CBSE SAMPLE QUESTION PAPERCLASS-XII
MATHS Section-A
1. If �⃗�=2𝑖̂ + 𝑗̂ + 3�̂� and �⃗⃗� = 3𝑖̂ + 5�̂� − 2�̂� , then find |�⃗� ×�⃗⃗�|.
2. Find the angle between the vectors 𝑖̂ − 𝑗 ̂ and 𝑗̂ − �̂�
3. Find the distance of a point (2,5,-3) from the plane 𝑟.⃗⃗⃗ (6𝑖̂ − 3𝑗̂ + 2𝑘)̂ = 4
4. Write the element a12 of the matrix A= [aij]2X 2 , whose elements aij are given by
aij = 𝑒2𝑖𝑥 sin 𝑗𝑥
5. Find the differential equation of the family of lines passing through origin.
6. Find the integrating factor for the following differential equation :
𝑥𝑙𝑜𝑔𝑥 𝑑𝑦
𝑑𝑥+ 𝑦 = 2 log 𝑥
Section-B
Q7 If A= [1 2 22 1 22 2 1
] Then show that 𝐴2 − 4𝐴 − 5𝐼 = 0and hence find 𝐴−1.
OR
If A[2 0 −15 1 00 1 3
] Then find 𝐴−1 using elementary row operations.
Q8 Using the properties of determinant s, solve the following for x:
𝑥 + 2 𝑥 + 6 𝑥 − 1 𝑥 + 6 𝑥 − 1 𝑥 + 2𝑥 − 1 𝑥 + 2 𝑥 + 6
=0
Q9∫𝑠𝑖𝑛2𝑥
𝑠𝑖𝑛𝑥+𝑐𝑜𝑠𝑥
𝜋/2
0𝑑𝑥
OR
Evaluate ∫ (𝑒3𝑥 + 7𝑥 − 5) 𝑑𝑥2
−1 as a limit of sums .
Q10 Evaluate: ∫𝑥2
𝑥4 +𝑥2 −2𝑑𝑥
Q 11 In a set of 10 coins,two coins are with heads on both the sides. A coin is selected at
random from this setand tossed five times . If all the five times, the result was heads, find
the probability that the selected coin had heads on both the sides.
OR
How many times must a fair coin be tossed so that the probability of getting atleast one
head is more than 80%?
Q 12 Find x such that the four points 𝐴 (4,1,2), 𝐵(5, 𝑥, 6), 𝐶 (5,1, −1) 𝑎𝑛𝑑 𝐷(7,4,0 )
are coplanar.
Q 13 A line passing through the point A with position vector �⃗� = 4𝑖̂ +2𝑗 ̂+ 2�̂� is parallel
to the vector �⃗⃗� = 2𝑖̂ + 3𝑗̂ +6�̂� .Find the length of the perpendicular drawn on this line from
a point P with position vector𝑟 =𝑖̂ + 2𝑗̂ + 3�̂� .
Q 14 Solve the following for 𝑥 :
sin−1(1 − 𝑥) - 2sin−1 𝑥 =𝜋
2
OR
Show that 2 sin−1 3
5− tan−1 17
31=
𝜋
4
15 . If 𝑦 = 𝑒𝑎𝑥 . cos 𝑏𝑥 , then prove that 𝑑2𝑦
𝑑𝑥2 − 2𝑎 𝑑𝑦
𝑑𝑥+ (𝑎2 + 𝑏2 ) 𝑦 = 0
16 If 𝑥𝑥 + 𝑥𝑦 + 𝑦𝑥 = 𝑎𝑏, then find 𝑑𝑦
𝑑𝑥.
17. If 𝑥 = 𝑎 sin 2𝑡 ( 1 + cos 2𝑡) 𝑎𝑛𝑑 𝑦 = 𝑏 cos 2𝑡( 1 − cos 2𝑡 ), 𝑡ℎ𝑒𝑛 𝑓𝑖𝑛𝑑 𝑑𝑦
𝑑𝑥 𝑎𝑡 𝑡 =
𝜋
4
18.Evaluate :∫( 𝑥+3)𝑒𝑥
(𝑥+5)3 𝑑𝑥
19. Three schools X, Y and Z organized a fete ( mela) for collecting funds for flood
victims in which they sold hand- held fans , mats and toys made from recycled material ,
the sale price of each being Rs 25 , Rs 100 and Rs 50 respectively. The following table
shows the number of articles of each type sold :
Articles School
X y Z
Hand held fans 30 40 35
Mats 12 15 20
Toys 70 55 75
Section C
20. Let 𝐴 = 𝑄 × 𝑄, where Q is a set of all rational numbers and * be the binary operation
on A defined by ( 𝑎, 𝑏) ∗ (𝑐, 𝑑) = ( 𝑎𝑐, 𝑏 + 𝑎𝑑) 𝑓𝑜𝑟 (𝑎, 𝑏), (𝑐, 𝑑) ∈ 𝐴.
Then find (i) The identity element of * in A
(ii) Invertible elements of A, and hence write the inverse of elements (5,3) 𝑎𝑛𝑑 (1
2, 4)
OR
Let 𝑓: 𝑊 → 𝑊 be defined as 𝑓(𝑛) = {𝑛 − 1, 𝑖𝑓 𝑛 𝑖𝑠 𝑜𝑑𝑑𝑛 + 1, 𝑖𝑓 𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛
Show that f is invertible and find the inverse of f. Here , W is the set of whole numbers.
21. Sketch the region bounded by the curve 𝑦 = √5 − 𝑥2 and |𝑥−1| and find its area
using integration.
22.Find the particular solution of the differential equation 𝑥2 𝑑𝑦 = (2𝑥𝑦 + 𝑦2)𝑑𝑥 , given
that 𝑦 = 1 𝑤ℎ𝑒𝑛 𝑥 = 1 .
23 Find the absolute maximum and absolute minimum values of the function f given by
𝑓(𝑥) = 𝑠𝑖𝑛2 𝑥 − 𝑐𝑜𝑠2 𝑥 , 𝑥 ∈ [0, 𝜋].
24 Show that the lines :
𝑟 =𝑖 ̂ + 𝑗̂ + �̂� + (𝑖 ̂ − 𝑗̂ + �̂�)
𝑟 = 4𝑗̂ + 2�̂� + µ (2𝑖 ̂ − 𝑗̂ + 3�̂�) are coplanar .also find the equation of plane containing
these lines
Q 25 Minimize and maximize z = 5x +2y subject to the following constraints :
x – 2y ≤ 2 , 3x +2y ≤ 12 , -3x + 2y ≤ 3 , x ≥ 0 , y ≥ 0
Q 26 Two numbers are selected at random ( without replacement ) from first six positive
integers.Let X denote the larger of the two numbers obtained. Find the probability
distribution of X . Find the mean and variance of this distribution.
SOLUTION
1 Given that 𝑟 ⃗⃗⃗ =2i + 3j +3k & b-= 3i 5j -2k, we need to find IaxbI
axb = 𝑖 𝑗 𝑘2 1 33 5 −2
=−17i +13j +7k ,thus IaxbI =√ 172 +132 +72 = √507
2 a-b = (i-j).(j-k) =-1 cos 𝛼=𝑎.𝑏
𝐼𝑎𝐼𝐼𝑏𝐼 =
−1
√2𝑋√2 =
−1
2 =cos 120 ° ; [a→ 𝑣𝑒𝑐𝑡𝑜𝑟 ′𝑎′]
𝛼 = 120°
3 consider the vector equation of the plane is
r.(6i-3j+2k)=4; (xi+yj+zk )(6i-3j+2k )=4
Equation of the plane is 6x-3y+2z-4=0; d=distance between the plane and the point (2,5,-3)
thus d =ax1 +by1 + cz1 +d
√a2 +b2 +c =
6𝑥2−3𝑥5+(−3)−4
√62+(−3)2+22 =13/7
4 aij =𝑒2𝑖𝑥) sin2x substitute i=1 and j =2 ;a12 =𝑒2𝑥𝑠𝑖𝑛2𝑥
5 Consider the equation y=mx;m= y/x ,diff. w.r.t.x the equation y=mx
𝑑𝑦
𝑑𝑥=m; dy/dx=y/x ;
𝑑𝑦
𝑑𝑥-−
𝑦
𝑥=0
6 logx𝑑𝑦
𝑑𝑥 +
𝑦
𝑥𝑙𝑜𝑔𝑥=
2𝑙𝑜𝑔𝑥
𝑥𝑙𝑜𝑔𝑥;
𝑑𝑦
𝑑𝑥 x +
𝑦
𝑥𝑙𝑜𝑔𝑥=
2
𝑥----1
𝑑𝑦
𝑑𝑥+ 𝑃𝑦 = 𝑄 ; P =1/xlogx and Q = 2/x
I.F =𝑒∫ 𝑝𝑑𝑥 =𝑒 ∫𝑑𝑥
𝑥𝑙𝑜𝑔𝑥=𝑒𝑙𝑜𝑔𝑥(𝑙𝑜𝑔𝑥) =logx
OR Here f(x) = 𝑒3𝑥 +7x – 5 a=-1 b=2 h=b-a/n =3
𝑛
By def.∫ (𝑒3𝑥2
−1+7x-5)dx =∑ ℎ. 𝑓(𝑎 + 𝑟ℎ)𝑛
𝑟=1 = lim𝑛→∞
∑ ℎ. 𝑓(−1 + 𝑟ℎ)𝑛𝑟=1
= lim𝑛→∞
∑ ℎ. (𝑛𝑟=1 𝑒3(−1+𝑟ℎ) +7(-1 +rh) -5)
= lim𝑛→∞
∑ [ℎ.𝑛𝑟=1 𝑒−3.𝑒3ℎ( 1+ 𝑒3ℎ + 𝑒6ℎ+----+𝑒3𝑛ℎ) + 7h2( 1+2+3+----+n)-12nh ]
= lim𝑛→∞
[ℎ𝑒3ℎ
𝑛𝑒3 𝑋𝑒3ℎ−1
𝑒3ℎ−1+7h2(
𝑛(𝑛+1)
2-12nh
lim [3𝑒
3𝑥3𝑛
𝑛𝑒3
𝑛→∞
𝑋( 𝑒3𝑛3
𝑛 -1)X(3ℎ
𝑒3ℎ −1)X
𝑛
3𝑋3) +
63
𝑛2X𝑛(𝑛+1)
2-12X3]
Now applying the limit we get =𝑒9−1
3𝑒3 +63
2-36
=𝑒9−1
3𝑒3 −9
2
Section – B
7 A = [1 2 22 1 22 2 1
]; A2 =[9 8 88 9 88 8 9
]
A2 -4A-5I ===[9 8 88 9 88 8 9
] - 4[1 2 22 1 22 2 1
] -5[1 0 00 1 00 0 1
] =9 − 9 8 − 8 8 − 88 − 8 9 − 9 8 − 88 − 8 8 − 8 9 − 9
=[0 0 00 0 00 0 0
]
Now A2-4A = 5I ; A2A-1 -4A-1A = 5A-1
A-4I = 5A-1; [1 2 22 1 22 2 1
] -[4 0 00 4 00 0 4
] =5A-1 ;−3 2 22 −3 22 2 −3
= 5A-1
A-1 =
−3/5 2/5 2/52/5 −3/5 2/52/5 2/5 −3/5
; IAI = 2 0 −15 1 00 1 3
=1 ; A-1 EXISTS
A-1 A = I ; A-1[2 0 −15 1 00 1 3
] = [1 0 00 1 00 0 1
]
APPLYING R1→1/2R1 ;A-1[1 0 −1/25 1 00 1 3
] =[1/2 0 0
0 1 00 0 1
]
Applying R2→R2 +(-5) R1
A-1[
1 0 −1/20 1 5/20 1 1/2
]1 0 −1/20 1 5/20 1 3
= = [1/2 0 0
−5/2 1 00 0 1
]
APPLYING R3→R3 + (-1)R2
A-1[
1 0 −1/20 1 5/20 1 1/2
] = [
1/2 0 0−5/2 1 05/2 −1 1
]
APPLYING R3→2R3
A-11 0 −1/20 1 5/20 1 1
= [1/2 0 0
−5/2 1 05 −1 1
]
APPLYING R1→R1+(1/2)R3 ; R2 → 𝑅2 + (−5/2)𝑅3
A-1[1 0 00 1 00 0 1
] = [3 −1 1
−15 6 −55 −2 2
]
A-1 =[3 −1 1
−15 6 −55 −2 2
]
8 Let∆=𝑋 + 2 𝑋 + 6 𝑋 − 1 𝑋 + 6 1 𝑋 + 2𝑋 − 1 𝑋 + 2 𝑋 + 6
APPLYINGC2→ C2-C1 AND C3 – C1
Let ∆=𝑋 + 2 4 −3𝑋 + 6 −7 −4𝑋 − 1 3 7
; APPLYING R2→R2-R1 AND R3→ 𝑅3 − 𝑅1 ; ∆=𝑋 + 2 4 −3
4 −11 −1−3 −1 10
APPLYING R2→ 𝑅2 + 𝑅3 ; ∆ = 𝑋 + 2 4 −3
1 −11 9−3 −1 10
; APPLYING R3→ 𝑅3 + (3)𝑅2;
∆ = 𝑋 + 2 4 −3
1 −11 90 −37 37
Expanding along C1
∆=( x+2 ) |−12 9−37 37
|-1|4 −3
−37 37| = ( x+2) ( -444 +333) -1(148-111) = -111x-259
x = −259
111=
−7
3
9 Let I =∫𝑠𝑖𝑛2𝑥
𝑆𝑖𝑛𝑥+𝑐𝑜𝑠𝑥
𝜋/2
0dx----------(1)
⇒ 𝐼 = ∫𝑠𝑖𝑛2(
𝜋
2−𝑥)
𝑆𝑖𝑛((𝜋
2−𝑥)+𝑐𝑜𝑠(
𝜋
2−𝑥)
𝜋/2
0dx using∫ 𝑓(𝑥) = ∫ 𝑓(𝑎 − 𝑥)𝑑𝑥
𝑎
0
𝑎
0
I =∫𝑐𝑜𝑠2𝑥
𝑆𝑖𝑛𝑥+𝑐𝑜𝑠𝑥
𝜋/2
0dx----------(2)
Adding 1 and 2
⇒2I =∫𝑑𝑥
𝑆𝑖𝑛𝑥+𝑐𝑜𝑠𝑥
𝜋/2
0=
1
√2∫
𝑑𝑥
𝑆𝑖𝑛𝑥1/√2+𝑐𝑜𝑠𝑥1/√2
𝜋/2
0=
1
√2∫
𝑑𝑥
𝑆𝑖𝑛𝑥𝑐𝑜𝑠𝜋/4+𝑐𝑜𝑠𝑥𝑠𝑖𝑛𝜋/4
𝜋/2
0
2I = 1
√2∫
𝑑𝑥
𝑆𝑖𝑛(𝜋
4+𝑥)
𝜋
20
= 1
√2∫ 𝑐𝑜𝑠𝑒𝑐 (
𝜋
4+ 𝑥) 𝑑𝑥
𝜋
20
= 1
√2[ 𝑙𝑜𝑔𝐼𝑐𝑜𝑠𝑒𝑐 (
𝜋
4+ 𝑥) − cot (
𝜋
4+ 𝑥)] 𝑤𝑖𝑡ℎ 𝑙𝑖𝑚𝑖𝑡𝑠 0 𝑎𝑛𝑑 𝜋/2
=1
√2[𝑙𝑜𝑔𝐼√2— 1𝐼 − 𝑙𝑜𝑔𝐼√2 − 1𝐼]
⇒ 𝐼 =1
2√2[InI
√2+1
√2−1I]
11. Let E1, E2 and A be the events defined as follows
E1 = selecting a coin having head on both the sides
2
1
1 1 0
1
( E )C
P
C
=2
10
There are eight coins not having heads on both the sides
8
1
2 1 0
1
5
1
2 5
1 1
1
1 1 2 2
8( E )
1 0
( A / E ) (1 ) 1
1 1( A / E )
( 2 ) 3 2
,
( ) ( / )( / A )
( ) ( / ) ( ) ( / )
21
81 0
2 8 1 91
1 0 1 0 3 2
CP
C
P
P
B y b a y e s t h e o r e m w e h a v e
P E P A EP E
P E P A E P E P A E
OR
Let p denotes probability of getting heads
Let q denotes probability of getting tails .
