Model Question Paper: 1
MATHS
CBSE  XII 
Q 1 If A = 
, B = 1 3 6 , verify that (AB)/ = B/ A/ .
Ans.1 Given A =
B = [1 3 6 ]
therefore AB =
[ 1 3 6] =
therefore (AB)^{/} =
also B^{/} = 

and A^{/} = [ 2 4 5] 
Q 2 Find Karl Pearson's coefficient of correlation between x and y
for the following
X 
2 
10 
8 
6 
8 
Y 
4 
6 
7 
10 
6 
Ans. 2
Prepare the table , here assumed me Ans are a = 8 and b = 7
X_{I} 
Y_{I} 
dX_{I}=(X_{I}a) 
dy_{I} =(y_{I}b) 
(dX_{I })^{2} 
(dy_{I })^{2} 
(dX_{I} ) (dy_{I }) 
2 
4 
6 
3 
36 
9 
18 
10 
6 
2 
1 
4 
1 
2 
8 
7 
0 
0 
0 
0 
0 
6 
10 
2 
3 
4 
9 
6 
8 
6 
0 
1 
0 
1 
0 
åX_{I}= 34 
åy_{I} =33 
ådX_{I} = 6 
ådy_{I} = 2 
å(dX_{I})^{2}=44 
å(dy_{I})^{2}=20 
å(dX_{I} dy_{I}) 
r = [ å(dX_{I} dy_{I})  ådX_{I}
ådy_{I} / n ] / Ö{å(dX_{I})2
 å(dX_{I})2/n}{å(dy_{I})2  å(dy_{I})2
/n}
r = {10  (6) (2)/5} / Ö {44  36/5}{20  4/5}
r = 0.28
Q 3 Find the magnitude of the vector a x b , if a = 2i + k and
b = i +j + k.
Ans.3 a x b = 

 i  j + 2k 
Therefore magnitude is given by Ö (1)^{2} + (1)^{2}
+ 2^{2} = Ö 6
Q 4 Verify LaGrange's mean value theorem for the function
f(x) = x + 1/x in the interval [1,3]
Ans.4
The given function is f(x) = (x^{2} + 1 ) / x .
It being a rational function such that x is not equal to 0, so continuous
in [1,3].
Also f (x) = [1  1/x^{2}] = [x^{2}1] / x^{2},which
clearly exists for all values of x in ]1,3[.
So f(x) is differentiable in ]1,3[.
So there must exist some c such that
F/( c ) = f(3)  f(1) / 3  1 = 10/3  2 / 2 = 2/3
Therefore c2  1 / c2 = 2/3 which gives c = Ö3.
Q 5 Find the centre and radius of the sphere
5x^{2} + 5y^{2} +5z^{2} +10x  6y + 8z + 5 = 0
Ans. 5
5x^{2} + 5y^{2} +5z^{2} +10x  6y + 8z + 5 = 0
in standard form
x^{2} + y^{2} +z^{2} +2x  6/5y + 8/5z + 1 = 0
therefore u = 1, v = 3/5, w = 4/5 and d = 1
therefore centre is ( u,v,w) = ( 1, 3/5, 4/5)
also radius is given by Ö u^{2} + v^{2} + w^{2}
 d = Ö1^{2} + (3/5)^{2} +(4/5)^{2}  =
1.
Q 6 Discuss the continuity of the function f(x) = { 3x  2, when x
£ 0
x + 1, when x 0 } at x = 0.
Ans. 6
Clearly f(0) = (3*0  2) =  2
Lim f(x) = lim f(0+h) = lim f(h) = lim (h+1) = 1
x ®0+ 
x ®0+ 
h®0 
h®0 
Lim f(x) 
=lim f(0h) = 
lim f(h) = 
lim [3(h)2) =  2 
x ®0 
x ®0 
h®0 
h®0 
since the limits are not equal therefore the function is discontinuous
at x = 0.
The equation of tangent to the curve y =  5x^{2}+ 6x + 7 at the
point
(1 / 2, 35 / 4 ) is given by
dy / dx =  10x + 6
thetrefore dy / dx at ( 1 / 2, 35 / 4 ) = 10 *1/2 + 6 = 1
therefore eq. Is
y  35/4 = 1 ( x  ½)
4y  35 = 4x 2
4y  4x = 33
Q 8 Evaluate 0^{òp/4} log ( 1 + tanx ) dx
Ans. 8 oò^{p/4} log ( 1 + tanx ) dx
let
I= oò^{p/4} log(1+ tan(p/4x)) dx
I= oò^{p/4} log( 1+ (1tanx)/(1+tanx)) dx
I=oò^{p/4} log(2 / 1+tanx) dx
I = oò^{p/4} log2 dx  0ò ^{p/2} log(1+tanx)
dx
I= log2 * p/4 I
I= p/8 log2
Q 9 Evaluate ò dx / (x+1)( x+2)(x+3) .
Ans.9 ò dx / (x+1)( x+2)(x+3) .
let
1 / (x+1)( x+2)(x+3) . = A / (x+1) +B / (x+2) + C / (x+3)
1 = A(x+2)(x+3) + B(x+1)(x+3) + C(x+1)(x+2)
which gives A = ½, B =  1, C = ½
therefore the given integral can be written as
½ ò dx / x+1  ò dx/ x+2 + ½ ò dx /
x+3
= ½ log (x+1)  log(x+2) + 1/2log(x+3) + c
Q 10Evaluate ò ex[ (1  sinx) / (1  cosx)] dx.
Ans. 10 ò ex[ (1  sinx) / (1  cosx)] dx.
ò e^{x} {1 / 1 cosx  sinx / 1 cosx } dx
ò e^{x} {1 / 2sin^{2}x/2  2sinx/2 cosx/2 / 2sin^{2}x/2}
dx
ò e^{x} {(1/2)cosec^{2}x/2  cotx/2} dx =  ò
e^{x} cotx/2 dx + 1/2ò e^{x} cosec^{2}x/2
dx
=  e^{x} cotx/2  ½ ò e^{x} cosec^{2}x/2
dx + C + ½ ò e^{x} cosec^{2}x/2 dx
=  e^{x} cotx/2 + C.
Q 11 ò [(x^{2}  1) / (1 + x^{4})] dx.
Ans. 11 ò [(x^{2}  1) / (1 + x^{4})] dx.
= divide Nr. And Dr. by x^{2} we have
= . ò [(1  1/ x^{2} ) / (1/x2 + x^{2})] dx
put x + 1/x = t
( 1 1/x^{2})dx = dt
also x^{2} + 1/x^{2} = t^{2} 2
therefore . ò dt / t^{2}  2
= 1/ 2Ö2 log (t  Ö2)/ (t+Ö2).
= 1/2Ö2 log [( x+1/x Ö2) / (x+1/x + Ö2)] + C
Q 12 One card is drawn from a set of 17 cards numbered 1 to 17. Find
the probability that the number drawn is divisible by 3 or 7.
Ans.12 single card can be drawn in 17 ways
no. divisible by 3 are 3,6,9,12,15.
No divisible by 7 are 7 and 14
Therefore total no. of favourable events are 7
Therefore the required probability is 7 / 17.
Q 13 Out of 9 outstanding students in a college, there are 4 boys
and 5 girls. A team of 4 Students are to be selected for a quiz programme.
Find the probability that two are boys and two girls.
Ans. 13 Students out of nine can be selected in 9C4 ways i.e.
126
also 2 out of 4 can be selected in 4C2 ways i.e. 6 and 2 out of 5 can
be selected in 5C2 ways i.e. 10.
Therefore the required probability is 10 * 6 / 126 = 10 / 21 .
Q 14 A particle follows the path S = 3t2 + 2t  4. Find the velocity
and acceleration of the Particle at t = 4sec. Assume S in meters.
Ans. 14 Given S = 3t^{2} + 2t  4, we know that velocity
is given by ds/ dt and acceleration is given by d^{2}s / dt^{2}
therefore we have
V = ds / dt = 6t + 2 therefore at t= 4 we have v = 6*4 + 2 = 26 m/s
Also A = d^{2}s/dt^{2} = 6m/s^{2}.
Q 15 Use differential to approximate the value of cube root of 0.009.
Ans. 15
let y = x^{1/3} and take x = 0.008, dx = + .001
so that x + dx = 0.009
now Dy = (x + Dx)^{1/3}  x^{1/3}
= (0.008 + .001)^{1/3}  0.008^{1/3}
therefore 0.009^{1/3} = Dy + 0.2
here Dy is approximately equal to dy and
dy = (dy/dx) dx = (1/3 x^{2/3}) dx
= 1/3 (0.008)^{2/3} (0.001) at x = 0.008
= (1/3 )(1/0.04)(0.001) = 0.00833
therefore 0.009^{1/3} = 0.00833 + 0.2 = 0.2083
Q 16 Solve the differential equation dy / dx = sin (x + y).
Ans. 16 dy / dx = sin(x + y)
put x + y = z
1 + dy/dx = dz /dx
dy/dx = dz/dx  1
substituting we get
dz/dx  1 = sin z
dz/sinz +1 = dx
òdz / sinz +1 = ò dx
x = ò (1/ 1 + sinz)* (1  sinz)/(1  sinz) dz
= ò (1  sinz)/cos^{2}z dz
x = òsec^{2}z dz  òsecz tanz dz
x = tanz  secz +C = tan(x+y)  sec(x+y) +C
Q 17 Solve (1 + x^{2}) dy/dx + 2xy  4x^{2} = 0.
Ans. 17 (1 + x^{2}) dy/dx + 2xy  4x^{2} = 0.
The above equation is in the form
dy/dx + Py = Q.
It can be rewritten as
dy/dx + {2x/(1+x2)} y = 4x^{2}.
Integrating factor
= e^{òpdx} = e^{ò 2x/(1+x2)dx}
= e^{ò (1/ t)dt} where t = 1+x2
= e^{log t} = t = 1 + x^{2}
Multiplying throughout by 1 + x^{2}
(1 + x^{2}) dy/dx + 2xy = 4x^{2}.
d/dx{(1+x^{2}).y} = 4x^{2}.
{(1+x^{2}).y} = ò 4x^{2} dx
= 4/3 x^{3} + c
y = (4/3 x^{3} + c )/(1 + x^{2})
(4/3 x^{3} + c)/(1 + x^{2}) Q 18 If y = log,
log log x^{3}, find dy/dx. Ans. 18 y = log log log
x^{3}
dy/dx = (1/ log log x^{3}) * (1/log x^{3})*(3x^{2})
= [3x^{2}/ {(log log x^{3})* (log x^{3})}]
Q 19 A ladder 13m long is leaning against a wall. The bottom of the
ladder is pulled along the ground, away from the wall, at the rate of
2m/s.How fast is its height on the wall decreasing when the foot of the
ladder is 5m away from the wall?
Ans. 19 Let the ladder is kept vertical initially.
When the horizontal distance is 5 m,
the vertical distance would be Ö(13^{2}  5^{2})
=12 m.
Speed when the ladder is pulled horizontally = 2m/s
Time taken for covering 5m = distance/ velocity = 5/2 =2.5 s
In this time the ladder is pulled 1 m vertically.
Therefore, vertical speed = 1/(2.5) = 0.4 m/s
Ans = 0.4 m/s
Q 20 Evaluate dx
as a limit of sums.
Ans. 20) Evaluate _{0}ò 2(1 + x^{2}) dx
_{a}ò ^{b} f(x)dx = (ba)lim 1/n[f(a) + f(a+h)
+……………..+f(a + (n1)h)]
nµ.............................................................................
where h = (ba)/n
Here a=0, b=2, f(x) = 1 + x^{2}, h= 2/n,
_{0}ò ^{2}(1 + x2) dx = 2 lim 1/n [f (0) + f(2/n)
+f(4/n)……………..+f((2n2)/n)]
= 2 lim 1/n [1 + ((2^{2}/n^{2})+1) +((4^{2}/n^{2})+1)……………..+((2n2)^{2}/n^{2}
+1 )]
= 2 lim 1/n[(1+1+….n times)+1/n^{2}(2^{2}+4^{2}+…………..+(2n2)^{2}]
= 2 lim 1/n[n + 2^{2}/n^{2}(1^{2}+2^{2}+…………..+(n1)^{2})]
= 2 lim 1/n[n + 4/n^{2}.{(n1)n(2n1)}/6]
= 2 lim 1/n[n + 2/3.{(n1)(2n1)}/n]
= 2 lim [1 + 2/3{(n1)(2n1)}/n^{2} ]
= 2 lim [1 + 2/3(11/n)(21/n)]
= 2 [1 +4/3]
= 14/3.
Ans. = 14/3
Q 21 Evaluate
Ans. 21
Expanding the determinant, we get
D= y[(2x+y)(x+y)2x^{2}]
= y[ 3xy + y^{2}]
= y^{2}[3xy + y]
Ans = y^{2}[3xy + y]
Q 22 Find the correlation coefficient between x and y, when the lines
of regression are: 2x  9y + 6 = 0 and x  2y + 1 = 0.
Ans.22) The regression line of y on x is
y = (2x+6)/9
thus b_{yx} =2/9
The regression line of x on y is
x = 2y 1
b_{xy} = 2
Correlation coefficient,
r = Ö (b_{yx} b_{xy})
= Ö (4/9)
=
Q 23 Using matrices solve the following
x + y  2z = 0;
2x + y  3z = 0;
5x + 4y  9z = 0
Ans. 23
Writing in the matrix form
lAl = 1(9+12) 1(18+15)2(85)
= 3 + 3 6 = 0
Q 24 Evaluate ^{p/
2 }(log sinx) dx
Ans. 24 p/2
Let I = 0ò (log sinx) dx ………………(1)
p/2
Then , I = oò log [ sin(p/2  x ) ]dx
[since oò^{a} f(x) dx = oò^{a}f(a x)dx
Or
I = oò^{p/2} log(cos x ) dx ……………………………….(2)
Adding equation (1) and (2),
2I = oò^{p/2} [log (sin x) +log (cos x)] dx
= oò^{p/2} log( sin 2x/2) dx
= oò^{p/2} log(sin2x)dx  oò^{p/2} log(2)
dx
= ½ oò^{p/2} log sin t dt  (log 2)* oò^{p/2}
dx { putting 2x = t in the 1st integral}
= ½*2 oò^{p/2} logsinx dx  p/2(log 2)
2I = I  p/2 (log 2)
or
I =  p/2 (log2)
Therefore
oò^{p/2} log sin x =  p/2 (log 2)
Q 25 Find the magnitude and direction of the resultant of two forces
14N and 22N inclined at an angle of 30^{o} to each other.
Ans. 25
Let the resultant force 'R', the magnitude of the resultant force can
be determined by
Given force1 A = 14 N, force B = 22 N,
R = