maths cbse 2012
TRANSCRIPT
-(1)-
Series : SMA/1
Roll No.
Code No. 65/1/1Candidates must write the Code onthe title page of the answer-book.
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Please check that this question paper contains 29 questions.
Please write down the Serial Number of the questions before attempting it.
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MATHEMATICS
[Time allowed : 3 hours] [Maximum marks : 100]
General Instructuions:
(i) All questions are compulsory.
(ii) Questions numbered 1 to 10 are very short-answer questions and carry 1 mark each.
(iii) Questions numbered 11 to 22 are short-answer questions and carry 4 marks each.
(iv) Questions numbered 23 to 29 are also short-answer questions and carry 6 marks each.
(v) Use of calculators is not allowed.
Studymate Solutions to CBSE Board Examination 2011-2012
-(2)-
STUDYmate
SECTION-A
Question numbers 1 to 10 carry 1 mark each.1. If a line has direction ratios 2, –1, –2, then what are its direction cosines?
Ans. d.r.’s = <2, –1, –2>
d.c.’s = 2 1 2
, ,3 3 3
2. Find ‘’ when the projection of 4
a i j k on 2 6 3
b i j k is 4 units.
Ans. Projection of on a b =
.
a b
b
4 = 2 6 12
4 36 9
4 = 2 18
7
28 – 18 = 2
5 =
3. Find the sum of the vectors 2 , 2 4 5 a i j k b i j k and 6 7
c i j k .
Ans. 0 4 a b c i j k
4. Evaluate:
3
2
1 dx
x
Ans. 3
3
22
1log | | dx x
x = log 3 – log 2 = 3
log2
5. Evaluate (1 ) x xdx
Ans. 32( ) x x dx
=3 5
2 2
3 52 2
C x x
=3 5
2 22 2
C3 5
x x
6. If =
5 3 8
2 0 1
1 2 3, write the minor of the element a
23.
Ans. Minor of a23
= 5 3
1 2 = 10 – 3 = 7.
-(3)-
STUDYmate
7. If 2 3 1 3 4 6
5 7 2 4 9
x
, write the value of x.
Ans.2 3 1 3 4 6
5 7 2 4 9
x
2 6 6 12 4 6
5 14 15 28 9
x
4 6 4 6
9 13 9
x
x = 13
8. Simplify: cos sin sin cos
cos sinsin cos cos sin
Ans. = 2 2
2 2
cos cos sin sin cos sin
cos sin cos sin cos sin
= 1 0
0 1
9. Write the principal value of 1 11 1cos 2sin
2 2
Ans. cos–112
– 2 sin–112
cos–1 cos3
+ 2 sin–1 sin6
23 6
23 3 3
10. Let * be a ‘binary’ operation on N given by a * b = LCM (a, b) for all a, b N. Find 5 * 7.
Ans. a * b = LCM of (a, b), a, b N
5 * 7 = LCM of (5, 7) = 35
SECTION-B
Question numbers 11 to 22 carry 4 marks each.
11. If (cos x)y = (cos y)x, find dydx
.
OR
If sin y = x sin (a + y), prove that 2sin ( )
sin
dy a y
dx a
Ans. Given (cos x)y = (cos y)x, taking logarithms on the two sides, we have
log (cos x)y = log (cos y)x
-(4)-
STUDYmate
or y log (cos x) = x log (cos y)
Differentiating both sides w.r.t. x, we get
1 1( sin ) log(cos ) ( sin ) log(cos ).1
cos cos
dy dyy x x x y y
x dx y dx
log (cos x) dydx
+ x tan ydydx
= log (cos y) + y tan x
log(cos ) tandy
x x ydx
= log (cos y) + y tan x
log(cos ) tanlog(cos ) tan
dy y y xdx x x y
OR
Ans. sin y = x sin (a + y)
sin
sin( )y
a y = x
2
sin( )cos sin cos( )sin ( )
a y y y a y dx
a y dy
sin( a y y
2
)
sin ( )
dx
a y dy
2
sinsin ( )
a dx
a y dy
2sin ( )sin
a y dya dx
12. How many times must a man toss a fair coin, so that the probability of having at least one head is more
than 80%?
Ans. Let the man throws the coin ‘n’ times.
Probability of getting a head in a single throw, 1
2p
Probability of not getting a head in a single throw, 1
2q
Given: Probability that the man gets atleast one head is more than 80%.
i.e. P(atleast 1 Head) > 80%
i.e. 1 – P (no head) 80
100
i.e.0 0
0
1 1 81
2 2 10
nnC
i.e.1 4
12 5
n
i.e.1 1
2 5
n
i.e.1 1
2 5
n
i.e. 2n > 5
-(5)-
STUDYmate
i.e. n = 3 or more.
The man must throw the coin atleast 3 times.
13. Find the Vector and Cartesian equations of the line passing through the point (1, 2, –4) and perpendicular
to the two lines 8 19 10
3 16 7
x y z
and 15 29 5
3 8 5
x y z
.
