(1) = 6(1) – 12 = -6<0
By second derivative test, x= 3 is a point of local minima and
x = 1 is a point of local maxima
local maximum value = f(1) = (1)3-6(1)2+9(1)+15 =19
Local minimum value = f(3) = (3)3-6(3)2+9(3)+15=15.
3) Prove that the function f(x) = logx do not have maxima or minima.
Ans: f(x) = logx
Differentiate w.r.t. x
f(x) = 1/x
f(x) = -1/x2
Now f(x) = 0 1/x = 0
x=
The function do not have maxima or minima.
4) Prove that the function f(x) = x3 + x2 + x+ 1 do not have maxima or
minima.
Ans: f(x) = x3+x2+x+1
Differentiate w.r.t. x
f(X) = 3X2+2X+1
f(x) = 6x+2
Now f(x) = 0 3x2+2x+1=0
X = [-2±4-4(3)(1)]/2(3)
X = [-2 ± -8] /6 which is imaginary
The given function do not have maxima or minima for all reals.
5) Find the absolute maximum value and the absolute minimum value of
the function f(x) = sinx + cosx, x[0,].
Ans: f(x) = sinx+cosx,
Differentiate w.r.t. x
f(x) = cosx - sinx
Now f (x) = 0 Cosx - sinx =0
sinx = cosx tanx = 1
x = /4 and 5/4
Now the value of the function f(x) at x= /4, 5/4 and end points
of intervals that is 0 and is f(0) = sin0+cos0 = 0+1 = 1
f(/4) = sin(/4) +cos (/4) = 1/2 + 1/2 =2/2 = 2
f(5/4) = sin(5/4) +cos (5/4) =(-1/2)+( - 1/2) =-2/2 = -2
f() = sin + cos = 0+(-1) = -1
Absolute maximum value of f on [ 0,] is 2 occurring at x=/4.
Absolute minimum value of f on [ 0,] is -2 occurring at
x=5/4.
6) Find two numbers whose sum is 24 and whose product is as large as
possible.
Ans: Let the numbers be ‘x’ & ‘y’
Given S = x+y = 24
y = 24-x
Product of numbers, P= x y is large
P = x(24-x) = 24x-x2
Differentiate w.r.t. x dP/dx = 24-2x
Differentiate w.r.t. x d2P/dx2 = -2<0 Product is maximum
For the product to be maximum dP/dx = 0
24-2x = 0 x = 12
The numbers are x & 24-x, 12 & 24-12 12 & 12
The numbers are 12 & 12
7) Find two positive numbers whose sum is 16 and the sum of whose cubes
is minimum.
Ans: Let numbers be x and y
Sum = x+y = 16 y = 16 –x
Given S = x3 +y3 is minimum
= x3 + (16-x)3
Differentiate w.r.t. x dS/dx = 3x2 + 3(16-x)2(-1)
d2S/dx2 = 6x - 3(2) (16-x) (-1)
=6x+6(16-x)
For S to be minimum dS/dx = 0
3x2-3(16-x)2 = 0
x2-(16-x)2 = 0
32x – 256 = 0
x = 8 y = 16-x = 16-8 = 8
Hence the numbers are 8 and 8.
8) Show that of all rectangles inscribed in a given fixed circle, the squares
has the maximum area.
Ans : Let ‘r’ be the radius of circle ABCD is a rectangle.
OA = r , OE = x , AE = y ,In le OAE ,
OA2 = OE2 + AE2
r2 = x2 +y2
y2 = r2 –x2 y = r2 - x2
Area of rectangle A = x. y
= xr2-x2
Squaring both sides A2 = x2(r2 –x2)
Let A2 = B B = x2(r2 –x2)
Differentiate w.r.t. x
dB/dx = x2(-2x)+(r2-x2)(2x)
= 2x(r2 – 2x2)
d2B/dx2 = 2x(-4x) + (r2-2x2)(2)
= 2r2 -12x2
For the area to be maximum dB/dx = 0
2x(r2-2x2) = 0 x = 0 & x2 = r2/2 x = r/2
d2B/dx2x=r/2 = 2r2 – 12(r2/2) = -4r2< 0
Area is maximum
Y2 = r2 - x2 = r2- r2/2 = r2/2
Since x= y = r/2 , ABCD is a square.
9) Find the equation of the normal at the a point (am2, am3) for the curve
ay2 = x3.
Ans: ay2 = x3
Y2 = x3/a
Differentiate w.r.t. x
2y dy/dx = [1/a] 3x2
dy/dx = 3x2/2ay
Slope of tangent = dy/dx(am2
,am3
) = 3(am2)2/ 2a(am3)
= 3a2m4/2a2m3 = 3m/2
slope of normal = -2/3m
Equation of normal at (am2, am3) having slope -2/3m is
Y - am3 = (-2/3m) (x - am2).
B
C D
O E
A
10) Find the equation of the normals to the curve y = x3 + 2x + 6 which are
parallel to the line x + 14y + 4 = 0.
Ans: y = x3+2x+6
Differentiate w.r.t. x
dy/dx = 3x2+2 = slope of tangent
slope of normal = -1/(3x2+2)
Normal is parallel to x +14y+4 = 0
Slope of normal = slope of x + 14y +4 = 0
-1/(3x2+2) = -1/14
3x2+2 = 14
3x2 = 12 x = ± 2
When x =2, y = (2)3 + 2(2) + 6 = 18, (2,18)
When x = -2, y =(-2)3 +2(-2)+6 = -6, (-2,-6)
Slope of normal = -1/14
equation of normal at (2,18) is y – 18 = (-1/14) (x – 2)
x+14y – 254 = 0
Also equation of normal at (-2,-6) is y+6 = (-1/14) (x+2)
x + 14y+86 = 0.
11) Find the points on the curve x2/9 +y2/16 = 1 at which the tangents are
parallel to y axis.
Ans: x2/9 +y2/16 = 1
Differentiate w.r.t. x
(1/9) 2x + (1/16) 2y (dy/dx) = 0
dy/dx = (-2x/9)/(y/8) = -16x/9y
Tangent parallel to y axis.
Slope of tangent = slope of y axis
-16x/9y = 1/0
y = 0
When y = 0 , x2/9 +0/16 = 1 x2= 9, x = ± 3
The points are (±3,0).
12) Find the equation of all lines having slope two which are tangents to the
curve y = 1/(x-3), x≠ 3.
Ans: y = 1/(x-3)
Differentiate w.r.t. x
dy/dx =
Given dy/dx = 2
-1/(x-3)2 = 2
2(x-3)2 =-1
2(x2-6x+9)=-1
2x2-12x+19 = 0
X = (12 ± 144-152) / 2(2) which is complex
No tangent to the curve which has slope two.
13) Prove that the function ‘f’ given by f(x) = log(sinx) is strictly increasing
on (0,/2) and strictly decreasing on (/2,).
Ans: f(x) = log(sinx)
Differentiate w.r.t. x
f(x) = (1/sinx) (cosx)
= cotx
Since for each x (0,/2), cotx > 0 f(x) > 0
So f is strictly increasing in (0,/2)
Since for each x (/2,), cotx < 0 f(x)< 0
So f is strictly decreasing in (/2,).
FIVE MARKS QUESTIONS:
1) The volume of a cube is increasing at the rate of 8 c.c/s. How fast is the
surface area increasing when the length of an edge is 12cm?
Ans: Let x , V, S be the length of side , volume and surface area of the
cube respectively.
Given dV/dt = 8c.c/s, dS/dt = ? when x = 12cm
Volume of cube = V = x3
Differentiate w.r.t. t
dV/dt = 3x2 . dx/dt
8 = 3(12)2 dx/dt
dx/dt = 8/3(144) = 1/54
Surface area of a cube S = 6x2
Differentiate w.r.t. t
dS/dt = 6(2x) ( dx/dt)
= 12(12) (1/54) = 24/9 = 2.6 sq.cm/s
surface area of a cube is increasing at the rate of 2.6 sq.cm/s.
2) A stone is dropped into a quiet lake and waves in circles at the speed of
5cm/s. At the instant when the radius of circular wave is 8 cm, how fast
is the enclosed area is increasing?
Ans: Let r , A be the radius and Area of a circle respectively
Given dr/dt = 5cm/s dA/dt = ? when r = 8cm
Area of a circle A = r2
Differentiate w.r.t. t
dA/dt = 2r dr/dt
= 2(8).(5)
= 80 cm2 /s
The enclosed area is increasing at the rate of 80 cm2/s
when r = 8cm.
3) The length ‘x’ of a rectangle is decreasing at the rate of 5cm/m and the
width ‘y’ increasing at the rate of 4cm/m. When x = 8cm and y = 6cm,
find the rates of changes of
(a) the perimeter and (b) the area of the rectangle.
Ans: Since the length ‘x’ is decreasing and width ‘Y ‘is increasing with
respect to time,
we have dx/dt = -5 cm / m , dy/dt = 4 cm /m
(a) The perimeter P of a rectangle is given by
P = 2(x+y)
Differentiate w.r.t. t
dP/dt = 2dx/dt + 2 dY/dt
= 2(-5) + 2 (4)
= -2 cm/min
(b) The area ‘A’ of the rectangle is given by A = x.y
dA/dt = dx/dt y + x dy/dt
= (-5) (6) + (8) (4)
= 2cm2/m
The perimeter and area of a rectangle is decreasing and
increasing at the rate of 2cm/m and 2 cm2/m respectively .
4) A balloon, which always remains spherical, has a variable diameter
(3/2)(2x+1). Find the rate of change of its volume w.r.t. x
Ans: Volume of a sphere = V = (4/3)r3
Given 2r = (3/2) (2x+1) r =( 3/4) (2x+1)
V = (4/3)[(3/4)(2x+1)]3
= (4/3) (27/64)(2x+1)3
V = (9/16)(2x+1)3
Differentiate w.r.t. x
dV/dx = (9/16) 3(2x+1)2(2)
= (27/8) (2x+1)2
volume of a sphere increases at the rate of (27/8) (2x+1)2
5) A balloon, which always remains spherical has a variable radius. Find
the rate at which its volume is increasing with the radius when the
radius is 10 cm.
Ans: Let ‘r’ and ‘ V’ be the radius and volume of a sphere.
To find dV/dr = ? when r = 10cm
Volume of a sphere V =( 4/3)r3
Differentiate w.r.t. r
dV/dr = (4/3) 3r2
= 4(10)2
= 400 cm3/cm
The volume of the spherical balloon is increasing with radius is
400 cm3/cm.
6) A water tank has the slope of an inverted right circular cone with its
axis vertical and lower most. Its semi-vertical angle is tan-1(0.5). Water
is poured into it at a constant rate of 5 cubicmeter per hour. Find the
rate at which the level of water is rising at the instant when the depth
of water in the tank is 4m.
Ans: Let r, h and be the radius , height and semi-vertical angle of
cone.
Tan = r/h = tan-1(r/h)
Given = tan -1(0.5)
r/h= 0.5 = ½ r = h/2 Given dV/dt = 5 c.m/h
volume of a cone V = (1/3)r2h : dh/dt = ? when h = 4 m
= (1/3) (h2/4) h
= (/12).h3 OA = h
Differentiate w.r.t. t AB = r
dV/dt = (/12).3h2(dh/dt)
5 = (/4)(4)2(dh/dt)
(dh/dt) = 5/4 = 35/88 m/h ( = 22/7)
Rate of change of water level = (35/88) m/h.
7) A ladder 5m long is leaning against a wall. The bottom of the
ladder is pulled along the ground, away from the wall at the rate
of 2cm/s. How fast is its height of the ladder decreasing when the
foot of the ladder is 4cm away from the wall?
Ans: Let AB be the ladder, AC wall, BC ground.
Let BC = x, AC = y
Given: AB = 5m, dx/dt = 2cm/s, dy/dt = ? when x = 4m.
From the fig, x2 + y2 = 52
(4)2 + y2 = 25
y2 =9, y =3.
Consider x2 + y2 = 52
Differentiate w.r.t ‘t’
2x(dx/dt) + 2y(dy/dt) = 0
2 (4)(2) + 2(3)(dy/dt) = 0
6(dy/dt) = -16
dy/dt = -8/3.
Height of the ladder is decreasing at the rate
of 8/3 cm/s.
O
B A
A
C B
8) A man 6ft tall moves away from a source of light 20ft above the ground
level, his rate of walking being 4 m.p.h. At what rate is the length of his
shadow changing? At what rate is the tip of his shadow moving?
Ans: At any time t, let AB = 6ft be the position of the man. Let C be the
source of light. OC = 20 ft. Then AD is the shadow and D is the tip
of the shadow.
Let OA = x and AD = y (be measured in miles)
Given: dx/dt = 4 m.p.h; dy/dt = ?; d(x+y)/dt = ?
From the figure,
,
20y = 6x+6y
14y = 6x ; y =
Differentiate w.r.t ‘t’
=
=
The shadow is changing at the rate of
m.p.h.
Now
Therefore tip of the shadow is changing at the rate of
m.p.h.
9) A stone is dropped into a pond, waves in the form of circles are
generated and the radius of the outer most ripple increases at the rate
of 2 inches/sec. How fast is the area increasing when the radius is 5
inches?
Ans: Let ‘r’ and ‘A’ be the radius and area of the circle respectively.
Given:
,
Area of circle, Differentiate w.r.t. t
sqinches/sec.
Therefore area of the circle increases at the rate of 20 sq. inches/sec.
C
O
B
A D
KHV’S PAGE 1 http://pue.kar.nic.in
MATHEMATICS: QUESTION BANK
CHAPTER 7: INTEGRALS(INDEFINITE)
Standard forms 1mark questions:
Write an antiderivative for each of the
following functions using differentiation
Question 1: i)sin 2x
Soln: The anti derivative of sin 2x is a
function of x whose derivative is
sin 2x.
It is known that,
Therefore, the anti derivative of
Question 2: Cos 3x
The anti derivative of cos 3x is a function of x
whose derivative is cos 3x.
It is known that,
Therefore, the anti derivative of
.
Question 3: e2x
The anti derivative of e2x
is the function of x
whose derivative is e2x
.
It is known that,
Therefore, the anti derivative of .
Evaluate the following integrals:
Question 4:
Question 5:
Question 6:
Question 7: Find an anti-derivative
of with respect to x.
Ans: cot2x=cosec2x-1; antiderivative
of cot2 x is -cotx-x+c
Question 8: Find an anti-derivative
of √ with respect to x.
√
√
√( ) = sinx+cosx
Antiderivative of sinx+cosx is
cosx-sinx+c
Question 9: Evaluate∫ (
) .
Ans: e5x+c
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TWO MARK QUESTIONS: Evaluate the following integrals:
Write an antiderivative for each of the
following functions using
differentiation :
Question 1:
The anti derivative of is the function
of x whose derivative is .
It is known that,
Therefore, the anti derivative of
.
Question 2:
The anti derivative of is the
function of x whose derivative is
.
It is known that,
Therefore, the anti derivative of
is .
Evaluate the following integrals:
Question 3:
Question 4:
Question 5:
Question 6:
Question 7:
On dividing, we obtain
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Question 8:
Question 9:
Question 10:
Question 11:
Question 12:
Question 13:
Question 14:
Question 15:Find the anti derivative of
Solution:
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3 mark questions:
If such that f(2) = 0, find f(x)
Solution: It is given that
∴Anti derivative of
∴
Also,
INTEGRATION BY SUBSTITUTION:
ONE MARK QUESTIONS:
1`. Evaluate 2tan (2 ).x dx
Solution:
2 2tan (2 ). (sec 2 1)
1tan 2
2
x dx x dx
x x c
2. Evaluate 2 xcosec dx
2
.
= -2cot
+c
TWO MARK QUESTIONS:
Integrate the following w.r.t x
1.
Hint: = t Ans: log(1+x2) +c
2.
Hint: log |x| = t ∴
Question 3:
Let 1 + log x = t
∴
Question 4:sin x ⋅ sin (cos x)
sin x ⋅ sin (cos x)
Let cos x = t
∴ −sin x dx = dt
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Question 5:
Let ax + b = t ⇒ adx = dt
Question 6:
Let 1 + 2x2 = t ∴ 4xdx = dt
Question 7:
Let ∴ (2x + 1)dx = dt
Question 8:
Let
∴
Question 9:
Let
∴ 2dx = dt
Question 10:
Let
∴ 9x2 dx = dt
Question 11:
Let log x = t ∴
Question 12:
Let
∴ −8x dx = dt
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Question 13:
Let ∴ 2xdx = dt
Question 14:
Let ∴ 2xdx = dt
Question 15:
Let ∴
Question 16:
Let ∴
Question 17:
Let sin 2x = t ∴
Question 18:
Let
∴ cos x dx = dt
Question 19: cot x log sin x
Let log sin x = t
Question 20:
Let 1 + cos x = t ∴ −sin x dx = dt
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Question 21:
Let 1 + cos x = t ∴ −sin x dx = dt
Question 22:
Let 1 + log x = t ∴
Question 23: equals
Let Also, let
⇒
Question 24: Find
Let ∴
Question 25: =
THREE MARKS QUESTIONS
Integrate the following :
Question 1:
Let ∴ 2adx = dt
Question 2:
Let ∴ dx = dt
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Question 3:
Let ∴ dx = dt
Question 4:
Let ∴
Question 5:
Dividing numerator and denominator by ex,
we obtain
Let ∴
Question 6:
Let ∴
Question 7:
Let 2x − 3 = t ∴ 2dx = dt
Question 8:
Let 7 − 4x = t ∴ −4dx = dt
Question 9:
Let ∴
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Question 10:
Let
∴
Question 11:
Let ∴
Question 12:
Let ∴
Question 13:
Let sin x + cos x = t ⇒ (cos x − sin x) dx = dt
Question 14:
Put cos x − sin x = t ⇒ (−sin x − cos x) dx = dt
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Question 15:
Question 16:
Let
∴
Question 17:
Let x4 = t
∴ 4x3 dx = dt
Let
∴ From (1), we obtain
******
INTEGRATION USING TRIGONOMETRIC IDENTITIES:
THREE MARKS QUESTIONS:
Integrate the following functions:
Question 1:
Question 2: It is known that,
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Question 3: cos 2x cos 4x cos 6x
It is known that,
Question 4: sin
3 (2x + 1)
Let
Question 5: sin
3 x cos
3 x
Question 6: sin x sin 2x sin 3x
It is known that
Question 7:sin 4x sin 8x
It is known that
Question 8:
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Question 9:
Question 10: sin
4 x
Question 11: cos
4 2x
Question 12:
Question 13:
Question 14:
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Question 15:
Question 16: tan
4x
From equation (1), we obtain
Question 17:
Question 18:
Question 19:
Question 20:
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Question 21: sin−1
(cos x)
It is known that,
Substituting in equation (1), we obtain
Question 22:
Question 23:
Question 24:
Let exx = t
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INTEGRALS OF SOME PARTICULAR FUNCTIONS
TWO MARK QUESTIONS: Integrate the following w.r.t x
Question 1:
Let x3 = t ∴ 3x
2 dx = dt
Question 2:
Let 2x = t ∴ 2dx = dt
Question 3:
Let 2 − x = t ⇒ −dx = dt
Question 4:
Let 5x = t ∴ 5dx = dt
Question 5:
Question 6:
Let x3 = t ∴ 3x
2 dx = dt
Question 7:
From (1), we obtain
Question 8:
Let x3 = t ⇒ 3x
2 dx = dt
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Question 9:
Let tan x = t ∴ sec2x dx = dt
Question 10:
Question 11:
Question 12:
Question 13:
Question 14:
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THREE MAKRS QUESTIONS:
Question 1:
Question 2:
Question 3:
Question 4:
Equating the coefficients of x and
constant term on both sides, we obtain
4A = 4 ⇒ A = 1
A + B = 1 ⇒ B = 0
Let 2x2 + x − 3 = t
∴ (4x + 1) dx = dt
Question 5:
Equating the coefficients of x and constant
term on both sides, we obtain
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From (1), we obtain
From equation (2), we obtain
Question 6: Integrate 2
x 2
x 2x 3
with respect to x.
2 2
12x 2 1
x 2 2I dx dxx 2x 3 x 2x 3
2 2
1 2x 2 1dx dx
2 x 2x 3 x 2x 3
2
2
1 12 x 2x 3 dx
2 x 1 2
2 2x 2x 3 log x 1 x 2x 3 c
ADDITIONAL 4 TO 5 MARK QUESTIONS: Integrate the following:
Question 1:
Equating the coefficient of x and constant term
on both sides, we obtain
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Substituting equations (2) and (3) in equation
(1), we obtain
Question 2:
Equating the coefficients of x and constant
term, we obtain 2A = 6 ⇒ A = 3
−9A + B = 7 ⇒ B = 34
∴ 6x + 7 = 3 (2x − 9) + 34
Substituting equations (2) and (3) in (1), we
obtain
Question 3:
Equating the coefficients of x and constant
term on both sides, we obtain
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Using equations (2) and (3) in (1), we obtain
Question 4:
Let x
2 + 2x +3 = t
⇒ (2x + 2) dx =dt
Using equations (2) and (3) in (1), we obtain
Question 5:
Equating the coefficients of x and constant
term on both sides, we obtain
Substituting (2) and (3) in (1), we obtain
Question 6:
Equating the coefficients of x and constant
term, we obtain
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Using equations (2) and (3) in (1), we obtain
INTEGRATION BY PARTIAL FRACTIONS
TWO MARK QUESTIONS:
Question 1:
Let
Equating the coefficients of x and constant
term, we obtain
A + B = 1 ; 2A + B = 0
On solving, we obtain A = −1 and B = 2
Question 2:
Let
Equating the coefficients of x and constant
term, we obtain
A + B = 0 ; −3A + 3B = 1
On solving, we obtain
THREE MARK QUESTIONS:
Question 1:
Let
Substituting x = 1, 2, and 3 respectively in
equation (1), we obtain
A = 1, B = −5, and C = 4
Question 2:
Let
Substituting x = 1, 2, and 3 respectively in
equation (1), we obtain
KHV’S PAGE 22 http://pue.kar.nic.in
Question 3:
Let
Substituting x = −1 and −2 in equation (1), we
obtain A = −2 and B = 4
Question 4:
It can be seen that the given integrand is not a
proper fraction. Therefore, on dividing (1 −
x2) by x(1 − 2x), we obtain
Let
Substituting x = 0 and in equation (1), we
obtain A = 2 and B = 3
Substituting in equation (1), we obtain
Question 4:
Let
Equating the coefficients of x
2, x, and constant
term, we obtain
A + C = 0 ;−A + B = 1 ; −B + C = 0
On solving these equations, we obtain
From equation (1), we obtain
Question 5:
Let
Substituting x = 1, we obtain
Equating the coefficients of x2 and constant
term, we obtain
A + C = 0 ;−2A + 2B + C = 0
On solving, we obtain
Question 6:
KHV’S PAGE 23 http://pue.kar.nic.in
Let
Substituting x = 1 in equation (1), we obtain
B = 4
Equating the coefficients of x2 and x, we
obtain A + C = 0; B − 2C = 3
On solving, we obtain
Question 7:
Let
Equating the coefficients of x
2 and x, we
obtain
Question 8:
Let
Substituting x = −1, −2, and 2 respectively in
equation (1), we obtain
Question 9:
It can be seen that the given integrand is not a
proper fraction.
Therefore, on dividing (x3 + x + 1) by x
2 − 1,
we obtain
Let
Substituting x = 1 and −1 in equation (1), we
obtain
Question 10:
Equating the coefficient of x and constant
term, we obtain A = 3
2A + B = −1 ⇒ B = −7
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Question 11:
[Hint: multiply numerator and denominator by
xn − 1
and put xn = t]
Multiplying numerator and
denominator by xn − 1
, we obtain
Substituting t = 0, −1 in equation (1), we
obtain
A = 1 and B = −1
Question 12: [Hint: Put
sin x = t]
Substituting t = 2 and then t = 1 in equation
(1), we obtain A = 1 and B = −1
Question 13:
Let x2 = t ⇒ 2x dx = dt
Substituting t = −3 and t = −1 in equation (1),
we obtain
Question 14:
Multiplying numerator and denominator by x
3,
we obtain
Let x
4 = t ⇒ 4x
3dx = dt
Substituting t = 0 and 1 in (1), we obtain
A = −1 and B = 1
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Question 15:
[Hint: Put ex = t]
Let ex = t ⇒ e
x dx = dt
Substituting t = 1 and t = 0 in equation (1), we
obtain A = −1 and B = 1
Question 16:
Substituting x = 1 and 2 in (1), we obtain
A = −1 and B = 2
Question 17:
Equating the coefficients of x
2, x, and constant
term, we obtain
A + B = 0; C = 0 ; A = 1
On solving these equations, we obtain
A = 1, B = −1, and C = 0
ADDITIONAL QUESTIONS: 4 TO 5 MARKS:
Question 1:
Equating the coefficient of x
2, x, and constant
term, we obtain
A − B = 0; B − C = 0; A + C = 2
On solving these equations, we obtain
A = 1, B = 1, and C = 1
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Question 2:
Equating the coefficient of x
3, x
2, x, and
constant term, we obtain
On solving these equations, we obtain
Question 18:
Equating the coefficients of x
3, x
2, x, and
constant term, we obtain
A + C = 0; B + D = 4 ;4A + 3C = 0
4B + 3D = 10
On solving these equations, we obtain
A = 0, B = −2, C = 0, and D = 6
INTEGRATION BY PARTS TWO MARKS QUESTIONS:
Question 1: x sin x
Let I =
Taking x as first function and sin x as second
function and integrating by parts, we obtain
Question 2:
Let I =
Taking x as first function and sin 3x as second
function and integrating by parts, we obtain
Question 3: ∫ . Given Integral
= ∫ ∫
.
=
KHV’S PAGE 27 http://pue.kar.nic.in
Question 4: x logx
Let
Taking log x as first function and x as second
function and integrating by parts, we obtain
Question 5: x log 2x
Let
Taking log 2x as first function and x as second
function and integrating by parts, we obtain
Question 6: x
2 log x
Let
Taking log x as first function and x2 as second
function and integrating by parts, we obtain
Question 7:
Let
Taking x as first function and sec2x as second
function and integrating by parts, we obtain
THREE MARKS QUESTIONS:
Integrate the following w.r.t. x
Question 1:
Let
Taking x2 as first function and e
x as second
function and integrating by parts, we obtain
Again integrating by parts, we obtain
Question 2:
Let
Taking as first function and x as
second function and integrating by parts, we
obtain
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Question 3:
Let
Taking as first function and x as
second function and integrating by parts, we
obtain
Question 4: Evaluate: ∫ .
∫ ∫
∫
∫ .
= ∫
∫
=
| |
Question 5:
Let
Taking cos−1
x as first function and x as second
function and integrating by parts, we obtain
Question 6:
Let
Taking as first function and 1 as
second function and integrating by parts, we
obtain
KHV’S PAGE 29 http://pue.kar.nic.in
Question 7:
Let
Taking as first function and
as second function and integrating by parts, we
obtain
*******
INTEGRAL OF THE FORM ∫ [ ( ) ( )]
TWO MARK QUESTIONS
Question 1:
Let
Let
⇒
∴ It is known that,
Question 2:
Also, let ⇒
It is known that,
2. Evaluate: x 1 sin xe dx
1 cos x
.
x
2
x
2
x 2 x
x
x x1 2sin cos
2 2I e dxx
2cos2
1 xe tan dx
x 22cos
2
1 x xe sec tan dx e f '(x) f (x) dx
2 2 2
xe tan
2
THREE MARK QUESTIONS
Integrate the following w.r.t x
Question 1:
Let
KHV’S PAGE 30 http://pue.kar.nic.in
Let ⇒
It is known that,
Question 2:
Let ⇒
It is known that,
From equation (1), we obtain
Question 3:
Let ⇒
It is known that,
Question 3:
Let
Integrating by parts, we obtain
Again integrating by parts, we obtain
Question 4:
Let ⇒
= 2θ
⇒
KHV’S PAGE 31 http://pue.kar.nic.in
Integrating by parts, we obtain
Question 5:
equals
Let
Also, let ⇒
It is known that,
INDEFINITE INTEGRALS
(FIVE MARK QUESTIONS)
1. Find the integral of
√ with
respect to x and hence evaluate
∫
√ .
Solution: Let then
∴ ∫
√ ∫
√
∫ (
)
Consider ∫
√
∫
√ ( )
(
)
2. Find the integral of
√ with
respect to x and hence evaluate
∫
√ .
Solution: Let
√
Let then
∴ ∫
√ ∫
| |
|
√
|
| √ | | |
| √ | ,
| |
Consider ∫
√
∫
√( )
|( ) √( ) | .
3. Find the integral of
with respect
to x and evaluate ∫
.
Solution: ∫
Consider
( )( )
[( ) ( )
( )( )]
[
( )]
[
( )]
∴ ∫
*
( )+ ∫
*
( )+
[ | | | |]
|
|
Consider∫
∫
( ) Let
then
∴
∫
( )
|
| .
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4. Find the integral of
with respect
to x and hence evaluate ∫
( ) .
Solution: ∫
Consider
( )( )
[( ) ( )
( )( )]
[
( )]
[
( )]
∴ ∫
*
( )+ ∫
*
( )+
[ | | | |]
|
|
Consider ∫
( )
Let ( ) then
∴
∫
|
| .
=
|
( )
( )|
5.Find the integral of 2 2
1
x awith
respect to and hence evaluate
2
1dx
5x 2x
Substituting
∫
√ =∫
√
=∫
√ ( ) ∫
√
=∫
=log|sec +tan |+C1
=log|
√
|+C1
= log| √ |-log|a|+C1
=log| √ |+C
where C=C1 –log|a|
Now ∫
√
=x2+2x+1=(x2+2x+1)+1
=(x+1)2+(1)2
∫
√ =∫
√( )
= | √ |
6.Find the integral of 2 2
1
x awith
respect to and hence evaluate
∫
Let I= 2 2
1dx;
x aPut x=atan ,
then dx=a sec2 ;
x2+a2 =a2 tan2 +a2
=a2(tan2 +1)=a2sec2
2
2 2 2 2
1
1 asec d 1I dx d
x a a sec a
1 1 xtan c
a a a
Consider ∫
7.Find the integral of √ with
respect to x and evaluate ∫ √ .
Solution: Let ∫√ applying
integration by parts
We get √
∫
√
√ ∫
√
√ ∫
√
KHV’S PAGE 33 http://pue.kar.nic.in
√ | √ |
√
| √ |
Consider ∫ √
√
| √ |
8. Find the integral of √ with
respect to x and evaluate
∫ √ .
Solution: Let ∫√ applying
integration by parts
We get √ ∫
√
√ ∫
√
√ ∫
√
√ (
)
√
(
)
Consider ∫ √
∫ √ ( )
( )
√ ( )
(
√ )
Find the integral of √ with
respect to x and hence evaluate
∫
Solution: Let ∫√ applying
integration by parts
We get √
∫
√
√ ∫( )
√
= √ ∫ *√
√ +
√ ∫
√
√ | √ |
√
| √ |
Now consider ∫
x2+2x+5=(x2+2x+1)+4=(x+1)2+4
I=∫[( ) ]
Put x+1=t dx=dt
√
√ +
| √ |
=
√
+ 2log| x+1+√ |
Note: The above questions is for 5 mark
questions in part D of the question paper for
second PUC.
Note: In this chapter “Indefinite Integrals”
Some of the solved examples given in the text
book are not included in the question bank.
The students are advised to go through the
those questions also.
*******
KHV’S PAGE 34 http://pue.kar.nic.in
Assignments (i) Integration by substitution
LEVEL I
1. 2. 3.
LEVEL II
1. 2. 3.
LEVEL III
1. 2. 3.
(ii) Application of trigonometric function in integrals LEVEL I
1. 2. 3.
LEVEL II
1. 2.
LEVEL III
1. 2.
(iii) Integration using standard results LEVEL I
1. 2. 3.
LEVEL II
1. 2. 3.
LEVEL III
1. 2. 3.
4. 5. [CBSE 2011]
(iv) Integration using Partial Fraction
LEVEL I
dxx
)x(logsec2
dxx1
e2
x1tanm
dx
x1
e
2
x1sin
dxxx
1
dx
1xx
1
6
dx
1e
1
x
dxxcos.xsin
xtan
dxxcosxsec
xtan dx
xcos.xsin
13
dx.xsin3 dx.x3cos2
dx.x3cos.x2cos.xcos
dx.xtan.xsec4 dx
xsin
x4sin
dx.xcos5 dx.xcos.xsin 32
9x4
dx
2
dx
10x2x
12
13x12x9
dx2
dx1xx
x24
dx
5xsin4xsin
xcos2
2xx67
dx
dxxx1
x2
42
dx
1xx
1xx2
2
dx
6x5x
2x
2
dx
x1
x1
4x5x
7x6
KHV’S PAGE 35 http://pue.kar.nic.in
1. 2. 3.
LEVEL II
1. 2. 3.
LEVEL III
1. 2. 3.
(v) Integration by Parts LEVEL I
1. 2. 3.
LEVEL II
1. 2. 3.
4. 5.
LEVEL III
1. 2. 3.
4. 5.
(vi) Some Special Integrals LEVEL I
1. 2.
LEVEL II
1. 2.
LEVEL III
1. 2.
(vii) Miscellaneous Questions LEVEL II
1. (Hint: Divide the Numerator and Denominator by cos2x and use the relation
sec2x=1+tan
2x; and put tanx=t
LEVEL III
1.∫
dx
)1x)(1x(
1x2
dx)3x)(2x)(1x(
x2
dx
)3x()1x(
2x32
dx)2x)(1x(
8x2x2
dx
)2x(x
1xx2
2
dx
)3x()1x(
1x2
2
dx)4x)(2x(
82 x2sinxsin
dx
dx
x1
13
dx.xsec.x 2
dx.xlog
dx)xseclogx(tanex
dx.xsin 1
dx.xsin.x 12
dx
x1
xsin.x
2
1
dx.x1
x1cos
2
21
dx.xsec3
dxxlogcos
dx)x2(
)x1(e2
x
dx)xlog1(
xlog2
dx.e
x2cos1
xsin2 x
dx.x3cos.e x2
dx.x4 2
dx.x41 2
dx.6x4x2
dx.xx41 2
dx.xx1)1x( 2 dxxx)5x( 2
xcos5xsin4
dx22
KHV’S PAGE 36 http://pue.kar.nic.in
SOLUTIONS: ASSIGNMENTS: INDEFINITE INTEGRALS
(i) Integration by substitution
LEVEL I 1. tan(logex) + C 2. 3.
LEVEL II 1. 2. 3.
LEVEL III 1. 2. 3.
(ii) ) Application of trigonometric function in integrals
LEVEL I 1. 2.
3.
LEVEL II 1. 2.
LEVEL III 1. 2.
(iii) Integration using Standard results
LEVEL I 1. 2. + C 3. + C
LEVEL II 1. + C 2. 3.
LEVEL III 1. 2.
3. C
4. [Hint: Put x=cos2 ]
5.
(iv) Integration using Partial Fraction
LEVEL I 1. 2.
3.
LEVEL II 1. x – 11log(x – 1) + 16log(x – 2) + C 2.
3.
Cem
1 xtanm 1
Ce xsin 1
Cx1log2 e Cxsec3
1 31 Ce1log x
e
Cxtan2 Cxcostan 1 Cxtanlog2
xtane
2
Cx3cos12
1xcos
4
3 C
6
x6sinx
2
1
Cx2sin8
1x4sin
16
1x6sin
4
1
4
x
C4
xtan
2
xtanORCxsec
4
1 424 Cxsin2x3sin
3
2
Cxsin5
1xsin
3
2xsin 53 C
5
xsin
3
xsin 53
C9x42
1xlog
2
1 2e
3
1xtan
3
1 1
3
2x3tan
9
1 1
3
1x2tan
3
1 21 C2xsintan 1 C
5
1x2sin 1
C5
1x2sin
21
C
3
1x2log
3
21xxlogx 2
6x5x
2
5xlog
2
16x5x 22
Cx1xsin 21
C20x9x2
9x2log3420x9x6 22
C)2xlog(3
5)1xlog(
3
1 C3xlog
2
32xlog21xlog
2
1
C
1x2
5
3x
1xlog
4
11
C2xlog4
3
x2
1xlog
4
1
C3xlog8
5
1x2
11xlog
8
3
KHV’S PAGE 37 http://pue.kar.nic.in
LEVEL III 1. log(x +2) 2.
3. + C [Hint: Partial fractions]
(v) Integration by Parts
LEVEL I 1.x.tanx + logcosx + C 2.xlogx – x + C 3.ex.logsecx + C
LEVEL II 1. 2.
3. 4.
5.
LEVEL III 1. 2. [Hint: =
3. 4. ex.tanx + C 5.
(vi) Some Special Integrals
LEVEL I 1. 2.
LEVEL II 1.
2.
LEVEL III 1.
2.
(vii) Miscellaneous Questions
LEVEL II 1.
LEVEL III 1.
+
log|cosx+sinx|+C
2
xtan4xlog
2
1 12
C
3
xcos21log2
2
xcos1log
6
xcos1log
3
1x2tan
3
1xx1log
6
1x1log
3
1 12
Cx1xsinx 21 C
9
x12xxsin
3
x 221
3
Cxxsinx1 12 Cx1logxtanx2 21
Cxtanxseclogxtan.xsec2
1
Cxlogsinxlogcos2
x C
x2
ex
Cxlog1
x
Cx3cos2x3sin3
13
e x2
Cx4xlog22
x4x 22
Cx2sin4
1
2
x41x 12
C6x4x2xlog
2
6x4x2x 22
C
5
2xsin
2
5
2
xx412x 12
C5
1x2sin
16
5xx11x2
8
1xx1
3
1 122/32
Cxx21x2log16
11xx1x2
8
11xx
3
1 222/32
C5
xtan2tan
52
1 1
KHV’S PAGE 38 http://pue.kar.nic.in
Detail of the concepts to be mastered by every student of class second PUC
with exercises and examples of NCERT Text Book.
Indefinit
e
Integrals
(i) Integration by substitution * Text book , Vol. II Examples 5&6
Page 300, 302,301,303
(ii) ) Application of trigonometric
function in integrals
** Text book , Vol. II Ex 7 Page 306,
Exercise 7.3 Q13&Q24
(iii) Integration of some particular
function
22 ax
dx, ,
, ,
, ,
*** Text book , Vol. II Exp 8, 9, 10
Page 311,312,313, Exercise 7.4 Q
3,4,8,9,13&23
(iv) Integration using Partial Fraction *** Text book , Vol. II Exp 11&12
Page 318 Exp 13 319,Exp 14 & 15
Page320
(v) Integration by Parts ** Text book , Vol. II Exp 18,19&20
Page 325 Exs 7.6 QNO ,10,11,
17,18,20
(vi)Some Special Integrals
,
*** Text book , Vol. II Exp 23 &24
Page 329
(vii) Miscellaneous Questions ** Text book , Vol. II Solved Ex. 40,
41
viii)Some special integrals Text book Supplimentary material
Page 614,615
SYMBOLS USED :
* : Important Questions, ** :Very Important Questions, *** : Very-Very Important Questions
22 ax
dx
dx
xa
1
22 cbxax
dx2
cbxax
dx
2
cbxax
dx)qpx(2
cbxax
dx)qpx(
2
dxxa 22
dxax 22
KHV’S PAGE 1 http://pue.kar.nic.in
MATHEMATICS: QUESTION BANK
CHAPTER 7: INTEGRALS(INDEFINITE)
Standard forms 1mark questions:
Write an antiderivative for each of the
following functions using differentiation
Question 1: i)sin 2x
Ans: The anti derivative of sin 2x is a
function of x whose derivative is
sin 2x.
It is known that,
Therefore, the anti derivative of
Question 2: Cos 3x
The anti derivative of cos 3x is a function of x
whose derivative is cos 3x.
It is known that,
Therefore, the anti derivative of
.
Question 3: e2x
The anti derivative of e2x is the function of x
whose derivative is e2x.
It is known that,
Therefore, the anti derivative of .
Evaluate the following integrals:
Question 4:
Question 5:
Question 6:
Question 7: Find an anti-derivative
of cot2 𝑥 with respect to x.
Ans: cot2x=cosec2x-1;
antiderivative of cot2 x is
-cotx-x+c
Question 8: Find an anti-derivative
of √1 + sin 2𝑥 with respect to x.
√1 + sin 2𝑥
= √sin2 𝑥 + cos2 𝑥 + 2𝑠𝑖𝑛𝑥 𝑐𝑜𝑠𝑥
= √(𝑠𝑖𝑛𝑥 + 𝑐𝑜𝑠𝑥)2= sinx+cosx
Antiderivative of sinx+cosx is
cosx-sinx+c
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TWO MARK QUESTIONS: Evaluate the following integrals:
Write an antiderivative for each of the
following functions using differentiation :
Question 1:
The anti derivative of is the function
of x whose derivative is .
It is known that,
Therefore, the anti derivative of
.
Question 2:
The anti derivative of is the
function of x whose derivative is
.
It is known that,
Therefore, the anti derivative of
is .
Evaluate the following integrals:
(Question 3 to 14)
Question 3:
Question 4:
Question 5:
Question 6:
Question 7:
Simplifying and dividing by x-1, we obtain
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Question 8:
Question 9:
Question 10:
Question 11:
Question 12:
Question 13:
Question 14:
Question 15:Find the anti derivative of
Solution:
KHV’S PAGE 4 http://pue.kar.nic.in
3 mark questions:
If such that f(2) = 0, find f(x)
Solution: It is given that
∴Anti derivative of
∴
Also,
INTEGRATION BY SUBSTITUTION:
ONE MARK QUESTIONS:
1`. Evaluate 2tan (2 ).x dx
Solution:
2 2tan (2 ). (sec 2 1)
1tan 2
2
x dx x dx
x x c
2. Evaluate2 x
cosec dx2
.
= -2cot𝑥
2+c
TWO MARK QUESTIONS:
Integrate the following functions w.r.t x
Question 1.
Hint: = t Ans: log(1+x2) +c
Question 2.
Hint: log |x| = t ∴
Question 3:
Let 1 + log x = t
∴
Question 4:sin x ⋅ sin (cos x)
sin x ⋅ sin (cos x)
Let cos x = t
∴ −sin x dx = dt
Question 5:
Let ax + b = t ⇒ adx = dt
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Question 6:
Let 1 + 2x2 = t ∴ 4xdx = dt
Question 7:
Let ∴ (2x + 1)dx = dt
Question 8:
Let
∴
Question 9:
Let
∴ 2dx = dt
Question 10:
Let
∴ 9x2 dx = dt
Question 11:
Let log x = t ∴
Question 12:
Let
∴ −8x dx = dt
Question 13:
Let ∴ 2xdx = dt
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Question 14:
Let ∴ 2xdx = dt
Question 15:
Let ∴
Question 16:
Let ∴
Question 17:
Let sin 2x = t ∴
Question 18:
Let
∴ cos x dx = dt
Question 19: cot x log sin x
Let log sin x = t
Question 20:
Let 1 + cos x = t ∴ −sin x dx = dt
Question 21:
Let 1 + cos x = t ∴ −sin x dx = dt
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Question 22:
Let 1 + log x = t ∴
Question 23: equals
Let Also, let
⇒
Question 24: Find
Let ∴
Question 25: =
THREE MARKS QUESTIONS
Integrate the following :
Question 1:
Let ∴ 2adx = dt
Question 2:
Let ∴ dx = dt
KHV’S PAGE 8 http://pue.kar.nic.in
Question 3:
Let ∴ dx = dt
Question 4:
Let ∴
Question 5:
Dividing numerator and denominator by ex,
we obtain
Let ∴
Question 6:
Let ∴
Question 7:
Let 2x − 3 = t ∴ 2dx = dt
Question 8:
Let 7 − 4x = t ∴ −4dx = dt
Question 9:
Let ∴
KHV’S PAGE 9 http://pue.kar.nic.in
Question 10:
Let
∴
Question 11:
Let ∴
Question 12:
Let ∴
Question 13:
Let sin x + cos x = t ⇒ (cos x − sin x) dx = dt
Question 14:
Put cos x − sin x = t ⇒ (−sin x − cos x) dx = dt
KHV’S PAGE 10 http://pue.kar.nic.in
Question 15:
Question 16:
Let
∴
Question 17:
Let x4 = t
∴ 4x3 dx = dt
Let
∴ From (1), we obtain
******
INTEGRATION USING TRIGONOMETRIC IDENTITIES:
THREE MARKS QUESTIONS:
Integrate the following functions:
Question 1:
Question 2: It is known that,
Question 3: cos 2x cos 4x cos 6x
It is known that,
KHV’S PAGE 11 http://pue.kar.nic.in
Question 4: sin3 (2x + 1)
Let
Question 5: sin3 x cos3 x
Question 6: sin x sin 2x sin 3x
It is known that
Question 7:sin 4x sin 8x
It is known that
Question 8:
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Question 9:
Question 10: sin4 x
Question 11: cos4 2x
Question 12:
Question 13:
Question 14:
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Question 15:
Question 16: tan4x
From equation (1), we obtain
Question 17:
Question 18:
Question 19:
Question 20:
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Question 21: sin−1 (cos x)
Sin-1(cosx)=sin-1[sin(2
-x)]= x
2
1sin (cos x) ( x)dx2
=
2xx
2 2
+C
Question 22:
Question 23:
Question 24:
Let exx = t
INTEGRALS OF SOME PARTICULAR FUNCTIONS
TWO MARK QUESTIONS: Integrate the following w.r.t x
Question 1:
Let x3 = t ∴ 3x2 dx = dt
Question 2:
Let 2x = t ∴ 2dx = dt
Question 3:
Let 2 − x = t ⇒ −dx = dt
Question 4:
Let 5x = t ∴ 5dx = dt
KHV’S PAGE 15 http://pue.kar.nic.in
Question 5:
Question 6:
Let x3 = t ∴ 3x2 dx = dt
Question 7:
From (1), we obtain
Question 8:
Let x3 = t ⇒ 3x2 dx = dt
Question 9:
Let tan x = t ∴ sec2x dx = dt
Question 10:
Question 11: 1
9𝑥2+6𝑥+5
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Question 12:
Question 13:
Question 14:
THREE MAKRS QUESTIONS:
Question 1:
Question 2:
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Question 3:
Question 4:
Equating the coefficients of x and
constant term on both sides, we obtain
4A = 4 ⇒ A = 1
A + B = 1 ⇒ B = 0
Let 2x2 + x − 3 = t
∴ (4x + 1) dx = dt
Question 5:
Equating the coefficients of x and constant
term on both sides, we obtain
From (1), we obtain
From equation (2), we obtain
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Question 6: Integrate 2
x 2
x 2x 3
with respect to x.
2 2
12x 2 1
x 2 2I dx dxx 2x 3 x 2x 3
2 2
1 2x 2 1dx dx
2 x 2x 3 x 2x 3
2
2
1 12 x 2x 3 dx
2 x 1 2
2 2x 2x 3 log x 1 x 2x 3 c
INTEGRATION BY PARTIAL FRACTIONS
TWO MARK QUESTIONS:
Question 1:
Let
Equating the coefficients of x and constant
term, we obtain
A + B = 1 ; 2A + B = 0
On solving, we obtain A = −1 and B = 2
Question 2:
Let
Equating the coefficients of x and constant
term, we obtain
A + B = 0 ; −3A + 3B = 1
On solving, we obtain
THREE MARK QUESTIONS:
Question 1:
Let
Substituting x = 1, 2, and 3 respectively in
equation (1), we obtain
A = 1, B = −5, and C = 4
Question 2:
Let
Substituting x = 1, 2, and 3 respectively in
equation (1), we obtain
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Question 3:
Let
Substituting x = −1 and −2 in equation (1), we
obtain A = −2 and B = 4
Question 4:
It can be seen that the given integrand is not a
proper fraction. Therefore, on dividing (1 −
x2) by x(1 − 2x), we obtain
Let
Substituting x = 0 and in equation (1), we
obtain A = 2 and B = 3
Substituting in equation (1), we obtain
Question 4:
Let
Equating the coefficients of x2, x, and constant
term, we obtain
A + C = 0 ;−A + B = 1 ; −B + C = 0
On solving these equations, we obtain
From equation (1), we obtain
Question 5:
Let
Substituting x = 1, we obtain
Equating the coefficients of x2 and constant
term, we obtain
A + C = 0 ;−2A + 2B + C = 0
On solving, we obtain
Question 6:
KHV’S PAGE 20 http://pue.kar.nic.in
Let
Substituting x = 1 in equation (1), we obtain
B = 4
Equating the coefficients of x2 and x, we
obtain A + C = 0; B − 2C = 3
On solving, we obtain
Question 7:
Let
Equating the coefficients of x2 and x, we
obtain
Question 8:
Let
Substituting x = −1, −2, and 2 respectively in
equation (1), we obtain
Question 9:
It can be seen that the given integrand is not a
proper fraction.
Therefore, on dividing (x3 + x + 1) by x2 − 1,
we obtain
Let
Substituting x = 1 and −1 in equation (1), we
obtain
Question 10:
Equating the coefficient of x and constant
term, we obtain A = 3
2A + B = −1 ⇒ B = −7
Question 11:
[Hint: multiply numerator and denominator by
xn − 1 and put xn = t]
KHV’S PAGE 21 http://pue.kar.nic.in
Multiplying numerator and
denominator by xn − 1, we obtain
Substituting t = 0, −1 in equation (1), we
obtain
A = 1 and B = −1
Question 12: [Hint: Put
sin x = t]
Substituting t = 2 and then t = 1 in equation
(1), we obtain A = 1 and B = −1
Question 13:
Let x2 = t ⇒ 2x dx = dt
Substituting t = −3 and t = −1 in equation (1),
we obtain
Question 14:
Multiplying numerator and denominator by x3,
we obtain
Let x4 = t ⇒ 4x3dx = dt
Substituting t = 0 and 1 in (1), we obtain
A = −1 and B = 1
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Question 15:
[Hint: Put ex = t]
Let ex = t ⇒ ex dx = dt
Substituting t = 1 and t = 0 in equation (1), we
obtain A = −1 and B = 1
Question 16:
Substituting x = 1 and 2 in (1), we obtain
A = −1 and B = 2
Question 17:
Equating the coefficients of x2, x, and constant
term, we obtain
A + B = 0; C = 0 ; A = 1
On solving these equations, we obtain
A = 1, B = −1, and C = 0
INTEGRATION BY PARTS TWO MARKS QUESTIONS:
Question 1: x sin x
Let I =
Taking x as first function and sin x as second
function and integrating by parts, we obtain
Question 2:
Let I =
Taking x as first function and sin 3x as second
function and integrating by parts, we obtain
Question 3: ∫ 𝑙𝑜𝑔𝑥 𝑑𝑥. Given Integral
=log 𝑥 ∙ ∫ 1 𝑑𝑥 − ∫ 1.𝑑
𝑑𝑥log 𝑥 𝑑𝑥.
== 𝑥 log 𝑥 − 𝑥 + 𝑐
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Question 4: x logx
Let
Taking log x as first function and x as second
function and integrating by parts, we obtain
Question 5: x log 2x
Let
Taking log 2x as first function and x as second
function and integrating by parts, we obtain
Question 6: x2 log x
Let
Taking log x as first function and x2 as second
function and integrating by parts, we obtain
Question 7:
Let
Taking x as first function and sec2x as second
function and integrating by parts, we obtain
THREE MARKS QUESTIONS:
Integrate the following w.r.t. x
Question 1:
Let
Taking x2 as first function and ex as second
function and integrating by parts, we obtain
Again integrating by parts, we obtain
Question 2:
Let
Taking as first function and x as
second function and integrating by parts, we
obtain
KHV’S PAGE 24 http://pue.kar.nic.in
Question 3:
Let
Taking as first function and x as
second function and integrating by parts, we
obtain
Question 4: Evaluate: ∫ tan−1 𝑥 𝑑𝑥.
∫ tan−1 𝑥 𝑑𝑥 = 𝑡𝑎𝑛−1 𝑥 . ∫ 1 𝑑𝑥
− ∫𝑑
𝑑𝑥𝑡𝑎𝑛− 1𝑥. ∫ 1 𝑑𝑥 . 𝑑𝑥.
=𝑥 tan−1 𝑥 − ∫𝑥
1+𝑥2 𝑑𝑥
= 𝑥 tan−1 𝑥 −1
2∫
2𝑥
1 + 𝑥2
= 𝑥 tan−1 𝑥 −1
2log|1 + 𝑥2| + 𝑐
Question 5:
Let
Taking cos−1 x as first function and x as second
function and integrating by parts, we obtain
Question 6:
Let
KHV’S PAGE 25 http://pue.kar.nic.in
Taking as first function and 1 as
second function and integrating by parts, we
obtain
Question 7:
Let
Taking as first function and
as second function and integrating by parts, we
obtain
*******
INTEGRAL OF THE FORM ∫ 𝒆𝒙[𝒇(𝒙) + 𝒇′(𝒙)]𝒅𝒙
TWO MARK QUESTIONS
Question 1:
Let
Let
⇒
∴ It is known that,
Question 2:
Also, let ⇒
It is known that,
2. Evaluate: x 1 sin x
e dx1 cos x
.
x
2
x
2
x 2 x
x
x x1 2sin cos
2 2I e dxx
2cos2
1 xe tan dx
x 22cos
2
1 x xe sec tan dx e f '(x) f (x) dx
2 2 2
xe tan
2
THREE MARK QUESTIONS
Integrate the following w.r.t x
Question 1:
Let
KHV’S PAGE 26 http://pue.kar.nic.in
Let ⇒
It is known that,
Question 2:
Let ⇒
It is known that,
From equation (1), we obtain
Question 3:
Let ⇒
It is known that,
Question 3:
Let
Integrating by parts, we obtain
Again integrating by parts, we obtain
Question 4:
Let ⇒
= 2θ
⇒
KHV’S PAGE 27 http://pue.kar.nic.in
Integrating by parts, we obtain
Question 5:
equals
Let
Also, let ⇒
It is known that,
INDEFINITE INTEGRALS
(FIVE MARK QUESTIONS)
1. Find the integral of 𝟏
√𝒂𝟐− 𝒙𝟐 with
respect to x and hence evaluate
∫𝟏
√𝟕−𝟔𝒙−𝒙𝟐 𝒅𝒙.
Solution: Let 𝑥 = 𝑎 sin 𝜃 then 𝑑𝑥 =
𝑎 cos 𝜃 𝑑𝜃
∴ ∫𝑑𝑥
√𝑎2−𝑥2= ∫
𝑎 cos 𝜃 𝑑𝜃
√𝑎2−𝑎2 sin2 𝜃
∫ 1 𝑑𝜃 = 𝜃 + 𝑐 = sin−1 (𝑥
𝑎) + 𝑐
Consider 𝐼 = ∫1
√7−6𝑥−𝑥2 𝑑𝑥
= ∫1
√16 − (𝑥 + 3)2 𝑑𝑥
= sin−1 (𝑥+3
4) + 𝑐
2. Find the integral of 𝟏
√𝒙𝟐+ 𝒂𝟐 with
respect to x and hence evaluate
∫𝟏
√𝒙𝟐+𝟐𝒙+𝟐 𝒅𝒙.
Solution: Let 𝐼 =1
√𝑥2+ 𝑎2 𝑑𝑥
Let 𝑥 = 𝑎 tan 𝜃 then
𝑑𝑥 = 𝑎 sec2 𝜃 𝑑𝜃
∴ 𝐼 = ∫𝑎 sec2 𝜃 𝑑𝜃
√𝑎2 tan2 𝜃+ 𝑎2= ∫ sec 𝜃 𝑑𝜃
= log|sec 𝜃 + tan 𝜃| + 𝑐1
= log | 𝑥
𝑎+ √
𝑥2
𝑎2+ 1 | + 𝑐1
= log|𝑥 + √𝑥2 + 𝑎2| − log|𝑎| + 𝑐1
= log|𝑥 + √𝑥2 + 𝑎2| + 𝑐 ,
𝑐 = 𝑐1 − log|𝑎|
Consider ∫1
√𝑥2+2𝑥+2 𝑑𝑥
= ∫1
√(𝑥 + 1)2 + 1 𝑑𝑥
= log |(𝑥 + 1) + √(𝑥 + 1)2 + 1| + 𝑐.
3. Find the integral of 𝟏
𝒙𝟐− 𝒂𝟐 with respect
to x and evaluate ∫𝒙
𝒙𝟒− 𝟏𝟔 𝒅𝒙.
Solution: Let 𝐼 = ∫1
𝑥2− 𝑎2 𝑑𝑥
Consider 1
𝑥2− 𝑎2 =1
(𝑥−𝑎)(𝑥+𝑎)
=1
2𝑎[(𝑥 + 𝑎) − (𝑥 − 𝑎)
(𝑥 − 𝑎)(𝑥 + 𝑎)]
=1
2𝑎[
1
(𝑥 − 𝑎)]
−1
2𝑎[
1
(𝑥 + 𝑎)]
∴ 𝐼 = ∫1
2𝑎[
1
(𝑥−𝑎)] 𝑑𝑥 − ∫
1
2𝑎[
1
(𝑥−𝑎)] 𝑑𝑥
=1
2𝑎[log|𝑥 − 𝑎| − log|𝑥 + 𝑎|] + 𝑐
=1
2𝑎log |
𝑥−𝑎
𝑥+𝑎| + 𝑐
Consider∫𝑥
𝑥4− 16 𝑑𝑥 = ∫
𝑥
(𝑥2)2− 16 𝑑𝑥 Let
𝑥2 = 𝑡 then 𝑥 𝑑𝑥 =𝑑𝑡
2
∴ 𝐼 =1
2∫
𝑑𝑡
(𝑡)2− 42 =1
2×
1
2×4log |
𝑡−4
𝑡+4| + 𝑐 .
KHV’S PAGE 28 http://pue.kar.nic.in
4. Find the integral of 𝟏
𝒂𝟐− 𝒙𝟐 with respect
to x and hence evaluate ∫𝟏
𝟏𝟔− (𝟐𝒙+𝟑)𝟐 𝒅𝒙.
Solution: Let 𝐼 = ∫1
𝑎2− 𝑥2 𝑑𝑥
Consider 1
𝑎2− 𝑥=
1
(𝑎−𝑥)(𝑎+𝑥)
=1
2𝑎[(𝑎 + 𝑥) + (𝑎 − 𝑥)
(𝑎 − 𝑥)(𝑎 + 𝑥)]
=1
2𝑎[
1
(𝑎 − 𝑥)]
+1
2𝑎[
1
(𝑎 + 𝑥)]
∴ 𝐼 = ∫1
2𝑎[
1
(𝑎−𝑥)] 𝑑𝑥 + ∫
1
2𝑎[
1
(𝑎+𝑥)] 𝑑𝑥
=1
2𝑎[−log|𝑎 − 𝑥| + log|𝑎 + 𝑥|] + 𝑐
=1
2𝑎log |
𝑎+𝑥
𝑎−𝑥| + 𝑐
Consider𝐼 = ∫1
16−(2𝑥+3)2 𝑑𝑥
Let (2𝑥 + 3) = 𝑡 then 𝑑𝑥 =𝑑𝑡
2
∴ 𝐼 =1
2∫
𝑑𝑡
42− 𝑡2 =1
2×
1
2×4log |
4+𝑡
4−𝑡| + 𝑐 .
=1
16log |
4+(2𝑥+3)
4−(2𝑥+3)| + 𝑐
5.Find the integral of 2 2
1
x awith
respect to 𝒙 and hence evaluate
2
1dx
5x 2x
Substituting 𝑥 = 𝑎 sec 𝜃
𝑑𝑥 = asec 𝑡𝑎𝑛𝑑𝜃
∫1
√𝑥2−𝑎2𝑑𝑥=∫
𝑎𝑠𝑒𝑐.𝑡𝑎𝑛
√𝑎2 sec2 −𝑎2𝑑
=∫𝑎𝑠𝑒𝑐.𝑡𝑎𝑛
√𝑎2(sec2 −1)𝑑 =
∫𝑎𝑠𝑒𝑐.𝑡𝑎𝑛
√𝑎2 tan2 𝑑
=∫ 𝑠𝑒𝑐 𝑑
=log|sec+tan|+C1
=log|𝑥
𝑎+ √
𝑥2
𝑎2 − 1|+C1
= log|𝑥 + √𝑥2 − 𝑎2|-log|a|+C1
=log|𝑥 + √𝑥2 − 𝑎2|+C
where C=C1 –log|a|
Now ∫1
√𝑥2 + 2𝑥 + 2𝑑𝑥
=x2+2x+1=(x2+2x+1)+1
=(x+1)2+(1)2
∫1
√𝑥2 + 2𝑥 + 2𝑑𝑥=∫
1
√(𝑥+1)2+1𝑑𝑥
=log|𝑥 + 1 + √𝑥2 + 2𝑥 + 2| + 𝑐
6.Find the integral of 2 2
1
x awith
respect to 𝒙 and hence evaluate
∫𝟏
𝒙𝟐+𝟐𝒙+𝟐𝒅𝒙
Let I= 2 2
1dx;
x aPut x=atan,
then dx=a sec2 ;
x2+a2 =a2 tan2 +a2
=a2(tan2 +1)=a2sec2
2
2 2 2 2
1
1 asec d 1I dx d
x a a sec a
1 1 xtan c
a a a
Consider ∫𝟏
𝒙𝟐+𝟐𝒙+𝟐𝒅𝒙
7.Find the integral of √𝒙𝟐 + 𝒂𝟐 with
respect to x and evaluate ∫ √𝒙𝟐 + 𝟗 𝒅𝒙.
Solution: Let 𝐼 = ∫ √𝑥2 + 𝑎2 𝑑𝑥 applying
integration by parts
We get 𝐼 = 𝑥 √𝑥2 + 𝑎2 −1
2∫
2𝑥2
√𝑥2+𝑎2 𝑑𝑥
= 𝑥 √𝑥2 + 𝑎2 − ∫𝑥2+𝑎2−𝑎2
√𝑥2+𝑎2 𝑑𝑥
KHV’S PAGE 29 http://pue.kar.nic.in
= 𝑥 √𝑥2 + 𝑎2 − 𝐼 + ∫𝑎2
√𝑥2+𝑎2 𝑑𝑥
2𝐼 = 𝑥 √𝑥2 + 𝑎2 + 𝑎2 log|𝑥 + √𝑥2 + 𝑎2| + 𝑐
𝐼 =𝑥
2 √𝑥2 + 𝑎2 +
𝑎2
2log|𝑥 + √𝑥2 + 𝑎2| + 𝑐
Consider ∫ √𝑥2 + 9 𝑑𝑥 =𝑥
2 √𝑥2 + 9
+9
2log |𝑥 + √𝑥2 + 9| + 𝑐
8. Find the integral of √𝒂𝟐 − 𝒙𝟐 with
respect to x and evaluate
∫ √𝟏 + 𝟒𝒙 − 𝒙𝟐 𝒅𝒙.
Solution: Let 𝐼 = ∫ √𝑎2 − 𝑥2 𝑑𝑥 applying
integration by parts
We get 𝐼 = 𝑥 √𝑎2 − 𝑥2 + ∫𝑥2
√𝑎2−𝑥2 𝑑𝑥
= 𝑥 √𝑎2 − 𝑥2 − ∫𝑎2−𝑥2−𝑎2
√𝑎2−𝑥2 𝑑𝑥
𝐼 = 𝑥 √𝑎2 − 𝑥2 − ∫𝑎2
√𝑎2−𝑥2 𝑑𝑥 − 𝐼
2𝐼 = 𝑥 √𝑎2 − 𝑥2 − 𝑎2 sin−1 (𝑥
𝑎) + 𝑐
𝐼 =𝑥
2 √𝑎2 − 𝑥2 +
𝑎2
2sin−1 (
𝑥
𝑎) + 𝑐
Consider ∫ √1 + 4𝑥 − 𝑥2 𝑑𝑥
= ∫ √5 − (𝑥 − 2)2 𝑑𝑥
=(𝑥−2)
2 √5 − (𝑥 − 2)2 −
5
2sin−1 (
𝑥−2
√5) + 𝑐
Find the integral of √𝒙𝟐 − 𝒂𝟐 with
respect to x and hence evaluate ∫ 𝒙𝟐 +
𝟐𝒙 + 𝟓 𝒅𝒙
Solution: Let 𝐼 = ∫ √𝑥2 − 𝑎2 𝑑𝑥 applying
integration by parts
We get 𝐼 = 𝑥 √𝑥2 − 𝑎2 −1
2∫
2𝑥2
√𝑥2−𝑎2 𝑑𝑥
= 𝑥 √𝑥2 − 𝑎2 − ∫(𝑥2−𝑎2)+𝑎2
√𝑥2−𝑎2 𝑑𝑥
= 𝑥 √𝑥2 − 𝑎2 − ∫ [√𝑥2 − 𝑎2 +𝑎2
√𝑥2−𝑎2] 𝑑𝑥
= 𝑥 √𝑥2 − 𝑎2 − 𝐼 − ∫𝑎2
√𝑥2−𝑎2 𝑑𝑥
2𝐼 = 𝑥 √𝑥2 − 𝑎2 − 𝑎2 log|𝑥 + √𝑥2 − 𝑎2| + 𝑐
𝐼 =𝑥
2 √𝑥2 + 𝑎2 −
𝑎2
2log|𝑥 + √𝑥2 + 𝑎2| + 𝑐
Now consider 𝐼 = ∫ 𝒙𝟐 + 𝟐𝒙 + 𝟓 𝒅𝒙
x2+2x+5=(x2+2x+1)+4=(x+1)2+4
I=∫[(x + 1)2 + 22] dx
Put x+1=t dx=dt
𝐼 = √𝑡2 + 22𝑑𝑡
=𝑡
2√𝑡2 + 22+
4
2 log |𝑡 + √𝑡2 + 22|
=𝑥+1
2√𝑥2 + 2𝑥 + 5
+ 2log| x+1+√𝑥2 + 2𝑥 + 1|
Note: The above questions is for 5 mark
questions in part D of the question paper for
second PUC.
Note: In this chapter “Indefinite Integrals”
Some of the solved examples given in the text
book are not included in the question bank.
The students are advised to go through the
those questions also.
*******
KHV’S PAGE 30 http://pue.kar.nic.in
Detail of the concepts to be mastered by every student of class second PUC
with exercises and examples of NCERT Text Book.
Indefinit
e
Integrals
(i) Integration by substitution * Text book , Vol. II Examples 5&6
Page 300, 302,301,303
(ii) ) Application of trigonometric
function in integrals
** Text book , Vol. II Ex 7 Page 306,
Exercise 7.3 Q13&Q24
(iii) Integration of some particular
function
22 ax
dx , ,
, ,
, ,
*** Text book , Vol. II Exp 8, 9, 10
Page 311,312,313, Exercise 7.4 Q
3,4,8,9,13&23
(iv) Integration using Partial Fraction *** Text book , Vol. II Exp 11&12
Page 318 Exp 13 319,Exp 14 & 15
Page320
(v) Integration by Parts ** Text book , Vol. II Exp 18,19&20
Page 325 Exs 7.6 QNO ,10,11,
17,18,20
(vi)Some Special Integrals
,
*** Text book , Vol. II Exp 23 &24
Page 329
(vii) Miscellaneous Questions ** Text book , Vol. II Solved Ex. 40,
41
viii)Some special integrals Text book Supplimentary material
Page 614,615
SYMBOLS USED :
* : Important Questions, ** :Very Important Questions, *** : Very-Very Important Questions
22 ax
dx
dx
xa
1
22 cbxax
dx2
cbxax
dx
2
cbxax
dx)qpx(2
cbxax
dx)qpx(
2
dxxa 22
dxax 22
Definite Integrals One marks questions :
Evaluate the following integrals :-
(1). ∫ sin5 𝑥 𝑑𝑥𝜋 2⁄
−𝜋2⁄
𝑆𝑜𝑙𝑛: ∫ sin5 𝑥 𝑑𝑥𝜋 2⁄
−𝜋2⁄
= 0 ∵ sin5 𝑥 𝑖𝑠 𝑎𝑛 𝑜𝑑𝑑 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛.
(2). ∫ 2𝑥
1+𝑥2 𝑑𝑥.1
0
𝑆𝑜𝑙𝑛: ∫ 2𝑥
1+𝑥2 𝑑𝑥
1
0 = [log(1 + 𝑥2)]0
1 = log 2
(3). ∫1
√1−𝑥2
1√2
⁄
0 𝑑𝑥 .
𝑆𝑜𝑙𝑛: ∫1
√1−𝑥2
1√2
⁄
0 𝑑𝑥 = (sin⊣ 𝑥)
0
1√2
⁄ = sin⊣ 1
√2− sin⊣ 0 =
𝜋
4 .
(4). ∫ (3 sin 𝑥 − 4 sin3 𝑥) 𝜋
3⁄
0 𝑑𝑥.
𝑆𝑜𝑙𝑛: ∫ (3 sin 𝑥 − 4 sin3 𝑥) 𝜋
3⁄
0 𝑑𝑥 = ∫ sin 3𝑥 𝑑𝑥 =
𝜋3⁄
0−
(cos 3𝑥)0
𝜋3⁄
3 =
−1
3 [−1 − 1] =
2
3
(5). ∫𝑥3
1+𝑥2 1
−1𝑑𝑥.
𝑆𝑜𝑙𝑛: ∫𝑥3
1+𝑥2 1
−1𝑑𝑥 = 0 because
𝑥3
1+𝑥2 𝑖𝑠 𝑎𝑛 𝑜𝑑𝑑 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛.
Two marks questions :
(1) Evaluate ∫ |𝑥 + 2|5
−5𝑑𝑥
Soln: Let 𝑥 + 2 = 𝑡 ⇒ 𝑑𝑥 = 𝑑𝑡 𝑤ℎ𝑒𝑛 𝑥 = 5 ⇒ 𝑡 = 7; 𝑤ℎ𝑒𝑛 𝑥 = −5 ⇒ 𝑡 = −3
∫ |𝑥 + 2|5
−5𝑑𝑥 ⇒ ∫ |𝑡|
7
−3𝑑𝑡 ⇒
1
2(𝑡|𝑡|)−3
7 ⇒1
2(7|7| − (−3)|−3|) ⇒ 29
(2) Evaluate ∫𝑥
𝑥2+1
3
2 𝑑𝑥
Soln: Let 𝑥2 + 1 = 𝑡 ⇒ 2𝑥 𝑑𝑥 = 𝑑𝑡 ⇒ 𝑥𝑑𝑥 =𝑑𝑡
2 𝑤ℎ𝑒𝑛 𝑥 = 2 ⇒ 𝑡 = 5; 𝑤ℎ𝑒𝑛 𝑥 = 3 ⇒ 𝑡 = 10
∫𝑥
𝑥2+1
3
2 𝑑𝑥 = ∫
1
𝑡
10
5
𝑑𝑡
2=
1
2(log 𝑡)5
10 =1
2 (log 10 − log 5) =
1
2log 2.
(3) Evaluate ∫ cos 2𝑥𝜋
20
𝑑𝑥.
Soln : ∫ cos 2𝑥𝜋
20
𝑑𝑥 =1
2(sin 2𝑥)
0
𝜋
2 =1
2(0 − 0) = 0
(4) Evaluate ∫sin 𝑥
1+cos2 𝑥
𝜋
20
𝑑𝑥
Soln: Let cos 𝑥 = 𝑡 ⇒ sin 𝑥 𝑑𝑥 = −𝑑𝑡; 𝑤ℎ𝑒𝑛 𝑥 = 0 ⇒ 𝑡 = 1; 𝑤ℎ𝑒𝑛 𝑥 =𝜋
2⇒ 𝑡 = 0
∫sin 𝑥
1+cos2 𝑥
𝜋
20
𝑑𝑥 = − ∫𝑑𝑡
1+𝑡2
0
1= −(tan−1 𝑡)1
0 = − (0 −𝜋
4) =
𝜋
4
(5) ∫ 𝑥 𝜋
0sin 𝑥 cos2 𝑥 𝑑𝑥.
𝑆𝑜𝑙𝑛: 𝐿𝑒𝑡 𝐼 = ∫ 𝑥 𝜋
0sin 𝑥 cos2 𝑥 𝑑𝑥 = ∫ (𝜋 − 𝑥) sin 𝑥 cos2 𝑥
𝜋
0𝑑𝑥.
∴ 2 𝐼 = ∫ (𝑥 + 𝜋 − 𝑥)𝜋
0 sin 𝑥 cos2 𝑥 𝑑𝑥 = 𝜋 ∫ cos2 𝑥.
𝜋
0 sin 𝑥 𝑑𝑥
= 𝜋 ∫ 𝑡2 −1
1 (−𝑑𝑡) here cos 𝑥 = 𝑡 ⇒ − sin 𝑥 𝑑𝑥 = 𝑑𝑡 when 𝑥 = 0 ⇒ 𝑡 = 1; 𝑥 = 𝜋 ⇒ 𝑡 = −1
2𝐼 = 𝜋 ∫ 𝑡2 𝑑𝑡
1
−1
= 𝜋.(𝑡3)−1
1
3 =
𝜋
3 (1 + 1) =
2𝜋
3 ⇒ 𝐼 =
𝜋
3 .
Six marks questions :
(1) Prove that ∫ 𝒇(𝒙)𝒃
𝒂 𝒅𝒙 = ∫ 𝒇(𝒂 + 𝒃 − 𝒙)
𝒃
𝒂𝒅𝒙 and evaluate ∫
𝒅𝒙
𝟏+√𝐭𝐚𝐧 𝒙
𝝅
𝟑𝝅
𝟔
.
Soln: Consider ∫ 𝑓(𝑎 + 𝑏 − 𝑥)𝑏
𝑎𝑑𝑥
𝑝𝑢𝑡 𝑎 + 𝑏 − 𝑥 = 𝑡
⤇ − 𝑑𝑥 = 𝑑𝑡. 𝐴𝑡 𝑥 = 𝑎. , 𝑡 = 𝑎 + 𝑏 – 𝑎 = 𝑏 𝑎𝑛𝑑 𝑎𝑡 𝑥 = 𝑏. , 𝑡 = 𝑎 + 𝑏 – 𝑏 = 𝑎
∴ 𝑅𝐻𝑆 = ∫ 𝑓(𝑎 + 𝑏 − 𝑥) 𝑑𝑥𝑏
𝑎= ∫ 𝑓(𝑡) (−𝑑𝑡)
𝑎
𝑏 [∵ ∫ 𝑓(𝑥) 𝑑𝑥
𝑏
𝑎= − ∫ 𝑓(𝑥)
𝑎
𝑏 𝑑𝑥 ]
= ∫ 𝑓(𝑡) 𝑑𝑡 𝑏
𝑎= ∫ 𝑓(𝑥) 𝑑𝑥
𝑏
𝑎= 𝐿𝐻𝑆. [Definite integrals are independent of variable]
Consider ∫𝑑𝑥
1+√tan 𝑥
𝜋
3𝜋
6
= ∫√cos 𝑥
√cos 𝑥+√sin 𝑥
𝜋 3⁄
𝜋 6⁄ 𝑑𝑥
Let I = ∫√cos 𝑥
√cos 𝑥+√sin 𝑥
𝜋 3⁄
𝜋 6⁄ 𝑑𝑥 = ∫
√cos(𝜋
6+
𝜋
3−𝑥)
√cos(𝜋
6+
𝜋
3−𝑥) + √sin(
𝜋
6+
𝜋
3−𝑥)
𝜋 3⁄
𝜋 6⁄ 𝑑𝑥 ..........(1)
⇒ 𝐼 = ∫√sin 𝑥
√sin 𝑥 + √cos 𝑥
𝜋 3⁄
𝜋 6⁄ 𝑑𝑥....................(2)
Adding (1) and (2)
2 I = I + I = ∫√cos 𝑥
√cos 𝑥+√sin 𝑥
𝜋 3⁄
𝜋 6⁄ 𝑑𝑥 + ∫
√sin 𝑥
√sin 𝑥 + √cos 𝑥
𝜋 3⁄
𝜋 6⁄ 𝑑𝑥 = ∫
√cos 𝑥+√sin 𝑥
√cos 𝑥+√sin 𝑥
𝜋 3⁄
𝜋 6⁄ 𝑑𝑥
2𝐼 = (𝑥)𝜋6⁄
𝜋3⁄
= 𝜋
3−
𝜋
6=
𝜋
6 ∴ I =
𝜋
12
(2) Prove that ∫ 𝒇(𝒙)𝒂
𝟎 𝒅𝒙 = ∫ 𝒇(𝒂 − 𝒙)
𝒂
𝟎 𝒅𝒙 and hence evaluate ∫ 𝐥𝐨𝐠(𝟏 + 𝐭𝐚𝐧 𝒙)
𝝅
𝟒𝟎
𝒅𝒙.
Soln : Consider ∫ 𝑓(𝑎 − 𝑥)𝑎
0 𝑑𝑥
Let 𝑎 − 𝑥 = 𝑡 ⇒ 𝑑𝑥 = −𝑑𝑡; 𝑤ℎ𝑒𝑛 𝑥 = 0 ⇒ 𝑡 = 𝑎; 𝑤ℎ𝑒𝑛 𝑥 = 𝑎 ⇒ 𝑡 = 0
∫ 𝑓(𝑎 − 𝑥)𝑎
0 𝑑𝑥 = ∫ 𝑓(𝑡)(−𝑑𝑡)
0
𝑎 [∵ ∫ 𝑓(𝑥) 𝑑𝑥
𝑏
𝑎= − ∫ 𝑓(𝑥)
𝑎
𝑏 𝑑𝑥 ]
= ∫ 𝑓(𝑡)𝑎
0𝑑𝑡 = ∫ 𝑓(𝑥)
𝑎
0𝑑𝑥 [Definite integrals are independent of variable]
Let I = ∫ log(1 + tan 𝑥) 𝜋
4⁄
0 𝑑𝑥 = ∫ log (1 + tan ⟨
𝜋
4− 𝑥⟩)
𝜋4⁄
0 𝑑𝑥 = ∫ log (1 +
1−tan 𝑥
1+tan 𝑥)
𝜋4⁄
0 𝑑𝑥
= ∫ log (2
1+tan 𝑥)
𝜋4⁄
0 𝑑𝑥
∴ 2I = I + I = ∫ log(1 + tan 𝑥)
𝜋4⁄
0
𝑑𝑥 + ∫ log (2
1 + tan 𝑥)
𝜋4⁄
0
𝑑𝑥 = ∫ log 2 𝑑𝑥 =
𝜋4⁄
0
log 2 (𝑥)0
𝜋4⁄
=𝜋
4 log 2
∴ I = 𝜋
8 log 2.
(3) Prove that ∫ 𝒇(𝒙) 𝒅𝒙𝟐𝒂
𝟎= {
𝟐. ∫ 𝒇(𝒙)𝒂
𝟎 𝐝𝐱 , 𝐢𝐟 𝐟(𝟐𝐚 − 𝐱) = 𝒇(𝒙),
𝟎 , 𝒊𝒇 𝒇(𝟐𝒂 − 𝒙) = −𝒇(𝒙) and hence evaluate
∫𝟏
𝒂𝟐 𝐜𝐨𝐬𝟐 𝒙+𝒃𝟐 𝐬𝐢𝐧𝟐 𝒙
𝝅
𝟎𝒅𝒙.
Soln : ∫ 𝑓(𝑥) 𝑑𝑥 =2𝑎
0 ∫ 𝑓(𝑥) 𝑑𝑥
𝑎
0+ ∫ 𝑓(𝑥) 𝑑𝑥
2𝑎
𝑎… . . (1)
𝑃𝑢𝑡 2𝑎 − 𝑥 = 𝑡 ⤇ 𝑑𝑥 = −𝑑𝑡
Consider ∫ 𝑓(𝑥) 𝑑𝑥.2𝑎
𝑎 𝐴𝑡 𝑥 = 𝑎. , 𝑡 = 𝑎, 𝑎𝑛𝑑 𝑎𝑡 𝑥 = 2𝑎, 𝑡 = 0
= ∫ 𝑓(2𝑎 − 𝑡)(−𝑑𝑡)0
𝑎
= ∫ 𝑓(2𝑎 − 𝑡) 𝑑𝑡 𝑎
0= ∫ 𝑓(2𝑎 − 𝑥) 𝑑𝑥
𝑎
0… . (2)
𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑔 (2) 𝑖𝑛 (1)., 𝑤𝑒 𝑔𝑒𝑡 ∶ ∫ 𝑓(𝑥) 𝑑𝑥 =2𝑎
0∫ 𝑓(𝑥) 𝑑𝑥
𝑎
0+ ∫ 𝑓(2𝑎 − 𝑥)
𝑎
0 𝑑𝑥
𝐼𝑓 𝑓(2𝑎 − 𝑥) = 𝑓(𝑥) the above result ∫ 𝑓(𝑥) 2𝑎
0𝑑𝑥 = 2 ∫ 𝑓(𝑥) 𝑑𝑥. ,
𝑎
0
𝐴𝑛𝑑 𝑖𝑓 𝑓(2𝑎 − 𝑥) = −𝑓(𝑥)., it reduces to: ∫ 𝑓(𝑥) 2𝑎
0𝑑𝑥 = 0.
Let I = ∫1
𝑎2 cos2 𝑥+ 𝑏2 sin2 𝑥
𝜋
0 𝑑𝑥 = 2 ∫
1
𝑎2 cos2 𝑥+ 𝑏2 sin2 𝑥
𝜋2⁄
0 𝑑𝑥
= 2 ∫sec2 𝑥
a2 + 𝑏2 tan2 𝑥
𝜋2⁄
0 𝑑𝑥
Put b tan 𝑥 = 𝑡, ⤇ 𝑏 sec2 𝑥 𝑑𝑥 = 𝑑𝑡, ∴ sec2 𝑥 𝑑𝑥 = 𝑑𝑡
𝑏, 𝐴𝑡 𝑥 = 0., 𝑡 = 0, 𝐴𝑡 𝑥 =
𝜋
2, 𝑡 → ∞.
𝐼 = 2 ∫𝑑𝑡
𝑏(𝑎2+𝑡2)
∞
0= 2
1
𝑏
1
𝑎 (tan⊣
𝑡
𝑎)
0
∞
= 2
𝑎𝑏 (
𝜋
2− 0) =
𝜋
𝑎𝑏
(4) Prove that ∫ 𝒇(𝒙) 𝒅𝒙𝒃
𝒂= ∫ 𝒇(𝒙)
𝒄
𝒂 𝒅𝒙 + ∫ 𝒇(𝒙)
𝒃
𝒄 𝒅𝒙 and hence evaluate ∫ |𝒙𝟑𝟐
−𝟏− 𝒙|𝒅𝒙
Soln: 𝑒𝑡 𝐹(𝑥)𝑏𝑒 𝑡ℎ𝑒 𝑎𝑛𝑡𝑖𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 𝑜𝑓 𝑓(𝑥).
𝑇ℎ𝑒𝑛. , 𝑅𝐻𝑆 = ∫ 𝑓(𝑥)
𝑐
𝑎
𝑑𝑥 + ∫ 𝑓(𝑥)𝑑𝑥
𝑏
𝑐
= 𝐹 (𝑐) − 𝐹 (𝑎) + 𝐹 (𝑏) – 𝐹 (𝑐)
∫ f(x) dxb
a= F(b) = F(a) = ∫ f(x)
b
adx = LHS.
𝐶𝑙𝑒𝑎𝑟𝑙𝑦, 𝑥3 − 𝑥 𝑥(𝑥 − 1) (𝑥 + 1) will be positive between −1 & 0, negative between 0 & 1 and again positive between 1 & 2.
∴ ∫(𝑥3 − 𝑥) 𝑑𝑥 =
2
−1
∫(𝑥3 − 𝑥) 𝑑𝑥 +
0
−1
∫(𝑥 − 𝑥3)
1
0
𝑑𝑥 + ∫(𝑥3 − 𝑥)
2
1
𝑑𝑥.
= (𝑥4
4−
𝑥2
2)
−1
0
+ (𝑥2
2−
𝑥4
4)
0
1
+ (𝑥4
4−
𝑥2
2)
1
2
= 0 − [1
4−
1
2] + [
1
2−
1
4] + {(4 − 2) − (
1
4−
1
2)}
= + 1
4 +
1
4 + 2 +
1
4= 2
3
4=
11
4
(5) Prove that ∫ 𝒇(𝒙)𝒂
−𝒂 𝒅𝒙 = {
𝟐 ∫ 𝒇(𝒙)𝒂
𝟎 𝐝𝐱. , 𝐢𝐟 𝐟 𝐢𝐬 𝐚𝐧 𝐞𝐯𝐞𝐧 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧
𝟎 , 𝒊𝒇 𝒇 𝒊𝒔 𝒂𝒏 𝒐𝒅𝒅 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏 and hence evaluate
∫ 𝐬𝐢𝐧𝟕 𝒙𝝅
𝟐
–𝝅
𝟐
𝒅𝒙.
Soln: consider ∫ 𝑓(𝑥) 𝑑𝑥𝑎
−𝑎= ∫ 𝑓(𝑥)
0
−𝑎 𝑑𝑥 + ∫ 𝑓(𝑥) 𝑑𝑥
𝑎
0… . (1)
𝑃𝑢𝑡 𝑥 = −𝑡 ⤇ 𝑥 = −𝑑𝑡
Consider ∫ 𝑓(𝑥) 𝑑𝑥0
−𝑎 𝐴𝑡 𝑥 = −𝑎. , 𝑡 = 𝑎 𝐴𝑛𝑑, 𝑎𝑡 𝑥 = 0., 𝑡 = 0.
= ∫ 𝑓(−𝑡)(−𝑑𝑡)0
𝑎 = ∫ 𝑓(−𝑡) 𝑑𝑡
𝑎
0= ∫ 𝑓(−𝑥) 𝑑𝑥
𝑎
0= {
∫ 𝑓(𝑥)𝑎
0dx if f is an even function, and,
− ∫ 𝑓(𝑥)𝑑𝑥𝑎
0 𝑖𝑓 𝑓 𝑖𝑠 𝑎𝑛 𝑜𝑑𝑑 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛.
… . (2)
𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑠 𝑒𝑞. (2) 𝑖𝑛 𝑒𝑞. (1). , 𝑤𝑒 𝑔𝑒𝑡 ∶
∫ 𝑓(𝑥) 𝑑𝑥
𝑎
−𝑎
= (∫ 𝑓(𝑥) 𝑑𝑥, 𝑖𝑓 𝑓 𝑖𝑠 𝑒𝑣𝑒𝑛
𝑎
0
− ∫ 𝑓(𝑥) 𝑑𝑥, 𝑖𝑓 𝑓 𝑖𝑠 𝑜𝑑𝑑𝑎
0
) + ∫ 𝑓(𝑥) 𝑑𝑥
𝑎
0
= {2. ∫ 𝑓(𝑥)
𝑎
0
dx , if f is an even function, and,
0 , 𝑖𝑓 𝑓 𝑖𝑠 𝑎𝑛 𝑜𝑑𝑑 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛.
Hence the value of ∫ sin7 𝑥 𝑑𝑥𝜋
2⁄−𝜋
2
= 0 [∴ sin7(−𝑥) = − sin7 𝑥 ⤇ 𝑖𝑡 𝑖𝑠 𝑎𝑛 𝑜𝑑𝑑 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛]
(6) Prove that ∫ 𝒇(𝒙)𝒂
𝟎 𝒅𝒙 = ∫ 𝒇(𝒂 − 𝒙)
𝒂
𝟎 𝒅𝒙 and hence evaluate ∫ 𝐥𝐨𝐠 𝐬𝐢𝐧 𝒙
𝝅𝟐⁄
𝟎𝒅𝒙.
Soln : Consider ∫ 𝑓(𝑎 − 𝑥)𝑎
0 𝑑𝑥
Let 𝑎 − 𝑥 = 𝑡 ⇒ 𝑑𝑥 = −𝑑𝑡; 𝑤ℎ𝑒𝑛 𝑥 = 0 ⇒ 𝑡 = 𝑎; 𝑤ℎ𝑒𝑛 𝑥 = 𝑎 ⇒ 𝑡 = 0
∫ 𝑓(𝑎 − 𝑥)𝑎
0 𝑑𝑥 = ∫ 𝑓(𝑡)(−𝑑𝑡)
0
𝑎 [∵ ∫ 𝑓(𝑥) 𝑑𝑥
𝑏
𝑎= − ∫ 𝑓(𝑥)
𝑎
𝑏 𝑑𝑥 ]
= ∫ 𝑓(𝑡)𝑎
0𝑑𝑡 = ∫ 𝑓(𝑥)
𝑎
0𝑑𝑥 [Definite integrals are independent of variable]
𝐿𝑒𝑡 𝐼 = ∫ log sin 𝑥 𝑑𝑥 =
𝜋 2⁄
0
∫ log sin (𝜋
2− 𝑥) 𝑑𝑥 =
𝜋 2⁄
0
∫ log cos 𝑥
𝜋 2⁄
0
𝑑𝑥
2 I = ∫ log sin 𝑥𝜋 2⁄
0 𝑑𝑥 + ∫ log cos 𝑥 𝑑𝑥
𝜋 2⁄
0= ∫ log(sin 𝑥 cos 𝑥)
𝜋 2⁄
0 𝑑𝑥
= ∫ log (sin 2𝑥
2) 𝑑𝑥 =
𝜋 2⁄
0∫ log sin 2𝑥 𝑑𝑥 −
𝜋 2⁄
0∫ log 2 𝑑𝑥
𝜋 2⁄
0… . (1)
Consider ∫ log sin 2𝑥 𝑑𝑥𝜋 2⁄
0 𝑃𝑢𝑡 2𝑥 = 𝑡 ⤇ 𝑑𝑥 =
𝑑𝑡
2
𝐴𝑡 𝑥 = 0, 𝑡 = 0. 𝐴𝑛𝑑,
= ∫ log sin 𝑡 𝑑𝑡
2
𝜋
0 𝐴𝑡 𝑥 = 𝜋
2,⁄ 𝑡 = 𝜋.
= ∫ log sin 𝑡 𝑑𝑡
𝜋 2⁄
0
[∴ ∫ 𝑓(𝑥)𝑑𝑥 = 2 ∫ 𝑓(𝑥)𝑑𝑥, 𝑖𝑓 𝑓(2𝑎 − 𝑥) = 𝑓(𝑥)
𝑎
0
2𝑎
0
]
= ∫ log sin 2𝑥
𝜋 2⁄
0
𝑑𝑥 = 𝐼 … … (2)
𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑠 𝑒𝑞 (2) 𝑖𝑛 𝑒𝑞. (1)., 𝑤𝑒 𝑔𝑒𝑡 ∶ 2 I = I − ∫ log 2 𝑑𝑥 𝜋 2⁄
0∴ I = − log 2 (𝑥)
0
𝜋2⁄
= − 𝜋
2 log 2
(7) Prove that ∫ 𝒇(𝒙)𝒂
𝟎𝒅𝒙 = 𝟐 ∫ 𝒇(𝒙)
𝒂
𝟎𝒅𝒙 when 𝒇(𝟐𝒂 − 𝒙) = 𝒇(𝒙) and hence evaluate ∫ |𝐜𝐨𝐬 𝒙|
𝝅
𝟎𝒅𝒙 .
Soln : ∫ 𝑓(𝑥) 𝑑𝑥 =2𝑎
0 ∫ 𝑓(𝑥) 𝑑𝑥
𝑎
0+ ∫ 𝑓(𝑥) 𝑑𝑥
2𝑎
𝑎… . . (1)
𝑃𝑢𝑡 2𝑎 − 𝑥 = 𝑡 ⤇ 𝑑𝑥 = −𝑑𝑡
Consider ∫ 𝑓(𝑥) 𝑑𝑥.2𝑎
𝑎 𝐴𝑡 𝑥 = 𝑎. , 𝑡 = 𝑎, 𝑎𝑛𝑑 𝑎𝑡 𝑥 = 2𝑎, 𝑡 = 0
= ∫ 𝑓(2𝑎 − 𝑡)(−𝑑𝑡)0
𝑎
= ∫ 𝑓(2𝑎 − 𝑡) 𝑑𝑡 𝑎
0= ∫ 𝑓(2𝑎 − 𝑥) 𝑑𝑥
𝑎
0… . (2)
𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑔 (2) 𝑖𝑛 (1)., 𝑤𝑒 𝑔𝑒𝑡 ∶ ∫ 𝑓(𝑥) 𝑑𝑥 =2𝑎
0∫ 𝑓(𝑥) 𝑑𝑥
𝑎
0+ ∫ 𝑓(2𝑎 − 𝑥)
𝑎
0 𝑑𝑥
𝐼𝑓 𝑓(2𝑎 − 𝑥) = 𝑓(𝑥) the above result becomes ∫ 𝑓(𝑥) 2𝑎
0𝑑𝑥 = 2 ∫ 𝑓(𝑥) 𝑑𝑥
𝑎
0
Consider 𝐼 = ∫ |cos 𝑥|𝜋
0𝑑𝑥 = 2 ∫ |cos 𝑥|
𝜋
20
𝑑𝑥 because |cos(𝜋 − 𝑥)| = |− cos 𝑥| = |cos 𝑥|
∴ 𝐼 = 2 ∫ cos 𝑥𝜋
20
𝑑𝑥 because cos 𝑥 is positive in 1st quadrant.
𝐼 = 2 (sin 𝑥)0
𝜋
2 = 2(1 − 0) = 2.
(8) Prove that ∫ 𝒇(𝒙)𝒂
−𝒂𝒅𝒙 = {
𝟐 ∫ 𝒇(𝒙)𝒂
𝟎 𝐝𝐱. , 𝐢𝐟 𝐟 𝐢𝐬 𝐚𝐧 𝐞𝐯𝐞𝐧 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧
𝟎 , 𝒊𝒇 𝒇 𝒊𝒔 𝒂𝒏 𝒐𝒅𝒅 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏 and hence evaluate
∫ (𝒙𝟑 + 𝒙 𝐜𝐨𝐬 𝒙 + 𝐭𝐚𝐧𝟓 𝒙 + 𝟏)𝝅
𝟐
−𝝅
𝟐
𝒅𝒙
Soln: consider ∫ 𝑓(𝑥) 𝑑𝑥𝑎
−𝑎= ∫ 𝑓(𝑥)
0
−𝑎 𝑑𝑥 + ∫ 𝑓(𝑥) 𝑑𝑥
𝑎
0… . (1)
𝑃𝑢𝑡 𝑥 = −𝑡 ⤇ 𝑥 = −𝑑𝑡
Consider ∫ 𝑓(𝑥) 𝑑𝑥0
−𝑎 𝐴𝑡 𝑥 = −𝑎. , 𝑡 = 𝑎 𝐴𝑛𝑑, 𝑎𝑡 𝑥 = 0., 𝑡 = 0.
= ∫ 𝑓(−𝑡)(−𝑑𝑡)0
𝑎 = ∫ 𝑓(−𝑡) 𝑑𝑡
𝑎
0= ∫ 𝑓(−𝑥) 𝑑𝑥
𝑎
0= {
∫ 𝑓(𝑥)𝑎
0dx if f is an even function, and,
− ∫ 𝑓(𝑥)𝑑𝑥𝑎
0 𝑖𝑓 𝑓 𝑖𝑠 𝑎𝑛 𝑜𝑑𝑑 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛.
… . (2)
𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑠 𝑒𝑞. (2) 𝑖𝑛 𝑒𝑞. (1). , 𝑤𝑒 𝑔𝑒𝑡 ∶
∫ 𝑓(𝑥) 𝑑𝑥
𝑎
−𝑎
= (∫ 𝑓(𝑥) 𝑑𝑥, 𝑖𝑓 𝑓 𝑖𝑠 𝑒𝑣𝑒𝑛
𝑎
0
− ∫ 𝑓(𝑥) 𝑑𝑥, 𝑖𝑓 𝑓 𝑖𝑠 𝑜𝑑𝑑𝑎
0
) + ∫ 𝑓(𝑥) 𝑑𝑥
𝑎
0
= {2. ∫ 𝑓(𝑥)
𝑎
0
dx , if f is an even function, and,
0 , 𝑖𝑓 𝑓 𝑖𝑠 𝑎𝑛 𝑜𝑑𝑑 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛.
Consider 𝐼 = ∫ (𝑥3 + 𝑥 cos 𝑥 + tan5 𝑥 + 1)𝜋
2
−𝜋
2
𝑑𝑥 since 𝑥3, 𝑥 cos 𝑥 , tan5 𝑥 odd functions
𝐼 = ∫ 1𝜋
2
−𝜋
2
𝑑𝑥 = (𝑥)−
𝜋
2
𝜋
2 =𝜋
2− (
𝜋
2) = 𝜋.
(9) Prove that ∫ 𝒇(𝒙)𝒂
𝟎 𝒅𝒙 = ∫ 𝒇(𝒂 − 𝒙)
𝒂
𝟎 𝒅𝒙 and hence evaluate ∫
𝐬𝐢𝐧𝟑𝟐 𝒙
𝐬𝐢𝐧𝟑𝟐 𝒙+𝐜𝐨𝐬
𝟑𝟐 𝒙
𝝅𝟐⁄
𝟎𝒅𝒙.
Soln: Consider ∫ 𝑓(𝑎 − 𝑥)𝑎
0 𝑑𝑥
Let 𝑎 − 𝑥 = 𝑡 ⇒ 𝑑𝑥 = −𝑑𝑡; 𝑤ℎ𝑒𝑛 𝑥 = 0 ⇒ 𝑡 = 𝑎; 𝑤ℎ𝑒𝑛 𝑥 = 𝑎 ⇒ 𝑡 = 0
∫ 𝑓(𝑎 − 𝑥)𝑎
0 𝑑𝑥 = ∫ 𝑓(𝑡)(−𝑑𝑡)
0
𝑎 [∵ ∫ 𝑓(𝑥) 𝑑𝑥
𝑏
𝑎= − ∫ 𝑓(𝑥)
𝑎
𝑏 𝑑𝑥 ]
= ∫ 𝑓(𝑡)𝑎
0𝑑𝑡 = ∫ 𝑓(𝑥)
𝑎
0𝑑𝑥 [Definite integrals are independent of variable]
Consider 𝐼 = ∫sin
𝟑𝟐 𝑥
sin𝟑𝟐 𝑥+cos
𝟑𝟐 𝑥
𝜋2⁄
0𝑑𝑥 ..................(1)
= ∫sin
𝟑𝟐(
𝜋
2−𝑥)
sin𝟑𝟐(
𝜋
2−𝑥)+cos
𝟑𝟐(
𝜋
2−𝑥)
𝜋2⁄
0𝑑𝑥
𝐼 = ∫cos
𝟑𝟐 𝑥
cos𝟑𝟐 𝑥+sin
𝟑𝟐 𝑥
𝜋2⁄
0𝑑𝑥 ....................(2)
Adding (1) and (2)
2𝐼 = ∫sin
𝟑𝟐 𝑥
sin𝟑𝟐 𝑥+cos
𝟑𝟐 𝑥
𝜋2⁄
0+
cos𝟑𝟐 𝑥
cos𝟑𝟐 𝑥+sin
𝟑𝟐 𝑥
𝑑𝑥 = ∫sin
𝟑𝟐 𝑥+cos
𝟑𝟐 𝑥
sin𝟑𝟐 𝑥+cos
𝟑𝟐 𝑥
𝜋2⁄
0 𝑑𝑥 = ∫ 𝑑𝑥
𝜋2⁄
0= (𝑥)
0
𝜋
2 =𝜋
2− 0 =
𝜋
2
Hence 𝐼 =𝜋
4.
(10) Prove that ∫ 𝒇(𝒙)𝒂
𝟎 𝒅𝒙 = ∫ 𝒇(𝒂 − 𝒙)
𝒂
𝟎 𝒅𝒙 and hence evaluate ∫ (𝟐 𝐥𝐨𝐠 𝐬𝐢𝐧 𝒙 − 𝐥𝐨𝐠 𝐬𝐢𝐧 𝟐𝒙)
𝝅𝟐⁄
𝟎𝒅𝒙.
Soln: Consider ∫ 𝑓(𝑎 − 𝑥)𝑎
0 𝑑𝑥
Let 𝑎 − 𝑥 = 𝑡 ⇒ 𝑑𝑥 = −𝑑𝑡; 𝑤ℎ𝑒𝑛 𝑥 = 0 ⇒ 𝑡 = 𝑎; 𝑤ℎ𝑒𝑛 𝑥 = 𝑎 ⇒ 𝑡 = 0
∫ 𝑓(𝑎 − 𝑥)𝑎
0 𝑑𝑥 = ∫ 𝑓(𝑡)(−𝑑𝑡)
0
𝑎 [∵ ∫ 𝑓(𝑥) 𝑑𝑥
𝑏
𝑎= − ∫ 𝑓(𝑥)
𝑎
𝑏 𝑑𝑥 ]
= ∫ 𝑓(𝑡)𝑎
0𝑑𝑡 = ∫ 𝑓(𝑥)
𝑎
0𝑑𝑥 [Definite integrals are independent of variable]
Consider 𝐼 = ∫ (2 log sin 𝑥 − log sin 2𝑥)𝜋
2⁄
0𝑑𝑥 = ∫ (log sin2 𝑥 − log sin 2𝑥)
𝜋2⁄
0𝑑𝑥
= ∫ log (sin2 𝑥
sin 2𝑥 )
𝜋2⁄
0 𝑑𝑥 = ∫ log (
sin2 𝑥
2sin 𝑥 cos 𝑥 )
𝜋2⁄
0𝑑𝑥 = ∫ log (
1
2tan 𝑥)
𝜋2⁄
0𝑑𝑥 ..........(1)
𝐼 = ∫ log (1
2(tan (
𝜋
2− 𝑥)) )
𝜋2⁄
0𝑑𝑥 = ∫ log (
1
2cot 𝑥)
𝜋2⁄
0...............(2)
Adding (1) and (2)
2𝐼 = ∫ log (1
2tan 𝑥 ) + log (
1
2cot 𝑥)
𝜋2⁄
0
𝑑𝑥 = ∫ log (1
2tan 𝑥 .
1
2cot 𝑥 )
𝜋2⁄
0
𝑑𝑥 = ∫ log (1
4)
𝜋2⁄
0
𝑑𝑥 = log1
4(𝑥)
0
𝜋2⁄
2𝐼 =𝜋
2log
1
4 ⇒ 𝐼 =
𝜋
4log
1
4.
1
CHAPTER 8:
APPLICATION OF INTEGRALS 3 mark questions
Question 1:
Find the area of the region bounded by the
curve y2 = x and the lines x = 1, x = 4 and the
x-axis.
Answer :
The area of the region bounded by the curve,
y2 = x, the lines, x = 1 and x = 4, and the x-
axis is the area ABCD.
Question 2:
Find the area of the region bounded by y2 =
9x, x = 2, x = 4 and the x-axis in the first
quadrant.
Answer :
The area of the region bounded by the curve,
y2 = 9x, x = 2, and x = 4, and the x-axis is the
area ABCD.
2
Question 3:
Find the area of the region bounded by x2 =
4y, y = 2, y = 4 and the y-axis in the first
quadrant.
Answer :
The area of the region bounded by the curve,
x2 = 4y, y = 2, and y = 4, and the y-axis is the
area ABCD.
Question 4:
Find the area of the region bounded by the
curve y2 = 4x, y-axis and the line y = 3 is
Answer : The area bounded by the curve, y
2 = 4x, y-
axis, and y = 3 is represented as
Question 5:
Find the area lying between the curve y2 =
4x and y = 2x is
Answer :
The area lying between the curve, y2 = 4x
and y = 2x, is represented by the shaded area
OBAO as
3
The points of intersection of these curves are
O (0, 0) and A (1, 2).
We draw AC perpendicular to x-axis such
that the coordinates of C are (1, 0).
∴ Area OBAO = Area (OCABO) – Area
(ΔOCA)
square units
Question 5. Find area enclosed by the Parabola y
2=4ax
and its latus rectum by integration Solution: y
2 = 4ax ---- (1) and
the equation of the Latus rectum is given by x = a ……… (2)
From (2) and (1) y2 =4a
2 y = 2a
Required area A = 2 [area of OSP
= 2 ∫
= 2 ∫ √
. √ . dx
= 4 √ (
)
= √
[ a√ ] =
Sq. units
5 MARK QUESTIONS:
Question 1:
Find the area of the region bounded by the
ellipse
Answer : The given equation of the ellipse,
, can be represented as
It can be observed that the ellipse is
symmetrical about x-axis and y-axis.
4
∴ Area bounded by ellipse = 4 × Area of
OAB
Therefore, area bounded by the ellipse = 4 ×
3π = 12π units
Question 2:
Find the area of the region bounded by the
ellipse
Answer : The given equation of the ellipse can be
represented as
It can be observed that the ellipse is
symmetrical about x-axis and y-axis.
∴ Area bounded by ellipse = 4 × Area OAB
Therefore, area bounded by the ellipse =
Question 3:
Find the area of the region in the first
quadrant enclosed by x-axis, line
and the circle
Answer : The area of the region bounded by the circle,
, and the x-axis is the
area OAB.
5
The point of intersection of the line and the
circle in the first quadrant is .
Area OAB = Area ΔOCA + Area ACB
Area of OAC
Area of ABC
Therefore, area enclosed by x-axis, the line
, and the circle in the
first quadrant =
Question 7:
Find the area of the smaller part of the circle
x2 + y
2 = a
2 cut off by the line
Answer :
The area of the smaller part of the circle, x2
+ y2 = a
2, cut off by the line, , is the
area ABCDA.
It can be observed that the area ABCD is
symmetrical about x-axis.
∴ Area ABCD = 2 × Area ABC
6
Therefore, the area of smaller part of the
circle, x2 + y
2 = a
2, cut off by the line,
, is units.
Question 8:
The area between x = y2 and x = 4 is divided
into two equal parts by the line x = a, find
the value of a.
Answer :
The line, x = a, divides the area bounded by
the parabola and x = 4 into two equal parts.
∴ Area OAD = Area ABCD
It can be observed that the given area is
symmetrical about x-axis.
⇒ Area OED = Area EFCD
From (1) and (2), we obtain
Therefore, the value of a is .
Question 9:
Find the area of the region bounded by the
parabola y = x2 and
Answer : The area bounded by the parabola, x
2 =
y,and the line, , can be represented as
7
The given area is symmetrical about y-axis.
∴ Area OACO = Area ODBO
The point of intersection of parabola, x2 = y,
and line, y = x, is A (1, 1).
Area of OACO = Area ΔOAM – Area
OMACO
Area of ΔOAM
Area of OMACO
⇒ Area of OACO = Area of ΔOAM – Area
of OMACO
Therefore, required area = units
Question 10:
Find the area bounded by the curve x2 = 4y
and the line x = 4y – 2
Answer : The area bounded by the curve, x
2 = 4y, and
line, x = 4y – 2, is represented by the shaded
area OBAO.
Let A and B be the points of intersection of
the line and parabola.
Coordinates of point .
Coordinates of point B are (2, 1).
We draw AL and BM perpendicular to x-
axis.
It can be observed that,
Area OBAO = Area OBCO + Area OACO
… (1)
Then, Area OBCO = Area OMBC – Area
OMBO
Similarly, Area OACO
= Area OLAC – Area OLAO
8
Therefore, required area =
Question 11:
Find the area of the region bounded by the
curve y2 = 4x and the line x = 3
Answer : The region bounded by the parabola, y
2 =
4x, and the line, x = 3, is the area OACO.
The area OACO is symmetrical about x-
axis.
∴ Area of OACO = 2 (Area of OAB)
Therefore, the required area is units.
Question 12:
Find the area of the circle 4x2 + 4y
2 = 9
which is interior to the parabola x2 = 4y
Answer : The required area is represented by the
shaded area OBCDO.
Solving the given equation of circle, 4x
2 +
4y2 = 9, and parabola, x
2 = 4y, we obtain the
point of intersection as
.
It can be observed that the required area is
symmetrical about y-axis.
9
∴ Area OBCDO = 2 × Area OBCO
We draw BM perpendicular to OA.
Therefore, the coordinates of M are .
Therefore, Area OBCO = Area OMBCO –
Area OMBO
Therefore, the required area OBCDO is
units
Question :13
Using integration finds the area of the region
bounded by the triangle whose vertices are
(–1, 0), (1, 3) and (3, 2).
Answer : BL and CM are drawn perpendicular to x-
axis.
It can be observed in the following figure
that,
Area (ΔACB) = Area (ALBA) + Area
(BLMCB) – Area (AMCA) … (1)
Equation of line segment AB is
Equation of line segment BC is
Equation of line segment AC is
Therefore, from equation (1), we obtain
Area (ΔABC) = (3 + 5 – 4) = 4 units
Question 14:
Using integration find the area of the
triangular region whose sides have the
equations y = 2x +1, y = 3x + 1 and x = 4.
Answer : The equations of sides of the triangle are y =
2x +1, y = 3x + 1, and x = 4.
On solving these equations, we obtain the
vertices of triangle as A(0, 1), B(4, 13), and
C (4, 9).
It can be observed that,
10
Area (ΔACB) = Area (OLBAO) –Area
(OLCAO)
Question 15:
Find the smaller area enclosed by the circle
x2 + y
2 = 4 and the line x + y = 2 is
Answer :
The smaller area enclosed by the circle, x2 +
y2 = 4, and the line, x + y = 2, is represented
by the shaded area ACBA as
It can be observed that,
Area ACBA = Area OACBO – Area
(ΔOAB)
1
Differential Equations
1 Mark questions:
1. Define a differential equation.
It is an equation containing derivatives.
2. Define order of a differential equation.
It is highest order of derivative appearing in the given equation.
3. Define degree of a differential equation.
It is the highest power of highest ordered derivative appearing in the given
equation.
4. Define general solution of a differential equation.
It is solution of given differential equation and it contains arbitrary constants.
5. Define particular solution of a differential equation.
It is that solution of given differential equation and is free from arbitrary constants.
2 Marks questions:
1. Form a differential equation of family of
i. Straight lines with slope = m and passing through origin.
Consider a straight line with slope = m and passing through origin
i.e. y = mx -----(1)
y1 = m -------(2) y = x y
1
ii. Circles with centre on y-axis and passing through origin.
022
0222102
122
1
1
1122
yyxyxfy
yyx
fyyyxfyyxConsider
2. Solve the following by using separation of variables.
i. dydxdxydyx
cyxxyyxdxydyxdxyd
ii. 01 2
2
x
ya
dx
dy
cxyx
dx
y
dy
11
22sinsin0
11
iii. 011 22 ydx
dyx
2
cxyx
dx
y
dy
11
22tantan0
11
iv. 2
1 1.1 yxy
cx
xydxxy
dy
2tan1
1
21
2
v. 03212 dyxdxy
cyxy
dy
x
dx
12log
2
132log
2
10
1232
vi. 22 1 ydx
dyx
cx
yx
dx
y
dx
1
tan01
1
22
vii. 23 2 xdx
dy
cxxxx
ydxxdy 223
323 3
22
viii. 11 xey
cxeydxedy xx 1
ix. 0y
dy
x
dx
cyxy
dy
x
dx loglog0
x. 0 dyydyx
cx
yxyd
x
dxydyx
00
2
xi. 0sincos dyydxx
cyxdyydxx cossin0sincos
xii. 0cos2 dxdyxy
cyy
dxxdyy tan2
0sec2
2
xiii. 011
y
y
x
x
e
dye
e
dxe
cee
e
ed
e
ed yx
y
y
x
x
1log1log0
1
1
1
1
xiv. 011 22 dyyxdyxy
3
cxy
x
dxy
y
dyydxxxdy
y
y
22
22
2
2
1log1log
01
2
1
201
1
xv. yxedx
dy
ceedxedyee
e
dx
dy xyxy
y
x
Problems on homogenous equations 3 mark question:
1. yx
yx
dx
dySolve
cyxxy
xx
yxxyx
V
dVV
V
dV
x
dx
V
dVV
V
VV
V
V
dx
dVx
V
V
Vxx
Vxx
dx
dVxV
dx
dVxV
dx
dy
xyx
yx
dx
dyGiven
221
22
1
22
2
2
logtan
loglogtanlog1
2.
2
1
1
1
1
1
1
1
1
1
11
Vyput,1,
Linear Differential Equations:
Solve the following:
1. yy
x
dy
dx2
cyy
xydydyyyxyisSolution
yePdyefI
yydyy
dyp
yQy
P
y
2222
..
loglog1
2;1
11
1log
1
1
2. 2
1 2 xyxy
4
cx
yxx
dxxdxxxxyisSolution
xeeisfIxxPdx
xQx
Pxx
yy
xPdx
44..
.loglog2
;22
42
4322
2log2
1
2
3. 02;cot2cot 2
1 yxxxxyy
2222
22
22
2
sinlog
2
sinsin02,
sinsinsin2sinsin2
sinsin2cossin2
sincotsin2sin.
sinsinlogcot
cot2;cot
xxxyccQygiven
cxxxydxxxxxdxxx
xdxdxxxdxxxdxxx
dxxxxxxxyisSolution
xeexdxxPdx
xxxQxP
xPdx
4. xeyy 2
1 3
ceeyedxedxeeeyisSolution
eefIxdxPdx
dxeQP
xxxxxxx
xPdx
x
3233
3
2
..
..33
;3
5. 211
2
30,3
yx
yyyyyyx
1
2
loglog3;1
33
yyPdyyQy
P
yy
x
dy
dx
y
yx
dy
dx
cyy
xdyy
yyxisSolution
yefI y
333.11
.
1..
1log
6. xxyy sintan21
cxxyxdx
dxxxxyisSolution
xefI
xxdxxPdx
xQxP
x
322
22
2coslog
2
coscoscoscos
cos.sincos.
cos..
coslogcoslog2tan2
sin;tan2
2
5
7. yxdy
dx
dx
dyyx 1
cyeexeyeeye
edydyyeexisSol
eefIydyPdy
yxQPyxdy
dx
yyyyyy
yyy
yPdy
1
...
..1
;1
8. 21
2
1
121
xxyyx
cxxy
dxx
dxx
xxyisSol
xeefIxx
xPdx
xQ
x
xP
xy
x
xy
xPdx
12
22
22
21log2
2
222221
tan1
1
1
1
1.11..
1..1log1
2
1
1;
1
2
1
1
1
2
2
9. xxyy 2sincot31
xdxxxx
dxx
dxxxxxyisSol
xeefI
xxxPdx
xQxP
xPdx
sec2tansec2sin
cos
cossin2.sinsin..
sin..
sinlogsinlog3cot3
2sin;cot3
2
113
3sinlog
3
3
10. 2
1
2 1
tan
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Statement problems:
1. Find equation of a curve which passes through origin given that slope at any point
on it = sum of coordinates.
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coordinates at any point exceeds slope at that point by 5.
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any point = sum of abscissa and product of coordinates.
2
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7
4. Find the equation of the curve which passes through (0, 1) given that slope at any
point on it
dx
dy satisfies (x – y) (dx + dy) = dx –dy.
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ordinate and slope at that point = abscissa.
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220,
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2
222,
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7. At any point P on a curve slope =2 (slope segment joining P & A (-4, -3). Find its
equation if it passes through (-2, 1)
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34
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8. In a bank principal, increases continuously at the rate of 5% per year. In how many years
of Rs.100 doubles itself? Use 6931.0log 2 e
8
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9. Find the equation of a curve whose differential equations is y1= ex sinx given that it
passes through origin.
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10. Find equation of a curve, whose differential equation 22 11 yxdx
dy
given that it passes through A (0, ½)
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1
MATHEMATICS II PUC
VECTOR ALGEBRA QUESTIONS & ANSWER
I One Mark Question
k.3j2i ofdirection in ther unit vecto theFind1) ++
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ba
abaarr
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But vectors are not equal since the corresponding components are distinct i.e. directions are different.
3) Find the values of x & y so that vectors 2i+3j and xi+4j are equal.
3,2
32
32
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∴=∴
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1) Find the vector parallel to the vector i - 2j and has magnitude 10 units.
( )
52
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=∴
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QIf vectors are parallel then unit vector along & parallel vectors are same
3
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3) Show the vectors 2i-3j+4k and -4i+6j-8k are collinear.
( )
collinear. areb &a
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∴
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III Three Marks Questions:
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4) Show that points A(1, 2, 7) , B(2, 6,3) & C(3, 10, -1) are collinear OR
Show that the points with position vectors i+2j+7k, 2i+6j+3k and 3i+10j-k are collinear.
( ) ( ) ( )( ) ( )( ) ( ) ( )( ) ( ) ( )
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332
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THREE DIMENSIONAL GEOMETRY.
One mark questions:
1) If a line makes angles 900, 1350 and 450 with the x, y and z axes respectively.
Find its direction cosines.
Solution:
Let = 900, = 135, = 450
Let l, m, n are the direction cosines of a line
l = cos =cos 900 = 0
m=cos = cos 1350 = 1
2 ,
n = cos = cos 450 1
2
2) If a line has direction ratio’s -18, 12, -4. Then what are its direction cosines.
Solution:
x = - 18 y = 12 z = -4
2 2 2
2 2 2 18 12 4 324 144 16 484 22r x y z
Direction cosines are 18 9
22 11
xl
r
12 6
22 114 2
22 11
ym and
rz
nr
3) Find the direction cosines of x, y and z axis.
Solution:
The x – axis makes angles 00, 900, 900 with the positive direction of
x, y and z – axis.
Direction cosines of x – axis are cos 00, cos 900, cos 900 i.e. 1, 0, 0.
Similarly direction cosines of y axis are cos 900, cos 00, cos 900 i.e. 0, 1, 0
and direction cosines of z – axis are cos 900, cos 900, cos 00, i.e. 0, 0, 1
4) Find the direction cosines of a line which makes equal angles with the co-
ordinate axes.
Solution:
Let , , be the angles made by the line with the positive direction of x-axis,
y –axis and z – axis
Also = = and cos2 + cos2 + cos2 = 1
cos2 + cos2 + cos2 = 1
3 cos2 = 1 cos2 = 1
3 cos =
1
3
The direction cosines are 1 1 1
, ,3 3 3
5) Find the equation of the plane having intercept 3 on the y-axis and parallel to
ZOX plane
Solution:
Y – intercept = b = 3
Any plane parallel to ZOX is y = b
The equation of the plane is y = 3
6) Find the distance of the plane 2x – 3y + 4z - 6 = 0 from the origin.
Solution:
Consider 2x – 3y + 4z – 6 = 0
2x – 3y + 4z = 6 – (1)
The Direction ratios are (2, -3, 4) = (x1, y1, z1)
22 22 3 4 4 9 16 29r
The Direction cosines are 1 2
29
xl
r
1 3
29
ym
r
1 4
29
zn
r
Divide equation (1) by 29
2 3 6
29 29 29 29
yx y z
and is of the form lx + my + nz = d
The distance of the plane from origin is 6
29d
7) Find the equation of the plane which makes intercepts 1, -1 and 2 on the x, y and
z axes respectively.
Solution:
a = x – intercept = 1, b = y – intercept = -1 and c = z – intercept = 2
The equation of the line is 1 . . 11 1 2
y yx z x zi e
a cb
8) Determine the direction cosines of the normal to the plane and the distance from
the origin is x + y + z = 1
Solution:
Consider x + y + z = 1 - (1)
Direction ratio’s of the plane are 1, 1, 1
2 2 2 1 1 11 1 1 1 1 1 3
3 3 3r l m n
Divide equation (1) by 1
33 3 3 3
yx z
It is of the form lx + my + nz = p
P = distance from origin = 1
3 9) Find the intercepts cut off by the plane 2x + y – z = 5
Solution:
Consider 2x + y – z = 5
2
1 . . 15 5 5 5 55/2
y yx z x zi e
a = x– intercept = 5/2 b = y – intercept = 5 c = z – intercept = -5
10) Show that the planes 2x + y + 3z – 2 = 0 and x – 2y + 5 = 0 are perpendicular.
Solution:
Consider 2x + y + 3z - 2 = 0 i.e. 2x + y + 3z = 2
And x – 2y + 5 = 0 i.e. x – 2y + 0.z = - 5
The normals to the plane are
1 22 3 2andP i j k P i j
1 2. 2 1 1 2 3 0 2 2 0 0P P
The planes 1 2P Pand are perpendicular
11) Show that the planes 2x – y + 3z – 1 = 0 and 2x – y + 3z + 3 = 0 are parallel.
Solution:
Consider 2x – y + 3z – 1 = 0 i.e. 2x – y + 3z = 1
And 2x – y + 3z + 3 = 0 i.e. 2x – y + 3z = -3
The normals to the plane are
1 22 3 2 3P i j k and P i j k
1 1 1
1 21
2 1 31, 1 1
2 1 3
a b cand
a cb
1 1 1
2 22
1a b c
a cb
The planes P1 and P2 are parallel.
12) Find the equation of the plane parallel to x – axis and passing through the origin.
Solution:
The direction ratio’s of x-axis is 1, 0, 0
The equation of the line through origin and parallel to x-axis
is 0 0 0
. .1 0 0 1 0 0
x y z x y zi e
13) Find the vector equation of the straight line passing through (1.2.3) and
perpendicular to the plane . 2 5 9 0 r i j k
Solution:
The required line passes through (1, 2, 3) and perpendicular to the plane
. 2 5 9 0r i j k is
2 3 2 5r i j k i j k
14) Find the equation of the plane passing through (a, b, c) and parallel to the plane
. 2 r i j k
Solution:
Consider . 2 2r i j k x y z
Any plane parallel to the given plane is x + y + z =
and is pass through (a, b, c) a + b + c =
Hence the equation of the plane parallel to the given plane is
x + y + z = a + b + c
15) Find the distance between the two planes 2x+3y+4z=4 and 4x+6y+8z= 12.
Solution:
Consider 2x + 3y + 4z = 4 - (1)
And 4x + 6y + 8z = 12
i.e. 2x + 3y + 4z – 6 = 0 - (2)
Distance from the point to the plane (2) = 2 2 2
2 3 4 6
2 3 4
x y z
4 6 2 2
4 9 16 29 29
Two mark questions:
1) Show that the points (2, 3, 4) (-1, -2, 1) and (5, 8, 7) are collinear.
Solution:
A = (2, 3, 4) B = (-1, -2, 1) and C = (5, 8, 7)
Direction ratio’s of the line joining A & B are, 2+1, 3+2, 4-1, i.e. 3, 5, 3
Direction ratio’s of the line joining B & C are -1-5, -2-8, 1-7, i.e. -6, -10, -6
The direction ratio’s of AB & BC are proportional & B is the common point
of AB & BC
The points A, B, C are collinear
2) Show that the line through the points (1, -1, 2) (3, 4, -2) is perpendicular to the
line through the points (0, 3, 2) and (3, 5, 6)
Solution:
Let A = (1, -1, 2) B = (3, 4, -2) C = (0, 3, 2) and D = (3, 5, 6)
Direction ratio’s of AB are, a1 = 3-1=2, b1 = 4- (-1) = 4+1=5 & C1 = -2-2 = -4
Direction ratio’s of CD are a2 = 3-0=3, b2 = 5-3=2, C2 = 6-2=4
Now a1a2 + b1b2 + c1c2 = 2 (3) + 5 (2) + (-4) 4
= 6+10 - 16 = 0
AB is perpendicular to CD
3) Show that the line through the points (4, 7, 8) (2, 3, 4) is parallel to the line
through the points (-1, -2, 1) (1, 2, 5).
Solution:
Let A = (4, 7, 8) B = (2, 3, 4) C = (-1, -2, 1) D = (1, 2, 5)
Direction ratio’s of AB are a1 = 2 – 4 = -2, b1 = 3-7=-4, c1 = 4 – 8 = -4
Direction ratio’s of CD are a2 = 1- (-1) =1+1=2, b2 = 2-(-2) = 2+2=4, c2 = 5-1 =4
1 1 1
2 22
2 4 41, 1, 12 4 2
a b c
a cb
1 1 1
2 22
a b c
a cb Hence AB is parallel to CD
4) The Cartesian equation of a line is 45 6
73 2yx z . Write its equation in
vector form.
Solution:
Consider 45 6
73 2yx z
a = (5, -4, 6) and b = (3, 7, 2) are the direction ratio’s
Vector equation of the line is r a b
5 4 6 3 7 2r i j k i j k
5) Find the distance of the point (2, 3, -5) from the plane 2 2 9r i j k
Solution:
Consider . 2 2 9 2 3 5r i j k and a i j k
and 2 2 9N i j k and d
. 2 1 3 2 5 2 2 6 10 18a N
2
2 21 2 2 1 4 4 9 3N
Distance of a point from the plane = d = .
18 9 93
3 3
a N d
N
6) Find the equation of the plane passing through the line of intersection of the
plane x + y + z = 6 and 2x + 3y + 4z – 5 = 0 and the point (1, 1, 1)
Solution:
Consider x + y + z = 6 x + y + z -6 = 0 and 2x + 3y + 4z – 5 = 0
The equation of the plane passing through the intersection of the two planes
is x + y + z – 6 + (2x + 3y + 4z – 5) = 0 and is pass through (1, 1, 1)
1 + 1 + 1 – 6 + (2+3+4-5) = 0
- 3 + 4 = 0 4 = 3 = ¾
The equation is (x + y + z – 6) + 3
4 (2x + 3y + 4z – 5) = 0 (multiply by 4)
4x + 4y + 4z – 24 + 3 (2x + 3y + 4z – 5) = 0
4x + 4y + 4z – 24 + 6x + 9y + 12z – 15 = 0
10x + 13y + 16z – 39 = 0
7) Derive the direction cosine of a line passing through two points.
Solution:
Let l, m, n be the direction cosines of a line PQ and the line PQ makes ,
and with positive directions of x, y and z axes respectively. Draw the
perpendiculars from P and Q to xy – plane to meet at R & S and draw PN
perpendicular to QS.
From the le PNQ, ˆPQN
2 1cosZ ZQN ON OQ
PQ PQ PQ
2 1 2 1cos cosx x y y
Similarly andPQ PQ
Where 2 2 2
2 1 2 1 2 1PQ x x y y z z
8) The Cartesian equation of a line is 53 6
2 4 2yx z
Find the vector equation of the line
Solution:
Consider 53 6
2 4 2yx z
3 652 4 2
x zy
x1 = -3 y1 = 5 z1 = -6 and a = 2 b = 4 and c = 2
1 1 1, , 3, 5, 6a x y z
, , 2, 4, 2b a b c are direction ratio’s
The vector equation of a line is r a b
3 5 6 2 4 2r i j k i j k
9) Find the vector equation of the plane which is at a distance of 7 units from the
origin and normal to the vector 3i + 5j – 6k
Solution:
2
2 23 5 6 3 5 6 9 25 36 70let n i j k and n
3 5 6 3 5 6ˆ
70 70 70 70
n i j kand n i j k
n
The equation of the plane ˆ. 7r n d and d
3 5 6. 7
70 70 70r i j k
10) Find the distance of the point (3, -2, 1) from the plane 2x – y + 2z + 3 = 0
Solution:
Consider 2x – y + 2z + 3 = 0
1 1 1
2 2 2 22 2
2 3 1 2 2 1 3
2 1 2
ax by cz dd
a b c
6 2 2 3 13
34 1 4
Three mark questions
1) Find the vector and Cartesian equations of the line that passes through the
points (3, -2, -5) and (3, -2, 6)
Solution:
Let A = (3, -2, -5) B = (3, -2, 6)
Direction ratio’s of AB are, a = 3 – 3 = 0
b = -2 – (-2) = - 2+2 = 0
c = 6 – (-5) = 6 + 5 = 11
0. 0. 11 11 3, 2, 5b ai bj ck i j k k and a = 3i – 2j – 5k
Vector equation of a line passing through two points is r a b
3 2 5 11r i j k k
Cartesian equation of a line is 1 1 1x x y y z z
a b c
2 53 3 2 5. .
0 0 11 0 0 11
y zx x y zi e
2) Show that three lines with direction cosine 12 3 4
, , ;13 13 13
4 12 3 3 4 12, , ; , ,
13 13 13 13 13 13
are mutually perpendicular.
Solution:
L1, L2, L3 are three lines.
The direction cosine of the line 1 1 1 1
12 3 4, , , ,
13 13 13L l m n
Direction cosines of the line 2 2 2 2
4 12 3, , , ,
13 13 13L l m n
Direction cosines of the line 3 3 3 3
3 4 12, , , ,
13 13 13L l m n
1 2 1 2 1 2
12 4 3 12 4 3 48 36 120
13 13 13 13 13 13 169l l m m n n
1 2L is perpendicular to L
2 3 2 3 2 3
4 3 12 4 3 12 12 48 36 48 48. . 0
13 13 13 13 13 13 169 169l l m m n n
2 3L is perpendicular to L
3 1 3 1 3 1
3 12 4 3 12 4 36 12 48 48 48. 0
13 13 13 13 13 13 169 169l l m m n n
3 1L is perpendicular to L
Hence the three lines are mutually perpendicular
3) Find the angle between the pair of lines 3 5r i j k i j k and
7 4 2 2 2r i k i j k
Solution:
Consider 3 5r i j k i j k 1b i j k
7 4 2 2 2r i k i j k 2 2 2 2b i j k
1 2
2 2 21 2
. 1 2 1 2 1 2 2 2 2 6
1 1 1 3 4 4 4 12 2 3
b b
b b
1 2 0
1 2
0
. 6 6 6cos 1 cos0
2 3 62 3. 3
0
b b
b b
4) Find the equation of the line which passes through the point (1, 2, 3) and is
parallel to the vector 3i + 2j – 2k, both in vector form and Cartesian form.
Solution:
Let 1, 2, 3 2 3 3 2 2a i j k and b i j k
The vector equation of the line is r a b
2 3 3 2 2r i j k i j k
Let r be the position vector of the point and r xi yj zk
2 3 3 2 2xi yj zk i j k i j k
2 3 3 2 2i j k i j k
1 3 2 2 3 2i j k
1 3 2 2 3 2
1 3 2 2 3 2
1 2 3
3 2 2
x y and z
x y z
x y z
1 2 3
3 2 2
x y z
is the equation of the line in Cartesian form.
5) Find the distance between parallel lines 2 4 2 3 6r i j k i j k and
3 3 5 2 3 6r i j k i j k
Solution:
Consider 2 4 2 3 6r i j k i j k
And 3 3 5 2 3 6r i j k i j k
1 12 4 2 3 6a i j k b i j k
and 2 23 3 5 2 3 6a i j k b i j k
1 2b b The lines are parallel
2 2 21 2 2 3 6 2 3 6b b b i j k and b
2 1 2 10 4 9 36 49 7a a i j k
2 1 2 3 6 3 6 2 12 2 6 9 14 4
2 1 1
i j k
b a a i j k i j k
2 2 22 1 9 4 4 81 196 16 293b a a
Distance between parallel lines = d = 2 1 293
7
b a a
b
6) Find the angle between the pair of lines 3 1 3
3 5 4
x y z
1 4 5
1 1 2
x y zand
Solution:
Consider 1
3 1 31 3, 5, 4
3 5 4
x y zDirection ratios of b
2
1 4 52 ' 1, 1, 2
1 1 2
x y zand Direction ratio s of b
1 2. 3 1 5 1 4 2 3 5 8 16b b
2 2 21
2 2 22
3 5 4 9 25 16 50 25 2 5 2
1 1 2 1 1 4 6
b
b
1 2
1 2
1
. 16 16 16 16 8cos
5 2 6 5 12 5 4 3 5 2 3 5 3
8cos
5 3
b b
b b
7) Find the shortest distance between the lines
1 1 1 3 5 7
7 6 1 1 2 1
x y z x y zand
Solution:
Consider 1 1 1 3 5 7
7 6 1 1 2 1
x y z x y zand
2
1 2
1
1 1 1. . 3 5 7
7 6 1
2
7 6
x y zi e a i j k
a i j k b i j k
b i j k
2 1 3 5 7 4 6 8a a i j k i j k i j k
1 2 7 6 1 6 2 7 1 14 6
1 2 1
4 6 8
i j k
b b i j k
i j k
2 2 2
1 2 4 6 8 16 36 64 116b b
Shortest distance = 1 2 2 1
1 2
. 16 36 64
116
b b a ad
b b
116
116 4 29 2 29116
8) Find the equation of the planes passing through three points (1, 1, 0)
(1, 2, 1) and (-2, 2, -1)
Solution:
Let a = (1, 1, 0) b = (1, 2, 1) and c = (-2, 2, -1) and r xi yj zk
1 1 0r a x i y j z k
0, 1, 1 3, 1, 1AB b a and AC c a
The vector equation of the plane is . 0r a AB AC
1 1
0 1 1 0
3 1 1
x y z
(x-1) (-1-1) – (y-1) (0+3) + z (0 + 3) = 0
-2(x-1) -3 (y-1) + 3z = 0
-2x + 2 – 3y + 3 + 3z = 0
-2x – 3y + 3z + 5 = 0
2x + 3y - 3z - 5 = 0
2x + 3y – 3z = 5 is the equation of the plane
9) Find the angle between the pair of lines given by 3 2 4 2 2r i j k i j k
and 5 2 3 2 6r i j i j k .
Solution: 1 22 2 3 2 6b i j k b i j k
1 2 1
1 2
. 3 4 12 19 19cos cos
21 219 49
b b
b b
10) Prove that if a plane has intercepts a, b, c and is at a distance of p units from the
origin then 2
2 2 2
1 1 1 1
a b c p
Solution:
Let a, b, c, are the intercepts of the plane
And the equation is 1 1x y z
a b c
P = The distance of the plane (1) from (0, 0, 0)
2 2 2 2 2 2 2 2 2
0 0 0 1 1 1
1 1 1 1 1 1 1 1 1P
a b c a b c a b c
2
2 2 2 2
2 2 2
1 1 1 1 1
1 1 1P
p a b c
a b c
Five mark questions:
1) Derive the equation of the line in space passing through a point and parallel to a
vector, both in the vector form and Cartesian form.
Solution:
Let a be the position vector of the given point A. w.r. to
the origin O of the rectangular co-ordinate system. Let l
be the line which passes through the point A and is
parallel to the given vector b . Let r be the position
vector of an arbitrary point P on the line. Then AP is
parallel to b .
i.e. AP b where is a real number
OP OA b
r a b
r a b is the vector equation of the line
Let A = (x1, y1, z1) be the co-ordinates of the given point and the direction ratio’s of the
line are a, b, c.
Let P = (x, y, z) be the co-ordinate of any point
Then 1 1 1r xi yj zk and a x i y j z k and b ai bj ck and r a b
xi + yj + zk = (x1i + y1j + z1k) + (ai + bj + ck)
= x1i + y1j + z1k + ai + bj + ck
= (x1 + a) i + (y1 + b) j + (z1 + c) k
Equating the coefficients of i, j and k we get
x = x1 + a y = y1 + b and z = z1 + c
these are the parametric equations of a line
x – x1 = a y – y1 = b and z – z1 = c
1 1 1x x y y z z
a b c
1 1 1x x y y z z
a b c
. This is the Cartesian equation of the line.
2) Derive the equation of a line in space passing through two given points both
invector form and Cartesian form.
Solution:
Let & &a b r are the position vectors of the two
points A (x1, y1, z1) is (x2, y2, z2) and p (x, y, z)
respectively.
AP OP OA r a and AB OB OA b a
If the point p lien on the line AB if and only if
AP and AB are collinear.
. .AP AB i e r a b a
r a b a is the vector equation of the line passing through two points.
Let 1 1 1 2 2 2, &r xi yj zk a x i y j z k b x i y j z k r a b a
2 1 2 1 2 1
1 2 1 1 2 1 1 2 1
1 2 1 1 2 1 1 2 1
1 2 1 1 2 1 1 2 1
1 1 1
2 1 2 1 2 1
2 1 1 1
, &
xi yj k x i y j z k x x i y y j z z k
x x x i y y y j z z z k
x x x x y y y y z z z z
x x x x y y y y z z z z
x x y y z z
x x y y z z
1 1 1
2 1 2 1 2 1
x x y y z z
x x y y z z
is the Cartesian equation of the line passing through
two points.
3) Derive the shortest distance between two skew lines both in vector form and
Cartesian form.
Proof:
Let l1 and l2 be the skew lines
Let 1 1 2 2r a b and r a b be the skew lines. Let s and T are any two points
on l1 and l2 with position vectors 1 2a and a respectively.
Then the magnitude of the shortest distance is equal to the projection of ST along the
direction of a line.
If PQ is the shortest distance between the lines l1 and l2 then it is perpendicular to
both 1 2b and b and n is the unit vector along PQ .
1 2
1 2
ˆb b
nb b
let be the angle between ST and PQ
Then 2 1
.cos cos
PQ STPQ ST and but PQ d and ST a a
PQ ST
2 1ˆcos
.
d n a a
d ST
1 2 2 1
1 2
cosb b a a
STb b
Shortest distance is d = PQ = 1 2 2 1
1 2
cosb b a a
STb b
is the Shortest distance of skew lines in vector form.
Let 1 :l 1 1 1
1 1 1
x x y y z z
a b c
and 2 2 2
2
2 2 2
x x y y z zl
a b c
be the equations of two
skew lines in Cartesian form.
The shortest distance between two skew lines is
2 2 2
1 2 2 1 1 2 2 1 1 2 2 1
db c b c c a c a a b a b
where
2 1 2 1 2 1
1 1 1
2 2 2
x x y y z z
a b c
a b c
4) Derive the equation of the plane in normal form both in the vector form and
Cartesian form.
Solution:
Consider a plane whose perpendicular distance from the origin is d. If ON is the
normal from the origin to the plane and n is the unit normal vector ON
Then ˆ.ON d n
Let P be any point on the plane then NP is perpendicular to ON
. 0 1NPON
Let r be the position vector of the point P
ˆ.Then NP OP ON r d n
From equation (1) ˆ ˆ. . . 0 0r d n d n But d
ˆ ˆ. 0
ˆ ˆ ˆ ˆ ˆ. . . 0 . 1.1 1
ˆ. 0
r d n n
r n d n n But n n
r n d
ˆ.r n d is the equation of the plane vector form
Let l, m, n be the direction cosines of n
Then n li mj nk and OP r xi yj zk
ˆ.r n d
.xi yj zk li mj nk d
Therefore lx + my + nz = d is the Cartesian equation of the plane in normal form
5) Derive the condition for the coplanarity of two lines in space both in the vector
form and Cartesian form.
Solution:
Let the given lines be 1 1 2 2 2r a b and r a b
The line (1) passes through the point A with the position vector 1a and parallel to 1b
and the line (2) passes through the point B with the position vector 2a and parallel to
2b
Thus 2 1AB B A a a
The given lines are coplanar if and only if AB is perpendicular to 1 2b b
i.e. 21. 0AB b b
1 22 1. 0a a b b is condition for the coplanarity of two lines in vector form.
Let A = (x1, y1, z1) and B = (x2, y2, z2) be the co-ordinates of the points A and B
respectively. Let a1, b1, c1 and a2, b2, c2 be the direction ratio’s of 21b and b
respectively.
Then 2 1 2 1 2 1AB B A x x i y y j z z k
1 21 1 1 2 2 2b a i b j c k and b a i b j c k
The given lines are coplanar if 21. 0AB b b
2 1 2 1 2 1
1 1 1
2 2 2
0
x x y y z z
a b c
a b c
is condition for coplanarity of two lines in cartestion form.
Topic :- Linear Programming
The running of any firm or of a factory involves many constraints like
financial,space,resources,power etc. The objective of any business person would be to make the
profit maximum (in the case of investment to make the cost a minimum) under all the
constraints.The linear programming problem is the problem of optimising an objective function
under a given set of constraints.When the objective function is profit function the optimisation is to
maximise the profit.In the case of cost function the optimisation is to minimise the cost.
All the constaints of the problem are linear inequalities and the objective function is linear.Hence, it
is called LPP.The following are a few illustrations of the LPP by Graphical method .
1. Maximise : P = .
Subject to : ≤ 20
≤ 6
0.
Now we draw the graphs of the equations = 20, and = 6 and recognise the
feasible region.
Y
X
O(0,0)
B(8,6)
C(10,5)
D(15,0)
A(0,6) =600
The shaded region OABCDO (bounded) represents the feasible region in which all the Constraints of
the problem are satisfied. Now, we evaluate the objective function at these corners of the region.
Vertex
0 0 0 0
A 0 6 3X0+5X6=30
B 8 6 3X8+5X6=54
C 10 5 3X10+5X5=55
D 15 0 3X15+5X0=45
From the above table we observe that Maximum value of P is 55 and corresponds to
X = 10 and y = 5
2). Minimise: C =
Subject to : ≥ 6
4.
≥ 6
.
we draw the graphs of the linear equations
O X
Y
B(1,3
)
C(3,1
)
D(
)
The shaded region BCD ( triangular region ,bounded) is the feasible region.
In this region all the constraints of the problem are satisfied.Now we evaluate the objective
function C = at the verticies.
Vertex x y C =
B 1 3 272
C 3 1 304
D 3/2 3/2 216
From the above table we observe that the minimum value of C is 216 and corresponds to
and
.
Note:- The co-ordinates of the points B, C and D can be read from the graph. They may
also be determined by solving the equations of the corresponding pair of lines
(3) Maximise :
Subject to :
Now we draw the graphs of the linear equations and recognise the feasible region.
Y
X
A(0,6) B(3,4)
C(5,0) O
The shaded region OABCO is the feasible region. Now we evaluate at the
vertices.
Vertex
0 0 0 0
A 0 12
B 3 4 23
C 5 0 25
From the table we find that the maximum value of Z = 25 and corresponds to .
Note:The above working is based on the result that the optimal solution of a LPP if exists
will occur at a corner point of the feasible region.
(4) Maximise
Subject to :
We draw the graphs corresponding to the linear equations and shade the feasible region.
O
Y
X
A(0,5)
B(2,4)
C(4,2)
D(4,0)
The shaded region OABCDO(bounded ) is the feasible region.
Now we evaluate at the vertices
.
Vertex
0 0 0 0
A 0 10
B 2 4 14
C 4 2 16
D 4 0 12
From the table we find that the maximum value of p = 16 and corresponds to
(5). Determine the maximum and minimum values of
Z = 4x + y
Subject to: x + 2y 4
x – 2y 0
6
x, y 0
solution: We draw the lines x+ 2y = 4 , x– 2y = 0 and x= 6.
D(6,0)
x+2y=4
x– 2y
x=6
C(6,3)
B(2,1)
X O
Y
A(4,0)
The feasible region is bounded region with A( 4, 0 ) , B ( 2 , 1 ) , C ( 6 , 3 ) and D( 6 , 0 )as
vertices.
Now we find the values of Z = 4x + 2y at these vertices.
Vertex X Y Z = 4x + 2 y
A 4 0 16
B 2 1 10
C 6 3 30
D 6 0 24
From the above table we find that the minimum value of Z is 10 and corresponds to x = 2 and
y = 1. The maximum value of Z is 30 and corresponds to x = 6 and y =3.
APPLICATIONS PROBLEMS
1). A factory manufactures two types of screws, A and B by using two
machines.The time required for the manufacture of one packet of each of the
two types of the screws on the two machines,the total time available of each of
the mechine and the profit obtained on the sale of each packet of the two types
of the screws are given below.
Machine
Screws
Time required in minutes Profit in Rs
Per packet. I II
A 4 6 8
B 6 3 5
Time available
in hours
4 4 --
Formulate this as a Lpp and determine the number of packets of each of the two
types of screws to be manufactured so as to get maximum profit assuming that
all the packets of the screws manufactured are sold.
Solution;-
Let x and y be the number of packets of the type A and type B screws to be
produced.
Then,the total time the mechine I works is 4 x + 6y.
The total time the machine II works is 6x + 3y.
The total profit in Rs. is p = 8x + 5y.
Since, the two machines are available at most for 4 hours = 240 minutes, we
have 4x + 6y 240. and 6x + 3y 240.
.`. the Lpp is to maximise p = 8x + 5y
Subject to:- 4x + 6y 240
6x + 3y 240
x, y 0
Now, we draw the graphs of the linear equations 4x + 6y = 240
and 6x +3y = 240.
The feasible region is bounded region OABCO with O(0,0), A( 0,40 ) , B ( 30,20) ,
and C(40 ,0 ) as vertices.
Now we find the values of p = 8x + 5y at these vertices.
Vertex X Y p = 8x + 5 y
O 0 0 0
A 0 40 200
B 30 20 340
C 40 0 320
Y
X O
B(30,20)
C(40,0)
(0,80)
A(0,40)
D(60,0)
From the table we find that the maximum value of p = 340 and corresponds to
.
`Hence, the maximum profit is Rs. 340 and corresponds to manufacture of 30 packets of A
type.and 20 packets of B type screws.
Suitcase
Machine
Time required in hours Time
available in
hours Type I Type II
A 3 3 18
B 2 4 16
Profit in Rs Per
suitcase
30 42 --
Determine the number of the two types of suitcases to be produced to get maximum profit ..
Solution;-
Let x and y be the number of type I and type II suitcases to be produced.
Then,the total time the mechine A works is 3 x + 3y.
The total time the machine B works is 2x + 4y.
The total profit in Rs. is p = 30x + 42y
Since, the two machines A and B are available at most for 18and 16 hours
respectively, we have 3x + 3y 18. and 2x + 4y 16.
.`. the Lpp is to maximise p = 30x + 42y
Subject to:- 3x + 3y 18
2x + 4y 16
x, y 0
Now, we draw the graphs of the linear equations 3x + 3y = 18
and 2x +4y=16.
2) The production of two types of suitcases requires processing and completion to be done on two machines A and B.The time required for processing and completion of each type of trunk on the two machines, the time available on each machine and profit on each type of the suit case is given below.
E(0,6)
The feasible region is bounded region OABCO with O(0,0), A( 0,4 ) , B ( 4,2) ,
and C(6 ,0 ) as vertices.All the condtions of the problem are satisfied within and
on the boundary of this
Region.
Now we find the values of p = 30 x + 42 y at these vertices.
Vertex x Y p = 30 x + 42 y
O 0 0 0
A 0 4 168
B 4 2 204
C 6 0 180
From the table we find that the maximum value of p = 204 and corresponds to
.
Hence, the maximum profit is Rs 204 and corresponds to manufacture of 4
suitcases of
type I.and 2 suitcases of type II..
Y
X O
B(4,2)
C(6,0)
A(0,4)
D(8,0)
3).At a cattle rearing , it is prescribed that the food for each animal contain at
most 16 units of nutrient A, and at least 24 and 48 units of nutrients B and C
respectively.Two types of fodders are available. The number of units of these
nutrients contained per Kg by the two types of fodders and their cost per Kg is
given below
Content of
nutrient/kg
A B C Cost per kg in
Rs.
Fodder- 1 1 3 2 12
Fodder-2 1 1 6 15
By graphical method determine the number of kg of the two types of the
fodder to be purchased so as to make the cost a minimum,yet meeting the
requirements.
Solution:
Let, x and y kg of fodder-1and fodder-2 be purchased.
Then, the cost in Rs. Is C = 12 x + 15 y.
Total content s of nutrients A, B and C are x + y , 3 x + y and 2 x + 6 y
respectively.
Since the content of A is to at most 16 while it has to be at least 24 for B and
48 for C, the constraints are
x + y , 3 x + y 24 , and 2 x + 6 y 48.
.`.The Lpp is To minimise; C = 12 x + 15 y
Subject to x + y
3 x + y 24
2 x + 6 y 48
X, y 0
Now ,we draw the graphs of the linear equations and recognise the feasible
region
The shaded region MNL ( triangular region ,bounded) is the feasible
region.
In this region all the constraints of the problem are satisfied.Now we evaluate
the objective function C = at the verticies.
Vertex x y C =
M 6 6 162
N 4 12 228
L 1/2 4 206
From the above table we observe that the minimum value of C is 162 and
corresponds to
and .
Therefore, in order to make the cost minimum 6kg each of fodder-1 and
fodder-2 are to be bought and the minimum cost is Rs.162.
Note:- The co-ordinates of the points M, N and L can be read from the graph.
They may also be determined by solving the equations of the corresponding pair
of lines.
O
X
Y
N(4,12
)
L(12,4
) )
A (0,8)
C(0,16
)
E(0,24
)
F (8,0)
D (16,0) B (24,0)
.