𝑝 = 1
2, 𝑞 = 1 −
1
2 =
1
2
Suppose the coin is tossed n times.
Let X denotes the number of times of getting heads in n trails .
0
1 1 1( ) ( ) ( ) ( ) , 0 , 1 , 2 , . . . .
2 2 2
8 0( 1 )
1 0 0
8 0( 1 ) ( X 2 ) . . . . . . P ( X n )
1 0 0
8 01 ( 0 )
1 0 0
1( 0 )
5
1 1( )
2 5
1 1( )
2 5
3 , 4 , 5 . . . . .
3
n r n r n r n r n n
r r r
n n
n
P X r C p q C C r n
P X
P X P
P X
P X
C
n
S o t h e f a i r c o i n s h o u l d b e t o s s e d f o r o r m o r e t i m
e s f o r g e t t i n g t h e r e q u i r e d p r o b a b i l i t y
12. Position vector of 𝑂𝐴 ⃗⃗ ⃗⃗ ⃗⃗ ⃗= 4𝑖̂ +𝑗̂ +2 �̂� ____
Position vector of 𝑂𝐵⃗⃗ ⃗⃗ ⃗⃗ = 5𝑖̂ + x𝑗̂ +6 �̂� ___
Position vector of 𝑂𝐶⃗⃗⃗⃗⃗⃗ =5𝑖̂ +𝑗̂ - �̂� ____
Position vector of 𝑂𝐷⃗⃗⃗⃗⃗⃗⃗ = 7𝑖̂ +4𝑗̂
𝐴𝐵⃗⃗⃗⃗ ⃗⃗ = 𝑂𝐵⃗⃗ ⃗⃗ ⃗⃗ -𝑂𝐴⃗⃗⃗⃗ ⃗⃗
= ( 5𝑖̂ + x𝑗̂ +6 �̂� ) – (4𝑖̂ +𝑗̂ +2 �̂� ) ____ ___ ____ = 𝑖̂ + ( 𝑥 − 1)𝑗 ̂+4 �̂� .
𝐴𝐶⃗⃗⃗⃗⃗⃗ = 𝑂𝐶⃗⃗⃗⃗⃗⃗ -𝑂𝐴⃗⃗⃗⃗ ⃗⃗
=( 5𝑖̂ +𝑗̂ - �̂� ) – (4𝑖̂ +𝑗̂ +2 �̂� )
= 𝑖̂ -3 �̂�
𝐴𝐷⃗⃗ ⃗⃗ ⃗⃗ = 𝑂𝐷⃗⃗⃗⃗⃗⃗⃗ -𝑂𝐴⃗⃗⃗⃗ ⃗⃗
=( 7𝑖̂ +4𝑗̂ )-( 4𝑖̂ +𝑗̂ +2 �̂� ).
= 3𝑖̂ + 3𝑗̂ -2𝑘 . The above three vectors are coplanar
____ ____
____
⇒𝐴𝐵⃗⃗⃗⃗ ⃗⃗ ∙ ( 𝐴𝐶⃗⃗⃗⃗⃗⃗ × 𝐴𝐷⃗⃗ ⃗⃗ ⃗⃗ )
=0.
⇒ 1(0 + 9)− (x − 1)(−2 + 9)+ 4(3) =0 ⇒ 9 − 7(x
−1)+ 12 = 0
⇒-7(x-1) = −21 ⇒ x − 1 = 3
∴x= 4
13. Let the equation of the line be 𝑟 = �⃗� + 𝜆�⃗⃗�_
Here, �⃗� = 4𝑖̂ +2 𝑗̂ + 2�̂� _
�⃗⃗� = 2𝑖̂ + 3 𝑗̂ +6�̂� ._
∴ Equation of the line is 𝑟 = ( 4𝑖̂ +2 𝑗̂ + 2�̂� ) +𝜆 (2𝑖̂ + 3 𝑗̂ +6�̂� )
Let L be the foot of the perpendicular and P be the required point from
which we have to find the length of the perpendicular__
P(�⃗� ) = 𝑖̂ + 2 𝑗 ̂+3�̂� ___
𝑃𝐿⃗⃗ ⃗⃗ ⃗=𝑂𝐿⃗⃗⃗⃗⃗⃗ - 𝑂𝑃⃗⃗⃗⃗ ⃗⃗ =(4𝑖̂ +2 𝑗̂+ 2�̂� )+𝜆 (2𝑖̂ +3 𝑗̂ +6�̂�)–(2𝑖̂ + 3 𝑗̂
+6�̂�)
= 3𝑖̂ -�̂� + 𝜆 (2𝑖̂ + 3 𝑗̂ +6�̂�) …………(1)
Now 𝑃𝐿⃗⃗ ⃗⃗ ⃗ ∙ �⃗⃗� = 0 (𝑃𝐿⃗⃗ ⃗⃗ ⃗ is perpendicular to�⃗⃗� )
[3𝑖̂ -�̂� + 𝜆 (2𝑖̂ + 3 𝑗̂ +6�̂�) ] ∙ (2𝑖̂ + 3 𝑗̂ +6�̂�) = 0
⇒( 3+2 𝜆 )2+3(3 𝜆 )+( -1 +6 𝜆 )6 =0
⇒6+4 𝜆+9 𝜆 -6 +36 𝜆 = 0
⇒ 49𝜆 = 0
m𝜆 = 0 𝑃𝐿⃗⃗ ⃗⃗ ⃗ =3𝑖̂-�̂�( using 1)
⇒ |𝑃𝐿⃗⃗ ⃗⃗ ⃗| = √10=================================⃗⃗ ⃗⃗
Length of the perpendicular drawn on the line from P
√10
14. sin-1(1 – x) – 2sin-1x =𝜋
2
⇒s
i
n
x − 1 4 1
⇒ 1 0 −3 = 0
3 3 −2
-1(1 – x)=𝜋
2+ 2sin x
⇒(1-x)= sin(𝜋
2+ 2sin-1x)
⇒1-x= cos(2 sin−1 𝑥)=cos[ cos−1( 1 − 2𝑥2)]⇒ 1-x= 1-2x2
⇒2x2 –x =0
⇒ 𝑥 = 0, 𝑥 =1
2
OR
2 sin−1 (3
5) − tan−1 (
17
31)
= 𝜋
4L.H.S=
cos−1(1 − 2 ×9
25)
− tan−1(17
31)
= cos−17
25− tan−1
17
31
= tan−1 24
7− tan−1 17
31
= tan−1 625
625 (
usingtan−1 𝐴 −
tan−1 𝐵 = tan−1 𝐴−𝐵
1+𝐴𝐵)
=𝜋
4 = R. H. S
Q15 y = 𝒆𝒂𝒙.𝐜𝐨𝐬 𝒃𝒙
𝒅𝒚
𝒅𝒙 = a𝒆𝒂𝒙.𝐜𝐨𝐬 𝒃𝒙 - b𝒆𝒂𝒙.𝐬𝐢𝐧 𝒃𝒙..........(1)
𝒅𝒚
𝒅𝒙 = ay -b𝒆𝒂𝒙.𝐬𝐢𝐧 𝒃𝒙
𝒅𝟐𝒚
𝒅𝒙𝟐 = a𝒅𝒚
𝒅𝒙- b(a𝒆𝒂𝒙 .𝐬𝐢𝐧 𝒃𝒙 + b𝒆𝒂𝒙.𝐜𝐨𝐬 𝒃𝒙 )
𝒅𝟐𝒚
𝒅𝒙𝟐 = a 𝒅𝒚
𝒅𝒙- (ay -
𝒅𝒚
𝒅𝒙 ) - 𝒃𝟐 y (using 1)
𝒅𝟐𝒚
𝒅𝒙𝟐 - 2a 𝒅𝒚
𝒅𝒙+ ( 𝒂𝟐 + 𝒃𝟐 )y = 0.
Hence Proved.
16. 𝒙𝒙 + 𝒙𝒚 + 𝒚𝒙 = 𝒂𝒃
Let u = 𝒙𝒙
𝐥𝐨𝐠 𝒖 = 𝒙 𝐥𝐨𝐠 𝒙
𝟏
𝒖
𝒅𝒖
𝒅𝒙 = x.
𝟏
𝒙 + 𝐥𝐨𝐠 𝒙
𝒅𝒖
𝒅𝒙 = 𝒙𝒙 (1+ 𝐥𝐨𝐠 𝒙 )
Let v =𝒙𝒚
𝐥𝐨𝐠 𝒗 = 𝒚 𝐥𝐨𝐠 𝒙
𝟏
𝒗
𝒅𝒗
𝒅𝒙 = (
𝒚
𝒙 + 𝐥𝐨𝐠 𝒙
𝒅𝒚
𝒅𝒙 )
𝒅𝒗
𝒅𝒙= 𝒙𝒚 (
𝒚
𝒙+𝐥𝐨𝐠 𝒙
𝒅𝒚
𝒅𝒙 )
Let w = 𝒚𝒙
𝐥𝐨𝐠 𝒘 = 𝒙 𝐥𝐨𝐠 𝒚
𝟏
𝒘.𝒅𝒘
𝒅𝒙= (
𝒙
𝒚
𝒅𝒚
𝒅𝒙+ 𝐥𝐨𝐠 𝒚 )
𝒅𝒘
𝒅𝒙= 𝒚𝒙 (
𝒙
𝒚
𝒅𝒚
𝒅𝒙+ 𝐥𝐨𝐠 𝒚 )
(1) Can be written as
u +v+ w = 𝒂𝒃
𝒅𝒖
𝒅𝒙+
𝒅𝒗
𝒅𝒙+
𝒅𝒘
𝒅𝒙= 𝟎
𝒙𝒙 (1+ 𝐥𝐨𝐠 𝒙 ) + 𝒙𝒚 (𝒚
𝒙+𝐥𝐨𝐠 𝒙
𝒅𝒚
𝒅𝒙) + 𝒚𝒙 (
𝒙
𝒚
𝒅𝒚
𝒅𝒙+ 𝐥𝐨𝐠 𝒚 ) = 0
𝒅𝒚
𝒅𝒙( 𝒙𝒚 𝐥𝐨𝐠 𝒙 + 𝒚𝒙.
𝒙
𝒚) = 𝒙𝒙 + 𝒙𝒙 𝐥𝐨𝐠 𝒙 𝒙𝒚 𝒚
𝒙+ 𝒚𝒙 𝐥𝐨𝐠 𝒚
𝒅𝒚
𝒅𝒙
𝒅𝒚
𝒅𝒙=
𝒙𝒙 + 𝒙𝒙 𝐥𝐨𝐠 𝒙 + 𝒚𝒙𝒚−𝟏 + 𝒚𝒙. 𝐥𝐨𝐠 𝒚
𝒙𝒚. 𝐥𝐨𝐠 𝒙 + 𝒙𝒚𝒙−𝟏
17. 𝒙 = 𝒂 𝐬𝐢𝐧 𝟐𝒕 ( 𝟏 + 𝐜𝐨𝐬 𝟐𝒕 )
𝒚 = 𝒃 𝐜𝐨𝐬 𝟐𝒕 (𝟏 − 𝐜𝐨𝐬 𝟐𝒕)
𝒅𝒙
𝒅𝒕= 𝟐𝒂 𝐜𝐨𝐬 𝟐𝒕 (𝟏 + 𝐜𝐨𝐬 𝟐𝒕) + 𝒂 𝐬𝐢𝐧 𝟐𝒕 (−𝟐 𝐬𝐢𝐧 𝟐𝒕 )
= 𝟐𝒂 𝐜𝐨𝐬 𝟐𝒕 + 𝟐𝒂 (𝐜𝐨𝐬 𝟐𝒕)𝟐 + 𝒂 𝐬𝐢𝐧 𝟐𝒕 (−𝟐 𝐬𝐢𝐧 𝟐𝒕)2
= 𝟐𝒂 𝐜𝐨𝐬 𝟐𝒕 + 𝟐𝒂 𝐜𝐨𝐬 𝟒𝒕
𝒅𝒚
𝒅𝒕= −𝟐𝒃 𝐬𝐢𝐧 𝟐𝒕 (𝟏 − 𝐜𝐨𝐬 𝟐𝒕) + 𝒃 𝐜𝐨𝐬 𝟐𝒕 (𝟐 𝐬𝐢𝐧 𝟐𝒕 )
= −𝟐𝒃 𝐬𝐢𝐧 𝟐𝒕 + 𝟐𝒃 𝐬𝐢𝐧 𝟐𝒕 𝐜𝐨𝐬 𝟐𝒕 + 𝟐𝒃 𝐜𝐨𝐬 𝟐𝒕 𝐬𝐢𝐧 𝟐𝒕
= −𝟐𝒃 𝐬𝐢𝐧 𝟐𝒕 + 𝟐𝒃 𝐬𝐢𝐧 𝟒𝒕
𝑑𝑦
𝑑𝑥=
−𝟐𝒃 𝐬𝐢𝐧 𝟐𝒕 + 𝟐𝒃 𝐬𝐢𝐧 𝟒𝒕
𝟐𝒂 𝐜𝐨𝐬 𝟐𝒕 + 𝟐𝒂 𝐜𝐨𝐬 𝟒𝒕
𝑨𝒕 𝒕 =𝝅
𝟒 ,
𝒅𝒚
𝒅𝒙 =
𝒃
𝒂
18. Let I = ∫( 𝒙+𝟑)𝒆𝒙
(𝒙+𝟓)𝟑 𝒅𝒙
= ∫( 𝒙 + 𝟓 − 𝟐)𝒆𝒙
(𝒙 + 𝟓)𝟑𝒅𝒙
= ∫( 𝒙 + 𝟓
(𝒙 + 𝟓)𝟑−
𝟐
(𝒙 + 𝟓)𝟑)𝒆𝒙 𝒅𝒙
= ∫( 𝟏
(𝒙 + 𝟓 )𝟐−
𝟐
(𝒙 + 𝟓)𝟑)𝒆𝒙𝒅𝒙
𝑻𝒉𝒊𝒔 𝒊𝒔 𝒐𝒇 𝒕𝒉𝒆 𝒇𝒐𝒓𝒎 ∫ 𝒆𝒙 (𝒇(𝒙) + 𝑓′(𝑥)) = 𝑒𝑥 . 𝒇(𝒙) + 𝑐
I = 𝒆𝒙
(𝒙+𝟓 )𝟐 + 𝒄
19 .[𝟑𝟎 12 𝟕𝟎𝟒𝟎 15 55𝟑𝟓 20 𝟕𝟓
]
2 5
1 0 0
5 0
=
3 0 2 5 1 2 1 0 0 7 0 5 0
4 0 2 5 1 5 1 0 0 5 5 5 0
3 5 2 5 2 0 1 0 0 7 5 5 0
=
5 4 5 0
5 2 5 0
6 6 2 5
X
Y
Z
Total fund collected = Rs 17325
SECTION – C
20. Let A = Q × Q, where Q is the set of rational numbers.
Given that * is the binary operation on A defined by (a, b) * (c, d) = (ac, b + ad) for (a,
b), (c, d) ∈ A.
(i)
We need to find the identity element of the operation * in A. Let(x, y) be the identity element in A.
Thus,
(a, b) * (x, y) = (x, y) * (a, b) = (a, b), for all (a, b) ∈ A
⇒(ax, b + ay) = (a, b)
⇒ax = a and b + ay =b ⇒
y = 0 and x = 1
Therefore, (1, 0) ∈ A is the identity element in A with respect to the operation *. (ii)
We need to find the invertible elements of A. Let
(p, q) be the inverse of the element (a, b) Thus,
(a, b) * (p, q) = (1, 0)
⇒(ap, b + aq) = (1, 0)
⇒ ap = 1 and b + aq = 0
⇒𝑝 =1
𝑎𝑎𝑛𝑑 𝑞 =
−𝑏
𝑎
Thus the inverse of (a, b) is (1
𝑎,
−𝑏
𝑎)
Therefore , inverses of (5, 3) and ( 1
2, 4) 𝑎𝑟𝑒 (
1
5,
−3
5) 𝑎𝑛𝑑 (2, −8)𝑟𝑒𝑠𝑝𝑒𝑐𝑡𝑖𝑣𝑒𝑙𝑦
OR
Let f: W→W be defined as
𝑓(𝑛) = {𝑛 − 1 , 𝑖𝑓 𝑛 𝑖𝑠 𝑜𝑑𝑑𝑛 + 1 , 𝑖𝑓 𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛
We need to prove that 'f' is invertible.
In order to prove that 'f' is invertible it is sufficient to prove that f is a bijection. A
function f: A→B is a one-one function or an injection, if f(x) = f(y) ⇒ x = y for all x, y ∈
A. Case i:
If x and y are odd. Let f(x) = f(y) ⇒x − 1 = y − 1
⇒x = y
Case ii:
If x and y are even, Let f(x) = f(y) ⇒x + 1 = y + 1 ⇒x = y
Thus, in both the cases, we have,
f(x) = f(y) ⇒ x = y for all x, y ∈W.
Hence f is an injection.
Let n be an arbitrary element of W.
If n is an odd whole number, there exists an even whole number n − 1 ∈ W such that f(n
− 1) = n − 1 + 1 = n.
If n is an even whole number, then there exists an odd whole number n + 1 ∈ W such that
f(n + 1) = n + 1 − 1 = n. Also, f(1) = 0 and f(0) = 1
Thus, every element of W (co-domain) has its pre-image in W (domain). So f is an onto function.
Thus, it is proved that f is an invertible function.
Thus, a function g: B→A which associates each element y ∈ B to a unique element x ∈ A
such that f(x) = y is called the inverse of f. That is, f(x) = y ⇔ g(y) = x The inverse of f is generally denoted by f -1.
Now let us find the inverse of f.
Let x, y ∈ W such that f(x) = y ⇒x + 1 = y, if x is even
x − 1 = y, if x is odd
⇒𝑥 = {𝑦 − 1 , 𝑖𝑓 𝑦 𝑖𝑠 𝑜𝑑𝑑𝑦 + 1 , 𝑖𝑓 𝑦 𝑖𝑠 𝑒𝑣𝑒𝑛
⇒𝑓−1(𝑦) = {𝑦 − 1 , 𝑖𝑓 𝑦 𝑖𝑠 𝑜𝑑𝑑𝑦 + 1 , 𝑖𝑓 𝑦 𝑖𝑠 𝑒𝑣𝑒𝑛
Therefore 𝑓−1(𝑥) = 𝑓(𝑥)
21. Consider the given equation
y= 5− x2
This equation represents a semicircle with centre at the
origin and radius =√5 𝑢𝑛𝑖𝑡𝑠
Given that the region is bounded by the above
semicircle and the line y = x − 1
Let us find the point of intersection of the
given curve meets the line y= x − 1
⇒√5 − 𝑥2 = 1x
Squaring both the sides, we have,
5 − 𝑥2 = 𝑥2 + 1 − 2𝑥 𝑠𝑜𝑙𝑣𝑖𝑛𝑔 𝑡ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑤𝑒 𝑔𝑒𝑡 𝑥 = −1 , 𝑥 = 2
When x= -1 , y= 2 and when x=2 , y=1
Consider the following figure.
Thus the intersection points are (−1,2) and (2,1)
Consider the following sketch of the bounded region.
Required Area, A=
CBSE XII | Mathematics
2
2 11
1 1 2
2 2
1 1 1
1 12
12 1
1
11
1 1
1 1
( )
5 ( 1 ) 5 ( 1 )
55 s i n
2 2 25
5 1 5 2 1s i n s i n
2 2 25 5
5 1 5 2 1s i n s i n
2 2 25 5
y y d x
x d x x d x x x d x
x x xx x
R e q u i r e d a r e a
Board Paper 2015 – All India Set – 1 Solution
22. 𝑥2 𝑑𝑦 = ( 2𝑥𝑦 + 𝑦2)𝑑𝑥
⇒𝑑𝑦
𝑑𝑥=
2𝑥𝑦+ 𝑦2
𝑥2 ……….(1)
Let y= vx 𝑑𝑦
𝑑𝑥= 𝑣 + 𝑥
𝑑𝑣
𝑑𝑥
(1)𝑏𝑒𝑐𝑜𝑚𝑒𝑠
𝑣 + 𝑥 𝑑𝑣
𝑑𝑥= 2𝑣 + 𝑣2
⇒𝑥 𝑑𝑣
𝑑𝑥= 𝑣2 + 𝑣
⇒𝑑𝑣
𝑣2+𝑣 =
𝑑𝑥
𝑥
Integrating we get
2
2
l o g l o g
1
. . . . . .
11 1 ,
2
2 ,
vc x
v
y yc x p u t t i n g v
y x x
y c x y c x
p u t t i n g y a n d x w e g e t c
y x y x w h i c h i s t h e p a r t i c u l a r s o l u t i o n
OR
(1 + 𝑥2)𝑑𝑦
𝑑𝑥= 𝑒𝑚 tan−1 𝑥 − 𝑦
𝑑𝑦
𝑑𝑥=
𝑒𝑚 tan−1 𝑥
(1 + 𝑥2)−
𝑦
(1 + 𝑥2)
𝑑𝑦
𝑑𝑥+
𝑦
(1 + 𝑥2)=
𝑒𝑚 tan−1 𝑥
(1 + 𝑥2)
𝑃 = 1
(1 + 𝑥2), 𝑄 =
𝑒𝑚 tan−1 𝑥
(1 + 𝑥2)
𝐼. 𝐹 = 𝑒tan−1 𝑥
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑡ℎ𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑖𝑠 𝑦. 𝑒tan−1 𝑥 = ∫𝑒(𝑚 +1)tan−1 𝑥
(1 + 𝑥2)𝑑𝑥
⇒𝑦. 𝑒tan−1 𝑥= 𝑒(𝑚 +1)tan−1 𝑥
𝑚+1+c
( substitutingtan−1 𝑥 = 𝑧)
Putting 𝑦 = 1 and x=1 in the general solution we get
𝑐 =𝑒(𝑚+1)
𝜋
4
(𝑚 + 1)− 𝑒
𝜋
4
Particular solution is 𝑦. 𝑒tan−1 𝑥=𝑒(𝑚 +1)tan−1 𝑥
𝑚+1+
𝑒(𝑚+1)
𝜋4
(𝑚+1)− 𝑒
𝜋
4
23
2
'
'
( ) s i n c o s ,
( ) 2 s i n c o s s i n
s i n ( 2 c o s 1 )
( ) 0
1s i n 0 c o s
2
50 ,
6
( 0 ) 1
5 1 3( )
6 4 2
( ) 1
m a x 1 0 m i n 1
f x x x
f x x x x
x x
f x
x o r x
x o r x
f
f
f
A b s o l u t e i m u m v a l u e i s a t x a n d a b s o l u t e i m u m v a l u e i s a t x
23. 𝑟 = 𝑖̂ + 𝑗̂ + �̂� +𝜎(𝑖̂ − 𝑗̂ + �̂�) ………(1)
Or 𝑥−1
1=
𝑦−1
−1=
𝑧−1
1
1 1 1
1 1 1
( , , z ) (1 , 1 , 1 )
1 , 1 , 1
x y
a b c
𝑟 = 4𝑖̂ + 2�̂� +𝜇(2𝑖̂ − 𝑗̂ + 3�̂�)
0𝑟 𝑥 − 0
2 =
𝑦 − 4
−1=
𝑧 − 2
3
2 2 2
2 2 2
( , , z ) ( 0 , 4 , 2 )
2 , 1 , 3
x y
a b c
Applying the condition of coplanarity we find that lines are coplanar.
Find the equation of required plane and the equation is 2𝑥 + 𝑦 − 𝑧 = 2
25. x – 2y ≤ 2
3x + 2y ≤ 12 −3x + 2y ≤ 3 x ≥ 0, y ≥ 0
Converting the inequations into equations, we obtain the lines x – 2y = 2…..(i)
3x + 2y = 12……(ii) −3x + 2y = 3……(iii) x = 0, y
= 0
From the graph, we get the corner points as
A(0, 5), B(3.5, 0.75), C(2, 0), D(1.5, 3.75), O(0, 0)
The values of the objective function are:
Point (x, y) Values of the objective function
Z = 5x + 2y
A(0, 5) 5 × 0 + 2 × 5 = 10
B(3.5, 0.75) 5 × 3.5 + 2 × 0.75 = 19 (Maximum)
C(2, 0) 5 × 2 + 2 × 0= 10
D(1.5, 3.75) 5 × 1.5 + 2 × 3.75 = 15
O(0, 0) 5 × 0 + 2 × 0 = 0 (Minimum)
The maximum value of Z is 19 and its minimum value is 0.
26. First six positive integers are {1, 2, 3, 4, 5, 6}
No. of ways of selecting 2 numbers from 6 numbers without replacement =6C2 = 15 X denotes the larger of the two numbers, so X can take the values 2,3, 4, 5, 6. Probability distribution of X:
X 2 3 4 5 6
p(x) 1 2 3 4 5
15 15 15 15 15
Computation of Mean and Variance:
xi P(X = xi) pixi pixi2
2 1 2 4
15 15 15
3 2 6 18
15 15 15
4 3 12 48
15 15 15
5 4 20 100
15 15 15
6 5 30 180
15 15 15
Σpixi = 70 = 14 Σ pixi2 = 350 = 70
15 3 15 3
Mean 4.67 and variance= 1.55