Ans. Any line through (1, 2, –4) can be written as
1 2 4
x y za b c
...(i)
(i) is at right angles to given lines with d.n. < 3, –16, 7 > and < 3, 8, –5 >
if 3a – 16b + 7c = 0 ...(ii)
and 3a + 8b – 5c = 0 ...(iii)
By cross-multiplication, we have
80 56 21 15 24 48
a b c
or24 36 72
a b c
or2 3 6
a b c...(iv)
From (i) and (iv), we find that required line is
1 2 4
2 3 6
x y z
In vector form, this line can be written as
2 4 2 3 6
r i j k i j k ( d.n. of the line are < 2, 3, 6 > and line passes through < 1, 2, –4 >)
14. If , , a b c are three vectors such that 5, 12, 13
a b c , and 0
a b c , find the value of
. . . a b b c c a .
Ans. 0 a b c
( ) · ( ) 0·0 a b c a b c
i.e.2
0 a b c
i.e. 2 2 2| | | | | | 2( · · · ) 0 a b c a b b c c a
i.e. 52 + 122 + 132 + 2( · · · ) 0 a b b c c a
i.e. 338 2( · · · ) 0 a b b c c a
i.e. · · · 169 a b b c c a
15. Solve the following differential equation:
2 22 2 0 dy
x xy ydx
Ans. 2 22 2 dy
x xy ydx
-(6)-
STUDYmate
2
2
22
dy xy ydx x
2
2 2
22 2
dy xy ydx x x
2
2
12
dy y ydx x x
...(i)
Put y = vx and dy dv
v xdx dx
in (i) we get,
v dv
x vdx
212
v
2 22
dv dxv x
2 1
2 log | | C2 1
vx
2
log | | C
xv
2
log | | C
x
xy
2log | | C
xx
y; (y 0)
16. Find the particular solution of the following differential equation;
2 2 2 21 , dy
x y x ydx
given that y = 1 when x = 0
Ans.2 2 2 21
dyx y x y
dx
2 2 2 2(1 ) (1 ) (1 )(1 ) dy
x y x x ydx
22
(1 )1
dy
x dxy
31tan C
3
xy x
Now, when x = 0, y = 1
3
1 0tan 1 0 C
3 C
4
Solution is 3
1tan3 4
xy x
17. Evaluate : sin sin 2 sin 3 x x x dx
OR
Evaluate : 2
2
(1 )(1 ) dxx x
Ans. Now, sin x sin 2x sin 3x
-(7)-
STUDYmate
=12
{2sin 3x sin x} sin 2x
=1
2{cos (2x) – cos (4x)} sin 2x
=1
4{2 sin 2x cos 2x – 2 cos 4x sin 2x}
=14
{sin 4 x – (sin 6x – sin 2x)}
sin sin 2 sin3 x x xdx
=1
(sin 4 sin 6 sin 2 )4
x x x dx
=1 cos4 cos6 cos2
C4 4 6 2
x x x
=cos4 cos6 cos2
C16 24 8
x x x
OR
Ans. Let 2 2
2 A B C(1 )(1 ) 1 1
x
x x x x ...(i)
Multiplying both sides by (1 – x) (1 + x2), we have
2 = A (1 + x2) + (Bx + C) (1 – x)
2 = x2 (A – B) + x (B – C) + (A + C) ...(ii)
Equating coefficients on the two sides of (ii), we get
A – B = 0, B – C = 0, A + C = 2
A = B = C = 1
2 2
2 1 1(1 )(1 ) 1 1
x
x x x x
2
2(1 )(1 ) dx
x x = 2
1 1
1 1
x
dx dxx x
= 2 2
log |1 | 1 2 11 2 1 1
x xdx dx
x x
= –log |x – 1| + 1
2log (1 + x2) + tan–1 x + C.
18. Find the point on the curve y = x3 – 11x + 5 at which the equation of tangent is y = x – 11.
OR
Using differentials, find the approximate value of 49.5
Ans. Given curve is y = x3 – 11x + 5 ...(i)
Given line is y = x – 11 ...(ii)
Slope of line (ii) = 1 ( (ii) is of the form y = mx + b)
From (i),dy
dx = 3x2 – 111
Since, slope of tangent = 1
-(8)-
STUDYmate
dydx
= 1, i.e., when 3x2 – 11 = 1
x = 2 ( 3x2 – 11 = 1 3x2 = 12 x2 = 4)
When x = 2, then
from (i), y = 23 – 11 × 2 + 5 = –9
When x = –2, then
from (i), y = (–2)3 – 11 (–2) + 5 = 19
Thus, we find that at the points (2, –9) and (–2, 19), the slope of tangent is 1.
OR
Ans. Let ( ) f x x so that 1
( ) .2
f xx
Now ( ) ( ) ( ) f x x f x xf x
2
xx x x
x
Taking x = 49 and x = 0.5, we obtain
49.50.5 0.5
49 7 7 0.036142 49
49.5 7.036
19. If y = (tan–1x)2, show that 2 2 22 1( 1) 2 ( 1) 2 x y x x y .
Ans. Given y = (tan–1 x)2 we get
12
12(tan )
1
dy
xdx x
or (1 + x2) y1 = 2 tan–1 x
Again differentiating w.r.t. x we get
22 1 2
2(1 ) (0 2 )
1
x y y x
x
2 2 22 1( 1) 2 ( 1) 2 x y x x y
20. Using properties of determinants, prove that
2
b c q r y z a p x
c a r p z x b q y
a b p q x y c r z
Ans. LHS =
b c q r y z
c a r p z x
a b p q x y
Interchanging rows and columns, we get,
b c c a a b
q r r p p q
y z z x x y
Operating C3 C
3 – C
1 – C
2, we have
-(9)-
STUDYmate
2
2
2
b c c a a b b c c a c
q r r p p q q r r p r
y z z x x y y z z x z(Take out –2 from C
3 i.e., operate C
3
12
C3)
= 2
b c c a c
q r r p r
y z z x zNow, Operate C
1 C
1 – C
3 and C
2 C
2 – C
3
= 2b a c
q p r
y x z
, operate C1 C
2
= 2
a b c
p q r
x y z
= 2
a p x
b q y
c r z
(by interchanging rows and columns again)
21. Prove that 1 costan , ,
1 sin 4 2 2 2
x xx
x
OR
Prove that 1 1 18 3 36sin sin cos
17 5 85
Ans.2
sin 2sin coscos 2 4 2 4 2
1 sin1 cos 2cos
2 4 2
x xx
xxx
x
sin4 2
tan4 2
cos4 2
xx
x
1 cos
tan .1 sin 4 2
x x
x Hence proved
OR
Ans. Let x = 1 8
sin17
and y =
1 3sin
5
sin x = 8
17 and sin y =
35
cos x = 21 sin x and cos y = 21 sin y
cos x = 15
17 and cos y =
4
5Now, cos (x + y) = cos x cos y – sin x sin y
=15 4 8 317 5 17 5
-(10)-
STUDYmate
=3685
x + y = cos–1 36
85
1 1 18 3 36sin sin cos
17 5 85 Hence Proved
22. Let A {3} and B {1} . Consider the function f : A A B defined by 2
( ) .3
xf x
x Show that
f is one-one and onto and hence find f –1.
Ans.2
( )3
x
f xx
Let 1 2, x x A
Let 1 2( ) ( )f x f x
i.e.1 2
1 2
2 2
3 3
x x
x x
i.e. (x1 – 2) (x
2 – 3) = (x
1 – 3) (x
2 – 2)
i.e. x1x
2 – 3x
1 – 2x
2 + 6 = x
1x
2 – 2x
1 – 3x
2 + 6
i.e. –x1 = – x
2
i.e. x1 = x
2
f is one - one. ... (i)
Also2
3
x
yx
i.e. y (x – 3) = (x – 2)
xy – 3y = x – 2
x (y – 1) = 3y – 2
3 2
1
y
xy
f is onto, y 1, y B ... (ii)
from (i) and (ii), f is bijective
it is invertible.
For inverse : Interchanging x and y in y = 2
3
x
x, we get
y = 3 2
1
x
x or f –1(x) =
3 2
1
x
x
SECTION-C
Question numbers 23 to 29 carry 6 marks each.23. Find the equation of the plane determined by the points A (3, –1, 2), B (5, 2, 4) and C (–1, –1, 6) and
hence find the distance between the plane and the point P (6, 5, 9).
Ans. Equation of plane in three point form is
-(11)-
STUDYmate
3 1 2
5 3 2 1 4 2 0
1 3 1 1 6 2
x y z
i.e.,
3 1 2
2 3 2 0
4 0 4
x y z
i.e. (x – 3) (12) – (y + 1) (8 + 8) + (z – 2) (12) = 0
i.e., 3(x – 3) – 4 (y + 1) + 3 (z – 2) = 0
i.e., 3x – 4y + 3z – 19 = 0 is the equation of the required plane.
and d = 18 20 27 19
9 16 9
d = 6 34
34 34
d = 3 34
17.
24. Of the students in a college, it is known that 60% reside in hostel and 40% are day scholars (no residing
in hostel). Previous year results report that 30% of all students who reside in hostel attain ‘A’ grade and
20% of day scholars attain ‘A’ grade in their annual examination. At the end of the year, one student is
chosen at random from the college and he has an ‘A’ grade, what is the probability that the student is a
hostlier?
Ans. Let E1 : ‘A student is residing in hostel’
and E2 : ‘A student is a day scholar’
then E1 and E
2 are mutually exclusive and exhaustive. Moreover,
P(E1) =
60 3100 5
and P(E2) =
40 2100 5
Let E : Student attains ‘A’ grade;
then P(E/E1) =
30 3
100 10 and P(E/E
2) =
20 2
100 10
Required probability = P (E1/E)
=1 1
1 1 2 2
P(E/E )P(E )P(E/E )P(E )+P(E/E )P(E ) [By Baye’s Theorem]
=
3 39 910 5
3 3 2 2 9 4 1310 5 10 5
25. A manufacturer produces nuts and bolts. It takes 1 hour of work on machine A and 3 hours on machine
B to produce a package of nuts. It takes 3 hours on machine A and 1 hour on machine B to produce a
package of bolts. He earns a profit of ` 17.50 per package on nuts and ` 7 per package of bolts. How
many packages of each should be produced each day so as to maximize his profits if he operates his
machines for at the most 12 hours a day? Form the above as a linear programming problem and solve it
graphically.
-(12)-
STUDYmate
Ans. Let the manufacturer produces x Packages of nuts and y Packages of bolts each day. We construct the
following table:
Item Number of packages Total Time on machine A Total Time on machine B Profits
Nuts 1 hrs 3 hrs 17.50
Bolts 3 hrs 1 hr 7
Total 3 hrs 3 hrs 17.50 7
x x x x
y y y y
x y x y x y
We have to maximize profit, i.e.,
P = 17.50x + 7y = 35
72
x y ...(i)
subject to the constraints
x + 3y 12 ...(ii)
Y
OX
D(0,12)
B(0,4) E(3,3)
C(4,0)
A(12,0)
3x + y 12 ...(iii)
x, y 0 ...(iv)
First of all, we locate the region represented by (2),
(3) and (4). For this, we consider the line x + 3y =
12, which passes through A (12, 0) and B (0, 4) and
the line 3x + y = 12, which passes through the points
C (4, 0) and D (0, 12). The two lines meet at E (3, 3).
The feasible region is shown shaded in the figure.
Note that O (0, 0) lies in this region.
The corner points, which are to be examined for optimum solution are C (4, 0), E (3, 3) and B (0, 4).
At C (4, 0) P = 35
4 7 0 702
At E (3, 3) P = 35
3 7 3 73.52
At B (0, 4) P = 35
0 7 4 282
Hence the profit is maximum equal to ̀ 73.50 when 3 packages of each of nuts and bolts are manufactured.
26. Prove that
/4
0
( tan cot ) 2.2
x x dx
OR
Evaluate : 3
2
1
(2 5 ) x x dx as a limit of a sum.
Ans. Let I = sin cos
cos sin
x xdx
x x
=sin cos
sin cos
x xdx
x x
Now let (sin x – cos x) = t ...(i)
(cos x + sin x)dx = dt
Squaring (i), sin2 x + cos2 x – 2 sin x cos x = t2
21
sin cos2
tx x
-(13)-
STUDYmate
When x = 0, t = 0 – 1 = –1
When x = 4
, t = 1 1
02 2
I =
0
21
21 dt
t
= 01
12 sin
t
= 1 12 sin (0) sin 1
= 2 02
= 22
OR
Ans. I =
32
1
(2 5 ) x x dx
a = 1, b = 3 h = 3 1
b a
n n nh = 2
Now, f(x) = 2x2 + 5x
f(a) = f(1) = 2(1)2 + 5(1) = 7
f(a + h) = f(1 + h) = 2(1 + h)2 + 5(1 + h)
= 2(1 + h2 + 2h) + 5 + 5h
= 2h2 + 9h + 7
f(a + 2h) = f(1 + 2h) = 2(1 + 2h)2 + 5(1 + 2h)
= 2(1 + 4h2 + 4h) + 5 + 10h
= 8h2 + 18h + 7
... ... ... ...
... ... ... ...
1 1 1 f a n h f n h = 22 1 1 5 1 1 n h n h
= 2 22 1 1 2 ( 1) 5 5 1 n h h n h n
= 2(n – 1)2 h2 + 9h (n – 1) + 7
Since ( )b
a
f x dx = 0
( ) ( ) ( 2 ) ... 1
hLt h f a f a h f a h f a n h
I = 2 2 2 2
07 {7 9 2 } {7 18 8 } ... {7 9 ( 1) 2( 1) }
h
Lt h h h h h h n n h
= 22 2 2
07 9 (1 2 ... 1) 2 1 2 ... 1
hLt h n h n h n
= 2
0
( 1) ( )( 1)(2 1)7 9 2
2 6
h
n n n n nLt h n h h
= 0
9 ( ) ( )(2 1)7
2 3
h
hn hn h hn hn h hnLt h hn
-(14)-
STUDYmate
= 9 2(2 0) 2(2 0)(4 0)
7 22 3
= 16
14 183
= 42 54 16
3
= 112
3
27. Using the method of integration, find the area of the region bounded by the lines 3x – 2y + 1 = 0, 2x + 3y
– 21 = 0 and x – 5y + 9 = 0.
Ans. L1: 3x – 2y + 1 = 0 y =
3 1
2
x1
30
0.5 0
x
y
L2: 2x + 3y – 21 = 0 y =
21 23 x 0 10.5
7 0
x
y
L3: x – 5y + 9 = 0 y =
9
5
x 0 9
1.8 0
x
y
L1
L2
L3
(–9, 0) (10.5, 0)
(3, 5)
AB
C
(1,2)
(6,3)
Point of Intersection:
L1 and L
2 :
3 1 21 22 3
x x
9x + 3 = 42 – 4x
13x = 39
x = 3
y = 5
L2 and L
3 :
3 1 9
2 5
x x
15x + 5 = 2x + 18
13x = 19
x = 1
y = 2
L2 and L
3 :
21 2 9
3 5
x x
-(15)-
STUDYmate
105 – 10x = 3x + 27
78 = 13x
x = 6
y = 3
A = (1, 2) ; B = (6, 3) ; C = (3, 5)
Thus, required area: A =
3 6 6
( 1) ( 2) ( 3)1 3 1
Line Line Lineydx ydx ydx
=
3 6 6
1 3 1
1 1 1(3 1) (21 2 ) ( 9)
2 3 5 x dx x dx x dx
=
3 62 2623
1 1
1 3 1 121 9
2 2 3 5 2
x xx x x x
= 1 27 3 1 1 13 1 126 36 63 9 18 54 9
2 2 2 3 5 2
= 1 1 114 36
2 3 5
12525
2
=25 25 38 25 13
7 12 192 2 2 2
sq. units.
28. Show that the height of a closed right circular cylinder of given surface and maximum volume, is equal
to the diameter of its base.
Ans. Let r be the radius of the circular base, h the height and S, the total surface area of a right circular
cylinder, then S = 2r2 + 2rh is given to be a constant.
Let V be the volume of the cylinder with the above dimensions, then
V = r2h = 2
2 2S 2(S 2 )
2 2
r rr r
r[ S = 2r2 + 2rh, h =
2S 2
2
r
r]
V = 3S S,0
2 2
r
r r ...(i) [ h = 2S 2
2
r
r > 0, r <
S2
]
Differentiating (i), twice w.r.t. r, we get
2V S3
2
dr
dr
and2
2
V6
dr
dr
Now Vd
dr exists at all points in
S0,
2
and
V0
d
dr 2S
3 02 r r2 =
S
6
r = S6
[ 0 < r < S2
]
Also, 2
2S/(6 )
V S6 0
6
r
d
dr
V has a local maximum value at r = S6
-(16)-
STUDYmate
Since V is continuous in S0,
2
and has only one extremum at S S
0,6 2
, therefore, V is
absolutely maximum for r = S6
.
When r = S6
, then h = 2
SS 2
S 2 2S 662 6 SS
26
rr
= i.e., h = S2
6 = 2 radius = diameter..
So, volume is maximum when the height is equal to the diameter.
29. Using matrices, solve the following system of linear equations:
x – y + 2z = 7
3x + 4y – 5z = – 5
2x – y + 3z = 12
OR
Using elementary operations, find the inverse of the following matrix.
1 1 2
1 2 3
3 1 1
Ans. The given system can be written as AX = B,
where
1 1 2
3 4 5
2 1 3
A
x
X y
z
and
7
5
12
B
Here
1 1 2
| | 3 4 5
2 1 3
A
= 1 (12 – 5) – (– 1) (9 + 10) + 2(–3 – 8)
= 7 + 19 – 22 = 4 0
A–1 exists
Therefore, the given system is consistent and has a unique solution given by
X = 1 1A B = ( . A) B
| |
adjA
-(17)-
STUDYmate
=
7 19 11 71
1 1 1 54
3 11 7 12
t
7 1 3 71
19 1 11 54
11 1 7 12
49 5 361
133 5 1324
77 5 84
8 21
4 14
12 3
x
y
z
x = 2 , y = 1, z = 3
OR
Ans. A = IA
1 1 2 1 0 0
1 2 3 0 1 0 A
3 1 1 0 0 1
R
1 R2
1 2 3 0 1 0
1 1 2 1 0 0 A
3 1 1 0 0 1
R
2 R
2 + R
1 ; R
3 R
3 – 3R
1
1 2 3 0 1 0
0 3 5 1 1 0 A
0 5 8 0 3 1
R
3 R
3 + 2R
2
1 2 3 0 1 0
0 3 5 1 1 0 A
0 1 2 2 1 1
R
2 R3
1 2 3 0 1 0
0 1 2 2 1 1 A
0 3 5 1 1 0
R
1 R
1 – 2R
2
R3 R
3 – 3R
2
1 0 1 4 3 2
0 1 2 2 1 1 A
0 1 1 5 4 3
R
3 – R
3
-(18)-
STUDYmate
1 0 1 4 3 2
0 1 2 2 1 1 A
0 0 1 5 4 3
R
1 R
1 + R
3
R2 R
2 – 2R
3
1 0 0 1 1 1
0 1 0 8 7 5 A
0 0 1 5 4 3
1
1 1 1
A 8 7 5
5 4 3
× · × · × · × · ×
-(19)-
STUDYmate
SECTION-A
Question numbers 1 to 10 carry 1 mark each.9. Find the sum of the following vectors:
ˆˆ ˆ ˆ ˆ ˆ2 , 2 3 , 2 3 a i j b i j c i k
Ans. a b c ˆˆ ˆ ˆ ˆ ˆ( 2 ) (2 3 ) (2 3 ) i j i j i k
ˆˆ ˆ5 5 3 i j k
10. If
5 3 8
2 0 1 ,
1 2 3
write the cofactor of the element a32
.
