5 maths cbse_2012-13_12th_20-03-13
Transcript of 5 maths cbse_2012-13_12th_20-03-13
-(1)-
Series : SKS/1
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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 2012-2013
-(2)-
STUDYmate
SECTION-A
Question numbers 1 to 10 carry 1 mark each.
1. Write the principal value of 1 1 1tan 1 cos
2
Ans. tan–1 (1) + cos–1 1
2
=1 1tan tan cos cos
4 3
=1cos cos
4 3
=2
4 3
=3 8
12
=
11
12
2. Write the value of 1 1tan 2tan
5
Ans.1 1
tan 2tan5
= 12
12
5tan tan1
15
=1 2 5
tan tan24
=1 5
tan tan12
=5
12
3. Find the value of a if 2 1 5
2 3 0 13
a b a c
a b c d
Ans. a – b = –1
2a – b = 0
–a = –1
a = 1
4. If 1 1 4 1
3 2 1 3
x x
x x
, then write the value of x.
Ans. (x + 1) (x + 2) – (x – 3) (x – 1) = 12 + 1
(x2 + 3x + 2) – (x2 – 4x + 3) = 13
7x – 1 = 13
7x = 14
x = 2
-(3)-
STUDYmate
5. If 9 1 4 1 2 1
2 1 3 0 4 9A
, then find the martix A.
Ans. A = 9 1 4 1 2 1
2 1 3 0 4 9
A = 8 3 5
2 3 6
6. Write the degree of the differential equation
2 423
20
d y dyx x
dxdx
Ans. Degree = 2
7. If ˆ ˆ ˆ2a xi j zk
and ˆ ˆ ˆ3b i yj k
are two equal vectors, then write the value of x + y + z.
Ans. ˆ ˆ ˆ ˆ ˆ ˆ2 3a xi j zk i yj k
x = 3 y = –2 z = –1
x + y + z = 0
8. If a unit vector a makes angles
3
with ,
4i
with j and an acute angle with k , then find
the value of .
Ans.2 2 2cos cos cos 1
3 4
21 1cos
4 2 = 1
cos2 = 1
4
cos = 1
2, as is an acute angle.
= 3
9. Find the Cartesian equation of the line which passes through the point (–2, 4, –5) and is
parallel to the line 3 4 8
3 5 6
x y z .
Ans. The required equation of the line
2 4 5
3 5 6
x y z
10. The amount of pollution content added in air in a city due to x-diesel vehicles is given by P(x)
= 0.005x3 + 0.02x2 + 30x. Find the marginal increase in pollution content when 3 dieselvehicles are added and write which value is indicated in the above question.
Ans. P(x) = 0.005x3 + 0.02x2 + 30x
P'(x) = 3 × 0.005x2 + 2(0.02)x + 30
P'(3) = 3 × 0.005 × 9 + 2(0.02)(3) + 30 = 30.255
-(4)-
STUDYmate
SECTION-B
Question numbers 11 to 22 carry 4 marks each.
11. Show that the function f in A = 2
3 defined as f(x) =
4 3
6 4
x
x
is one-one and onto. Hence
find f–1.
Ans.
4 3( )
6 4
xf x
x
Let
1 2 1 22
, ;3
x x A x x
Consider, f (x1) = f (x
2)
i.e.
21
1 2
4 34 3
6 4 6 4
xx
x x
i.e. (4x1 + 3) (6x
2 – 4) = (4x
2 + 3) (6x
1 – 4)
i.e. 24x1x
2 – 16x
1 + 18x
2 – 12 = 24 x
1x
2 – 16x
2 + 18 x
1 – 12
i.e. –34x1 = –34x
2
i.e. x1 = x
2
f is one - one.
Let
4 3
6 4
xy
x
6xy – 4y = 4x + 3
i.e. (6y – 4) x = 3+ 4y
3 4
6 4
yx
y
4 2
. .6 3
y i e
2
3y
f is onto
Since f is one-one and onto
1 3 4
6 4
yx f y
y
i.e. y (6x – 4) = 4x + 3
1 3 4
6 4
xf x
x
12. Find the value of the following:
21 1
2 2
1 2 1tan sin cos
2 1 1
x y
x y
, |x| < 1, y > 0 and xy < 1.
OR
Prove that: 1 1 11 1 1
tan tan tan2 5 8 4
Ans.2
1 12 2
1 2 1 1tan sin cos
2 1 2 1
x y
x y
-(5)-
STUDYmate
= 1 11 1
tan (2tan ) (2tan )2 2
x y
= tan (tan–1 x + tan–1 y)
= 1
x y
xy
OR
Ans. 1 1
1 112 5tan tan
1 8110
1 1 1[using tan tan tan , 1]1
x yx y xy
xy
= 1 17 1tan tan
9 8
=1
7 19 8tan
71
72
=1 56 9
tan72 7
=1 65
tan65
= tan–1 (1)
=4
13. Using properties of determinants, prove the following:
2
22 3
2
1
1 1
1
x x
x x x
x x
Ans. Operating C1 C
1 + C
2 + C
3, we obtain
L.H.S. =
2 2
2
2 2
1
1 1
1 1
x x x x
x x x
x x x
Take out 1 + x + x2 from C1
= (1 + x + x2)
2
2
1
1 1
1 1
x x
x
x
Operate R2 R
2 – R
1 : R
3 R
3 – R
1
= (1 + x + x2)
2
2
1
0 1 (1 )
0 ( 1) 1
x x
x x x
x x x
Expand with C1
= (1 + x + x2) 2
1 (1 )
(1 ) (1 )
x x x
x x x ,
Take out 1 – x from C1 and same from C
2
-(6)-
STUDYmate
= (1 + x + x2) (1 – x)21
1 x
x x
= (1 + x + x2) (1 – x)2 (1 + x + x2)
= {(1 + x + x2) (1 – x)}2
= (1 – x3)2 = R.H.S.
14. Differentiate the following function with respect to x:
(log x)x + xlog x
Ans. Let y = (log x)x + xlog x,
Thenlog{(log ) } x xdy d
x xdx dx
=log(log ) ( )x xd d
x xdx dx
=log(log ) {( log(log )} (log log )x xd d
x x x x x xdx dx
( ) ( log )
v vd du u v u
dx dx
= log1 1 1(log ) log(log ) 2(log )
logx xx x x x x
x x x
2(log log ) ((log ) )
d dx x x
dx dx
=log1 log
(log ) log(log ) 2log
x xxx x x
x x
15. If 2 2log[ ],y x x a show that 2
2 22
( ) 0.d y dy
x a xdxdx
Ans. 2 2log [ ]y x x a
2 2 2 2
1 21
2
dy x
dx x x a x a
2 2
12 2 2 2
____1____
x a xy
x x a x a
2 21( ) 1y x a
2 2 1
2 2 2
1.2 .( ) 0
2
x yy x a
x a
y2 (x2 + a2) + xy
1 = 0
16. Show that the function ( ) | 3|, ,f x x x is continuous but not differentiable at x = 3.
OR
If x = a sin t and y = a (cot t + log tan t/2), find 2
2.
d y
dx
Ans. f(x) = |x – 3|; x = 3
LHD = f '(3–) =
00
3 3limhh
f h f
h
=00
|3 3| |3 3|limhh
h
h
-(7)-
STUDYmate
=00
| |limhh
h
h
= 00
limhh
h
h
= –1
RH D = f '(3+) =
00
3 3limhh
f h f
h
=00
|3 3| |3 3|limhh
h
h
=00
| |limhh
h
h
= 00
limhh
h
h
= 1
As LHD RHD f is not differentiable.
Again, LHL =3
lim | 3|x
x
(at x = 3)
= 0lim|3 3|h
h
= 0lim| | 0h
h
RHL =3
lim | 3|x
x
(at x = 3)
= 0lim|3 3|h
h
= 0lim| | 0h
h
Since, LHL = RHL
f is continuous at x = 3.
OR
Ans. y = cos log tan2
ta t
; x = a sin t
dy
dt= 21 1
sin sec2 2
tan2
ta t
t
= 2
cot1 12sin2
sin cos2 2
t
a tt t
= 1
sinsin
a tt
= 21 sin
sin
ta
t
= 2cos
sin
ta
t
dx
dt= a cos t
-(8)-
STUDYmate
2cossincos
a tdy tdx a t
= cot t
2
2
d y d dy
dx dxdx
= 2cosecdt
tdx
= 2
2 1 cosec .seccosec .
cos
t tt
a t a
= 2cosec
cos
t
a t
17. Evaluate :
sin( )
sin(
x adx
x a
OR
Evaluate :
2
5 2
1 2 3
xdx
x x
Ans.sin( )
Isin( )
x a
dxx a
sin( 2 )I
sin
t a
dxt
sin cos2 cos sin2sin sin
t a t a
dt dtt t
= t cos 2a – sin 2a . cot dt
= t cos 2a – sin 2a ln | sin t | + C
= (x + a) cos 2a – sin 2a ln | sin (x + a) | + C
OR
Ans. 2
5 2I
1 2 3
xdx
x x
25 2 A (1 2 3 ) B d
x x xdx
5x – 2 = A (2 + 6x) + B
Comparing the co-efficients we get
5 = 6A
A = 56
– 2 = 2A + B
B = – 2A – 2
= 5
23
= 113
-(9)-
STUDYmate
21
2 2
5 11(2 6 )
6 3I1 2 3 1 2 3
II
x
dx dxx x x x
1 2
5 2 6I
6 1 2 3
xdx
x x
Let 1 + 2x + 3x2 = t
(2 + 6x)dx = dt
1
5 5I ln
6 6
dtt
t
= 2
1
5ln|1 2 3 | C
6 x x
2 2
11I
3 3 2 1
dx
x x
2
112 1 1 193 3 9 9
dx
x x
2
119 1 2
3 9
dx
x
12
111 1 3. tan C9 2 2
3 3
x
= I = I1 – I
2
= 2 15 11 3 1
ln|1 2 3 | tan C6 3 2 2
xx x
18. Evaluate :
2
2 2( 4) ( 9)
xdx
x x
Ans. I = 2
2 2
1 2 4 9 4 9
2 4 9
xdx
x x
= 2 2 2
2 2 2 2 2 2 2 2
1 4 1 9 1 4 1 13
2 2 2 24 9 4 9 4 9 4 9
x x x dxdx dx dx dx
x x x x x x x x
= 1 1
2 2
1 1 1 1 1 3 1 1. tan . tan .
2 3 3 2 2 2 2 5 4 9
x xdx
x x
= 1 1 1 11 1 13 13
tan tan tan tan6 3 4 2 20 2 30 3
x x x xC
= 1 11 13 1 13
tan tan3 6 30 2 4 20
x xC
= 1 13 2
tan tan5 3 5 2
x xC
-(10)-
STUDYmate
19. Evaluate : (| | | 2| | 4|)x x x dx
Ans. 4
0
| | | 2| | 4|x x x dx I
0 < x < 4 |x| = + x
0 < x < 2 |x – 2| = – (x – 2)
0 < x < 4 |x – 2| = + (x – 2)
0 < x < 4 |x – 4| = – (x – 4)
I = 4 4 4 4
0 0 0 0
2 ( 2) 4x x x x
4 2 4 42 2 2 2
0 0 2 0
2 2 42 2 2 2
x x x xx x x
= (8) + [(4 – 2) – 0] + [(8 – 8) – (2 – 4)] + [16 – 16/2]
= 8 + 2 + 2 + 16/2
= 20
20. If a and b are two vectors such that
| | | |,a b a then prove that vector
2a b is
perpendicular to vector .b
Ans. 2 2| | | |a b a
2 2 2| | 2 . | | | |a a b b a
2 . . 0a b b b
(2 ). 0a b b
2a b b
21. Find the coordinates of the point, where the line
2 1 2
3 4 2
x y z intersects the plane
x – y + z – 5 = 0. Also, find the angle between the line and the plane.
OR
Find the vector equation of the plane which contains the line of intersection of the planes
ˆ ˆ ˆ. ( 2 3 ) 4 0r i j k and
ˆ ˆ ˆ. (2 ) 5 0r i j k and which is perpendicular to the plane
ˆ ˆ ˆ. (5 3 6 ) 8 0.r i j k
Ans.2 1 2
3 4 2
x y z [x = 2 + 2, y = 4 – 1; z = 2 + 2]
This point lies on the plane x – y + z – 5 = 0
(3 + 2) – (4 – 1) + (2 + 2) – 5 = 0
3 + 2 – 4 + 1 + 2 + 2 – 5 = 0
= 0
-(11)-
STUDYmate
Point of Intersection = [2, –1, 2]
sin = 3 1 4 1 2 1
9 16 4 1 1 1
sin = 3 4 2
29 3
sin = 1
87
= 1 1
sin87
OR
Ans. Equation of plane through 1 &
2
(x + 2y + 3z – 4) + k(2x + y – z + 5) = 0
x (1 + 2k) + y(2 + k) + z (3 – k) – 4 + 5k = 0
This plane is perpendicular to 5x + 3y – 6z + 8 = 0
5 (1 + 2k) + 3 (2 + k) – 6 (3 – k) = 0
5 + 10 k + 6 + 3k – 18 + 6k = 0
19 k – 7 = 0
719
k
Required plane is will be
14 7 71 2 3
19 19 19
x y z
854 0
19
33x + 45y + 50z + 9 = 0
22. A speak truth in 60% of the cases, while B in 90% of the cases. In what percent of cases are
they likely to contradict each other in stating the same fact? In the cases of contradiction doyou think, the statement of B will carry more weight as he speaks truth in more number of
cases than A?
Ans. P (AT) =
60
100P (A
C) =
40
100
P (BT) =
90
100P (B
C) =
10
100
P (contradiction) 60 10 400 90
100 100 100 100
600 3600 4200
10000 10000
% P (contradiction) = 42 %
In general parlace we understand who speaks truth, will have a better say and accepted
more.
-(12)-
STUDYmate
SECTION-C
Question numbers 23 to 29 carry 6 marks each.
23. A school wants to award its students for the values of Honesty, Regularity and Hard work witha total cash award of ` 6,000. Three times the award money for Hardwork added to that given
for honesty amounts to ` 11,000. The award money given for Honesty and Hard work together
is double the one given for Regularity. Represent the above situation algebraically and findthe award money for each value, using matrix method. Apart from these values, namely,
Honesty, Regularity and Hard work, suggest one more value which the school must include
for award.
Ans. Let Honesty award = ` x
Regularity award = ` y
Hard work award = ` z
According to question;
x + y + z = 6000
3z + x = 11000
x + z = 2y
x + y + z = 6000
x + 0y + 3z = 11000
x – 2y + z = 0
We can represent these equations using Matrices as:
1 1 1 6000
1 0 3 11000
1 2 1 0
x
y
z
AX = B
X = A–1B
Now |A| =
1 1 1
1 0 3
1 2 1
= 6 + 2 – 2 = 6 0
A–1 exists
adj. A =
6 3 3
2 0 2
2 3 1
A–1 = 1adj. A
|A|
=
6 3 31
2 0 26
2 3 1
X = A–1B =
6 3 3 60001
2 0 2 110006
2 3 1 0
=
36000 33000 01
12000 0 06
12000 33000 0
-(13)-
STUDYmate
=
30001
120006
21000
=
500
2000
3500
Award for honesty = ` 500
Award for regularity = ` 2000
Award for hardwork = 3500
One more value can be punctuality.
24. Show that the height of the cylinder of maximum value, that can be inscribed in a sphere of
radius R is 3R
.3
Also find the maximum volume.
OR
Find the equation of the normal at a point on the curve x2 = 4y which passes through thepoint(1, 2). Also find the equation of the corresponding tangent.
Ans. Let x be the radius of the base and h be the height of the cylinder ABCD which is inscried ina sphere of radius R. It is obvious that for maximum volumke the axis of the cylinder must be
along OA2 = OL2 + AL2
AL = 2 2R x
Let V be the volume of the cylinder. Then, D M C
O
A L B
a
V = (AL)2 × LM
V = (AL)2 × 2 (OL)
V = (R2 – x2) × 2x
V = 2 (R2x – x3)
2
2 22
V V2 ( 3 ) 12
d dR x and x
dx dx
For maximum or minimum values of V, we must have
V0
d
dx 2(R2 – 3x2) = 0
x = 3
R
Clearly,
2
2
3
V12 0
3
Rx
d R
dx
Hence, V is maximum when 3
R
x and hence LM = 2x = 2
3
R.
OR
Differentiating x2 = 4y with respect to x, we get
2
dy x
dx
Let (h, k) be the coordinates of the point of contact of the normal to the curve x2 = 4y. Now, slope
of the tangent at (h, k) is given by
-(14)-
STUDYmate
( , ) 2 h k
dy h
dx
Hence, slope of the normal at (h, k) = 2
h
Therefore, the equation of normal at (h, k) is
2( )
y k x h
h… (i)
Since it passes through the point (1, 2), we have
22 (1 )
k h
h
or2
2 (1 ) k hh
… (ii)
Since (h, k) lies on the curve x2 = 4y, we have
h2 = 4k … (iii)
From (ii) and (iii), we have h = 2 and k = 1. Substituting the values of h and k in (i), we get the
required equation of normal as
21 ( 2)
2
y x
or x + y = 3
25. Using integration, find the area bounded by the curve x2 = 4y and the line x = 4y – 2.
OR
Using integration, find the area of the region enclosed between the two circles
x2 + y2 = 4 and (x – 2)2 + y2 = 4.
Ans. The equations of the given curves are
x2 = 4y … (i)
and x = 4y – 2 … (ii)
Eq. (i) represents a parabola with vertex at the originand axis along positive direction of y-axis. Eq. (ii)
represents a straight line which meets the
coordinate axes at (– 2, 0) and (2, 0) respectively.
To find the points of intersection of the given parabola
and the line, we solve (i) and (ii) simultaneously.Solving the two equations simultaneously.
We obtain that the points of intersection of the givenparabola and the line are (2, 1) and (–1, 1/4).
The region whose area is to be found out is shaded in figure.
Let us slice the shaded region into vertical strips. We find that each vertical strip runs from
parabola (i) to the line (ii). So, the approximating rectangle shown in figure has
Width = x, Length = (y2 – y
1), and the Area (y
2 – y
1) x.
Since the approximating rectangle can move from x = – 1 to x = 2.
Required area A is given by
2
2 1
1
A ( )
y y dx
2 2
1
2A
4 4
x xdx
2 1
2
2 1
P( , ) and Q( , ) lie on (ii) and (i) respec.
2and
4 4
x y x y
x xy y
-(15)-
STUDYmate
22 31A
8 2 12
x xx
4 2 8 1 1 1 9
A8 2 12 8 2 12 8
sq. units
OR
Ans. Equations of the given circles are
x2 + y2 = 4 ...(1)
and (x – 2)2 + y2 = 4 ...(2)
Equation (1) is a circle with centre O at the origin
and radius 2. Equation (2) is a circle with centre C(2, 0) and radius 2. Solving equations (1) and (2), we
have
(x –2)2 + y2 = x2 + y2
or x2 – 4x + 4 + y2 = x2 + y2
or x = 1 which gives y = 3
Thus, the points of intersection of the given circles are A 1, 3 and A' 1, 3 as shown in
the figure.
Required area of the enclosed region OACA'O between circles
= 2 [area of the region ODCAO] (Why?)
= 2 [area of the region ODAO + area of the region DCAD]
=
1 2
0 1
2 ydx ydx
= 1 2
2 2
0 1
2 4 2 4x dx x dx
(Why?)
= 1 2
2 1 2 1
10
1 1 2 1 12 2 4 2 4sin 2 4 4sin
2 2 2 2 2 2
x xx x x x
= 1 2
2 1 2 1
10
22 4 2 4sin 4 4sin
2 2
x xx x x x
= 1 1 1 11 13 4sin 4sin 1 4sin 1 3 4sin
2 2
= 3 4 4 4 3 46 2 2 6
= 2 2
3 2 2 33 3
= 8
2 33
26. Show that the differential equation 2yex/y dx + (y – 2x ex/y) dy = 0 is homogeneous. Find the
particular solution of this differential equation, given that x = 0 when y = 1.
Ans. The given differential equation can be written as
2
2
x
y
x
y
dx xe y
dyye
...(i)
-(16)-
STUDYmate
Let F(x, y) = 2
2
x
y
x
y
xe y
ye
Then F(x, y) = 02F ,
2
x
y
x
y
xe yx y
ye
Thus, F(x, y) is a homogeneous function of degree zero. Therefore, the given differentialequation is a homogeneous differential equation.
To solve it, we make the substitution
x = vy ...(ii)
Differentiating equation (ii) with respect to y, we get
dx dvv y
dy dy
Substituting the value of x and dx
dy in equation (i), we get
2 1
2
v
v
dv vev y
dy e
or2 1
2
v
v
dv vey v
dy e
or1
2 v
dvy
dy e
or 2evdv = dy
y
or 2 .v dye dv
y
or 2ev = –log |y| + C
and replacing v by x
y , we get
2 log| | C
x
ye y ...(iii)
Substituting x = 0 and y = 1 in equation (iii), we get
2e0 + log |1| = C = C = 2
Substituting the value of C in equation (iii), we get
2 log| | 2
x
ye y
which is particular solution of the given differential equation.
27. Find the vector equation of the plane passing thorugh the three points with position vectors
ˆ ˆ ˆ ˆ2 , 2i j k i j k and ˆ ˆ ˆ2 .i j k Also find the coordinates of the point of intersection of this
plane and the line ˆ ˆ ˆ ˆ ˆ ˆ3 (2 2 ).r i j k i j k
Ans.
( ).( ) 0r a b c
ˆ ˆ ˆ
ˆ ˆ ˆ2 1 1 3 5
1 2 1
i j k
n b c i j k
-(17)-
STUDYmate
ˆ ˆ ˆ ˆ ˆ ˆ( 2 ) . ( 3 5 ) 0r i j k i j k
Equation of plane is
ˆ ˆ ˆ.(3 5 ) 14r i j k
or ˆ ˆ ˆ.(3 5 ) 14r i j k
Now ˆ ˆ ˆ ˆ ˆ ˆ3 (2 2 )r i j k i j k
This line intersects the plane
[(3 + 2l) ˆ ˆ ˆ ˆ ˆ ˆ[(3 2 ) ( 1 2 ) ( 1 ) ]. (3 5 ) 14)i j k i j k
– = 1
= – 1
Point of intersection is
ˆ ˆ ˆ2 .i j k
28. A cooperative society of farmers has 50 hectares of land to grow two crops A and B. The profits
from crops A and B per hectare are estimated as ` 10,500 and ` 9,000 respectively. To control
weeds, a liquid herbicide has to be used for crops A and B at the rate of 20 litres and 10 litresper hectare, respectively. Further not more than 800 litres of herbicide should be used in
order to protect fish and wildlife using a pond which collects drainage from this land. Keeping
in mind that the protection of fish and other wildlife is more important than earning profit,how much land should be allocated to each crop so as to maximize the total profit? Form an
LPP from the above and solve it graphically. Do you agree with the message that the protection
of wildlife is utmost necessary to preserve the balance in environment?
Ans. Let x hectares for crop A and y hectares for crop B be allocated.
A B
x
1050020 L/H
y
900010 L/H
50 hectaresProfit800 litres atmost
Y
C50
O A40
X
B(30, 20)
50x y + = 50
2x y + = 80
-(18)-
STUDYmate
A.T.Q.
We need to maximise profit given by
Z = 10500x + 9000y subject to
x + y 50 (Area constraint)
20x + 10y 800 (Herbicide constraint)
x, y 0 (Non-negative constraint)
After plotting graph, the corner points are
Corner Point Value of Z
O(0, 0)A(40, 0)B(30, 20)C(0, 50)
Z = 0Z = 420000Z = 315000 + 180000 = 495000Z = 450000
Maximum
30 hectares for crop A and 20 hectares for crop B.
29. Assume that the chances of a patient having a heart attack is 40%. Assuming that a meditationand yoga course reduces the risk of heart attack by 30% and prescription of certain drug
reduces its chance by 25%. At a time a patient can choose any one of the two options with
equal probabilities. It is given that after going through one of the two options, the patientselected at random suffers a heart attack. Find the probability that the patient followed a
course of meditation and yoga. Interpret the result and state which of the above stated methods
if more beneficial for the patient.
Ans. Let E1: the patient follows meditation and yoga
E2: tha patient uses drug
then E1 and E
2 are mutually exclusive and P(E
1) = P(E
2) =
1
2
Let E: the selected patient suffers a heart attack
then P(E/E1) =
40 30 281
100 100 100
and P(E/E2) =
40 25 301
100 100 100
Hence, P (patient who suffers heart attack follows meditation and yoga).
= P(E1/E) =
1 1
1 1 2 2
P E/E P E
P E/E P E P E/E P E (Using Bayes Theorem)
=
28 128 14100 2
28 1 30 1 58 29100 2 100 2
× · × · × · × · ×
-(19)-
STUDYmate
SECTION-A
9. Write the degree of the differential equation
4 2
23 0.
dy d yx
dx dx
Ans.
4 2
23 0
dy d yx
dx dx
Degree = 1
16. P speaks truth in 70% of the cases and Q in 80% of the cases. In what percent of cases are
they likely to agree in stating the same fact?
Do you think, when they agree, means both are speaking truth?
Ans. P(PT) =
70
100P(P
L) =
30
100
P(QT) =
80
100P(Q
L) =
20
100
P(agreement) = (PL & Q
T) or (P
L & Q
L)
= 5600 600
10000 10000
= 6200
62%10000
No, agreement does not mean they are speaking truth.
18. If ˆ ˆ ˆa i j k and
ˆ ˆ ˆ,b j k k find a vector
,c such that
a c b and
. 3a c
Ans. Let ˆ ˆ ˆc xi yj zk
ˆ ˆ ˆ
1 1 1
i j k
a c
x y z
ˆ ˆ ˆ( ) ( ) ( )i z y j z x k y x
Now,
a c b (given)
ˆ ˆ( )b j k
z – y = 0, x – z = 1, y – x = – 1
y = z and x – y = 1 ... (i)
Again, . 3a c x + y + z = 3
x + y + y = 3
x + 2y = 3 ... (ii)
Code No. 65/1/2
Studymate Solutions to CBSE Board Examination 2012-2013
UNCOMMON QUESTIONS ONLY
Series : SKS/1
-(20)-
STUDYmate
Solving (i) and (ii)
2 3
1
3 2/3
x y
x y
y
2
3y z
x = 1 + y = 5
3
So, 5 2 2ˆ ˆ ˆ.
3 3 3c i j k
19. Evaluate : 3
1
| 1| | 2| | 3| .x x x dx
Ans. I = 3
1
| 1| | 2| | 3|x x x dy
2 3
1 2
| 1 2 3 ( 1 2 3 )x x x dx x x x dx
2 3
1 2
(4 )x dx x dx
2 32 2
1 2
42 2
x xx
1 96 4 2
2 2
7 5
62 2
5 5
52 2
20. Evaluate :
2
2 2
1
( 4)( 25)
xdx
x x
Ans. I
2
2 2
1
( 4)( 25)
xdx
x x2
2 2 2 2
1
( 4) ( 25) ( 4) 25
x A B
x x x x
x = 0
1
25 4 14 25 100
A BA B ... (i)
x = 1
2
5 26 5 26
A B
1
5 26 65
A B
13A + 5
2 B = 1
-(21)-
STUDYmate
26A + 5B = 2 ... (ii)
1 8
,7 7
A B
I = 2 2
1 8
7 74 25
dx dx
x x
1 11 1 8 1. tan . tan
7 2 2 7 5 5
x xc
28. Show that the differential equation
sin sin 0dy y y
x x ydx x x is homogeneous. Find the
particular solution of this differential equation, given that x = 1 when
.2
y
Ans. sin sindy y y
x y xdx x x
1
sin
dy yydx xx
Let F (x, y) = 1
sin
yyxx
F (x, y) =
1
sin
y
yxx
1
sin
yyxx
= ° F (x, y)
Given d.e. is homogeneous
Let y = vx
dy dv
v xdx dx
Now
sin sin 0dv
x v x v x vx vdx
2sin sin sin 0
dvvx v x v x vx v
dx
sindx
v dvx
– cos v = – ln |x| + c
– cos
y
x = – ln |x| + c
1,
2x y
0 = c
cos ln| |y
xx
cos( / )y xx e
cos logy
xx
-(22)-
STUDYmate
29. Find the vector equation of the plane determined by the points A (3, –1, 2), B (5, 2, 4) andC (–1, –1, 6). Also find the distance of Point P (6, 5, 9) from this plane.
Ans.
3 1 2
5 3 2 1 4 2 0
1 3 1 1 6 2
x y z
3 1 2
2 3 2 0
4 0 4
x y z
(x – 3) (12 – 0) – (y + 1) (8 + 8) + (z – 2) (0 + 12) = 0
12x – 36 – 16y – 16 + 12 z – 24 = 0
12x – 16y + 12z – 52 – 24 = 0
12x – 16y + 12z – 76 = 0
3x – 4y + 3z – 19 = 0
ˆ ˆ ˆ.(3 4 3 ) 19r i j k
Now, distance of plane from Points P (6, 5, 9)
Distance of plane from (6, 5, 9)
(3 6) ( 4)(5) (3)(9) 19
34
18 20 27 19 6
34 34
× · × · × · × · ×
-(23)-
STUDYmate
SECTION-A
2. Write a unit vector in the direction of the sum of vectors ˆ ˆ ˆ2 2a i j k and
ˆ ˆ 3 .b i j k
Ans. ˆ ˆ ˆ0 5c a b i j k
2 2| | 1 5 26c a b
^ ˆ 5 ˆ
2 2
ia b k
b b
11. A speaks truth 75% of the cases, while B in 90 % of the cases. In what percent of cases arethey likely to contradict each other in stating the same fact? Do you think that statement of
B is true?
Ans. P(AT) =
75100
P(AL) =
25100
P(BT) =
90100
P(BL) =
10100
P (contradiction) = 75 10 25 90
100 100 100 100
= 750 2250
10000
= 3000
10000
= 3
10
% of P (Contradiction) = 3
10 × 100 = 30%
14. Evaluate : 5
2
| 2| | 3| || 5| .x x x dx
Ans. 2 < x < 5 |x – 2| + (x – 2)
2 < x < 3 |x – 3| – (x – 3)
3 < x < 5 |x – 3| x – 3
2 < x < 5 |x – 5| 5 – x
5 3 5 5
2 2 3 2
( 2) (3 ) ( 3) (5 )x x x x
5 3 5 52 2 2 2
2 2 3 2
2 3 3 52 2
x x x xx x x x
x x
25 9 25 9 2510 2 4 9 (6 2) 15 9 25 (10 2)
2 2 2 2 2
Code No. 65/1/3
Studymate Solutions to CBSE Board Examination 2012-2013
UNCOMMON QUESTIONS ONLY
Series : SKS/1
-(24)-
STUDYmate
5 9 5 9 252 4 8
2 2 2 2 2
9 7 4 9 39
2 2 2 2 2
15. Evaluate :
2
2 2
2 1
( 4)
xdx
x x
Ans. I =
2
2 2
2 1
( 4)
xdx
x x
Put x2 = t for PFD
2 1
( 4) 4
t A B
t t t t
2t + 1= A (t + 4) + Bt
Comp : t = 2 = A + B where 1 7
4 4A B
Const 1 = 4 A
I = 2 2
1 7
4 4 4
dx dx
x x
11 7 1tan
4 4 2 2
xC
x
11 7tan
4 8 2
xC
x
25. Find the coordinates of the point where the line through (3, –4, –5) and (2, –3, 1) crosses theplane, passing through the point (2, 2, 1) (3, 0, 1) and (4, –1, 0).
Ans. Equation of line
3 4 5
1 1 6
x y z
General Point [x = – + 3, y = + 4, z = 6 – 5]
Equation of plane
2 2 1
3 2 0 2 1 1 0
4 2 1 2 0 1
x y z
2 2 1
1 2 0 0
2 3 1
x y z
(x – 2) {2 – 0} – (y – 2) {– 1 – 0} + {z – 1} {– 3 + 4} = 0
2x – 4 + y – 2 + z – 1 = 0
2x + y + z – 7 = 0 _______________ ,
General point of line is a point on ,
2 ( – + 3) + ( – 4) + (6 – 5) – 7 = 0
–2 + 6 + – 4 + 6 – 5 – 7 = 0
5 – 10 = 0
= 2
Point of intersection of line and plane (1, –2, 7)
× · × · × · × · ×