Ans. 3 232
5 8M ( 1) (5 16) 11
2 1
SECTION-B
Question numbers 11 to 22 carry 4 marks each.19. Using properties of determinants, prove the following
3 3 3
1 1 1
( )( )( )( ) a b c a b b c c a a b c
a b c
Ans.3 3 3
1 1 1
a b c
a b c
Applying C1 C
1 – C
3(Given)
C2 C
2 – C
3
3 3 3 3 3
0 0 1
a c b c c
a c b c c
= 2 2 2 2 3
0 0 1
( )( ) ( ) ( )
a c b c c
a c a ac c b c b bc c c
Code No. 65/1/2
Studymate Solutions to CBSE Board Examination 2011-2012
UNCOMMON QUESTIONS ONLY
Series : SMA/1
-(20)-
STUDYmate
=
2 2 2 2 3
0 0 1
( ) ( ) 1 1
( ) ( )
a c b c c
a ac c b bc c c
= 2 2 2 2( ) ( ) ( ) (a c b c b bc c a ac c Expanding along R1
= (a – c) (b – c) {(b2 – a2) + (bc – ac)}
= (a – c) (b – c) {(b2 – a2) + c(b – a)}
= (a – c) (b – c) (b – a) {(b + a) + c}
= (a – b) (b – c) (c – a) (a + b + c)
= RHS. Hence proved
20. If y = 3 cos (log x) + 4 sin (log x), show that
22
20
d y dyx x y
dxdx
Ans. Given y = 3 cos (log x) + 4 sin (log x) … (i)
Differentiating w.r.t. x, we get
dy
dx = – 3 sin (log x)
1
x + 4 cos (log x)
1
xMultiplying by x, we get
xy1 = – 3 sin (log x) + 4 cos (log x) … (ii)
Again differentiating w.r.t. x, we obtain
xy2 + y
1 . 1 = – 3 cos (log x)
1
x – 4 sin (log x)
1
xMultiplying throughout by x, we have
x2y2 + xy
1 = – (3 cos (log x) + 4 sin (log x))
or x2y2 + xy
1 = – y (using (i))
or x2y2 + xy
1 + y = 0
21. Find the equation of the line passing through the point (–1, 3, –2) and perpendicular to the lines
1 2 3
x y z and
2 1 1
3 2 5
x y z
Ans. Let the equation of line passing through (–1, 3, –2) be
( 1) 3 ( 2)
x y z
a b c... (i) where (a, b, c) are dr’s of the line
Since, the required line is to the lines:
1 2 3
x y z and
2 1 1
3 2 5
x y z
a + 2b + 3c = 0 and – 3a + 2b + 5c = 0 (using a1a
2 + b
1b
2 + c
1c
2 = 0)
Solving, 2 3 3 1 1 2
2 5 5 3 3 2
a b c
4 14 8
a b c
-(21)-
STUDYmate
or2 7 4
a b c
a = 2, b = –7, c = 4
Putting the values of a, b and c in equation (i) we get
1 3 2
2 7 4
x y z
or1 3 2
2 7 4
x y z
[Note: Using is not compulsory]
22. Find the particular solution of the following differential equation
( 1) 2 1; 0 ydyx e y
dx when x = 0
Ans. Given differential equation is
( 1) 2 1 ydyx e
dx...(i)
12 1
y
dy dx
xe
,12
y
y
e dy dx
xe
Integrating, we obtain
12
y
y
e dy dxC
xe
–12
y
y
e dy dxC
xe
– log |2 – ey| = log | x + 1| + C
log (x + 1) (2 – ey) = –C
|(x + 1) (2 – ey) | = e–c
(x + 1) (2 – ey) = ± e–c = A (say)
(x + 1) (2 – ey) = A ... (ii)
Also, when x = 0, y = 0,
(0 + 1) (2 – e0) = A
1(2 – 1) = A
A = 1
Substituting this value of A in (ii), we obtain, the required particular solution as
(x + 1) (2 – ey) = 1.
SECTION-C
Question numbers 23 to 29 carry 6 marks each.28. A girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she
gets 1, 2, 3 or 4, she tosses a coin two times and notes the number of heads obtained. If she obtained
exactly two heads, what is the probability that she threw 1, 2, 3 or 4 with the die?
Ans. Let E1 : ‘1, 2, 3, or 4 is shown on die’, and
-(22)-
STUDYmate
E2 : ‘5 or 6 is shown on die’,
then E1 and E
2 are mutually exclusive and exhaustive. Moreover,
1
4 2P(E )
6 3
and 2
2 1P(E )
6 3
Let E : ‘exactly one head shows up’,
then1
EP P
E
(exactly one head shows up when coin is tossed twice) = P({HT, TH}) =
2 1
4 2
and2
EP P
E
(exactly one head shows up when coin is tossed thrice)
= P ({HTT, THT, TTH})
= 38
Required probability = 1EP
E
11
1 21 2
EP P(E )
E
E EP P(E ) P P(E )
E E
(using Bayes Theorem)
1 2 18 82 3 3
1 2 3 1 1 1 8 3 112 3 8 3 3 8
29. Using the method of integration, find the area of the region bounded by the following lines:
3x – y – 3 = 0, 2x + y – 12 = 0, x – 2y – 1 = 0
Ans. 3x – y – 3 = 0 ... (i)
2x + y – 12 = 0 ... (ii)
x – 2y – 1 = 0 ... (iii)
Solving (i) and (ii) 5x – 15 = 0 i.e. x = 3 i.e. y = 9 – 3 = 6
Point of intersection (3, 6)
Solving (ii) and (iii)
2 12 0
2 4 2 0
5 10 0
2
x y
x y
y
y
x = 5
Point of intersection (5, 2)
-(23)-
STUDYmate
Solving (iii) and (i)
3 3 0
3 6 3 0
0 and 15 0
x y
x y
y xy
Point of intersection is (1, 0)
0
y
xA(1, 0) E(3,0)
B(3, 6)
C(5, 2)
( )i( )ii
( )iii
D(5,0)
Required area
= Area of region ABCA
= area of region ABCDA – Area of ACDA
3 5 5
( ) ( ) ( )1 3 1
i ii iiiy dx y dx y dx
3 5 5
1 3 1
( 1)(3 3) ( 2 12)
2
xx dx x dx dx
3 5
2 252
31 1
33 12
2 4 2
x x xx x x
27 3 25 5 1 19 3 ( 25 60) ( 9 36)
2 2 4 2 4 2
24 24 46 [35 27]
2 4 2
= 10 sq. units
× · × · × · × · ×
-(24)-
STUDYmate
SECTION-A
Question numbers 1 to 10 carry 1 mark each.9. Find the sum of the following vectors:
ˆ ˆ ˆˆ ˆ ˆ ˆ3 , 2 , 2 3 2 a i k b j k c i j k
Ans. ˆˆ ˆ3 2 a b c i j k
10. If
1 2 3
2 0 1 ,
5 3 8
write the minor of the element a22
.
Ans. 22
1 31 8 3 5 8 15 7
5 8 M
SECTION-B
Question numbers 11 to 22 carry 4 marks each.19. Using properties of determinants, prove the following
1 1 1
1 1 1
1 1 1
a
b ab bc ca abc
c
Ans. Taking out factors a, b, c common from R1, R
2 and R
3, we get
L.H.S. =
1 1 11
1 1 11
1 1 11
a a a
abcb b b
c c c
Applying R1 R
1 + R
2 + R
3, we have
1 1 1 1 1 1 1 1 11 1 1
1 1 11
1 1 11
a b c a b c a b c
abcb b b
c c c
Code No. 65/1/3
Studymate Solutions to CBSE Board Examination 2011-2012
UNCOMMON QUESTIONS ONLY
Series : SMA/1
-(25)-
STUDYmate
1 1 1
1 1 1 1 1 11 1
1 1 11
abca b c b b b
c c c
Now applying C2 C
2 – C
1, C
3 C
3 – C
1, we get
1 0 0
1 1 1 11 1 0
10 1
abca b c b
c
1 1 11 [1(1 0)]
abca b c
1 1 11 R.H.S.
abc abc bc ca aba b c
20. If y = sin–1 x, show that 2
22
(1 ) 0. d y dy
x xdxdx
Ans. We have y = sin–1x. Then
2
1
(1 )
dy
dx x
or 21 1 dy
xdx
So 21 ). 0
d dyx
dx dx
2
2 22
1 ). . 1 ) 0d y dy d
x xdx dxdx
22
2 2
21 ). . 0
2 1
d y dy xx
dxdx x
or
22
2 21 ). – . 0
1
d y dy xx
dxdx x
Hence2
22
(1 ) 0d y dy
x xdxdx
21. Find the particular solution of the following differential equation
( 2)( 2); 1 dy
xy x y ydx
when x = 1
Ans. Given differential equation is
( 2) ( 2)dy
xy x ydx
...(i)
-(26)-
STUDYmate
or2
,2
y xdy dx
y x
integrating, we obtain
2
2
y x
dy dx Cy x
or2 2
1 12
dy dx Cy x
or y – 2 log | y + 2| = x + 2 log | x| + C ...(ii)
But, (1, –1) lies on this curve, therefore,
–1 – 2 log|–1 + 2| = 1 + 2 log 1 + C
–1 – 2 log 1 = 1 + 2 log 1 + C
C = –2 ( log 1 = 0)
Hence, from (ii), we find the required curve as
y – 2 log |y + 2| = x + 2 log |x| – 2
or y = x + 2 log | x (y + 2)| – 2.
22. Find the equation of a line passing through the point P (2, –1, 3) and perpendicular to the lines
ˆ ˆˆ ˆ ˆ ˆ( ) (2 2 ) r i j k i j k and ˆ ˆˆ ˆ ˆ ˆ(2 3 ) ( 2 2 ).
r i j k i j k
Ans. Let, the equation be
1 1 1
x x y y z z
a b cIt passes through (2, –1, 3)
2 1 3
x y z
a b c... (i)
Also (i) is to ˆ ˆˆ ˆ ˆ ˆ( ) (2 2 ) r i j k i j k
2a – 2b + c = 0 ... (ii)
Also (i) is to ˆ ˆˆ ˆ ˆ ˆ(2 3 ) µ( 2 2 ) r i j k i j k
a + 2b + 2c = 0 ... (iii)
Solving (ii) and (iii)
4 2 4 1 4 2
a b c
6 3 6
a b c
2 1 2
a b c
Taking a = 2, b = 1, c = –2
Required equation lie is
2 1 3
2 1 2
x y z
or ˆ ˆˆ ˆ ˆ ˆ(2 3 ) (2 2 ) r i j k i j k
-(27)-
STUDYmate
SECTION-C
Question numbers 23 to 29 carry 6 marks each.28. Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls. Two balls are transferred
at random from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be red
in colour. Find the probability that the transferred balls were both black.
Ans. Bag I contains 3 red, 4 black balls
Bag II contains 4 red, 5 black balls
2 balls transferred from Bag I to Bag II
Case 1 : Event A = Both black balls transferred
Case 2 : Event B = One black balls and One red ball transferred
Case 3 : Event C = Both red balls transferred.
4
27
2
2( ) ,
7
CP A
C
3 41 17
2
4( ) ,
7
C CP B
C
32
72
1( )
7
CP C
C
Let event E = 1 ball drawn from bag II is red.
4
( / )11
P E A (Since bag II contains 4R, 7B balls now)
5( / )
11P E B (Bag II contains 5R, 6B balls now)
and6
( / )11
P E C (Bag II contains 6R, 5B balls now)
Required Probability = P (A/E)
( ) ( / )
( ) ( / ) ( ) ( / ) ( ) ( / )
P A P E A
P A P E A P B P E B P C P E C
2 4
7 112 4 4 5 1 67 11 7 77 7 11
8 8 4
8 20 6 34 17
29. Using the method of integration, find the area of the region bounded by the following lines
5x – 2y – 10 = 0, x + y – 9 = 0, 2x – 5y – 4 = 0.
Ans. 5x – 2y – 10 = 0 ... (i)
x + y – 9 = 0 ... (ii)
2x – 5y – 4 = 0. ... (iii)
Solving (i) and (ii)
5 2 10 0
9 0
7 28 0
x y
x y
x
x = 4 ; y = 5; Point of intersection is (4, 5) 0 E(4,0)
y
xA(2, 0)
B(4, 5)
C(7, 2)
( )i
( )ii
( )iii
D(7,0)
-(28)-
STUDYmate
Solving (i) and (iii)
10 4 20 0
10 25 20 0
21 0
x y
x y
y
i.e., y = 0; x = 2, therefore, Point of intersection is (2, 0)
Solving (ii) and (iii)
2 2 18 0
2 5 4 0
7 14 0
x y
x y
y
i.e., y = 2; x = 7, therefore, Point of intersection is (7, 2)
Required Area = Area of Region ABCA
= Area of Region ABCDA – Area of Region ACDA
4 7 7
( ) ( ) ( )2 4 2
i ii iiiy dx y dx y dx
4 7 7
2 4 2
5 10 2 4(9 )
2 5
x xdx x dx dx
4 7 72 2 2
2 4 2
5 45 9
4 2 5 5
x x x xx x
80 20 49 16 49 28 4 820 10 63 36
4 4 2 2 5 5 5 5
21 4[5] [38.5 28]
5 5
= 43.5 – 28 – 5
= 10.5
21
2 sq. units
× · × · × · × · ×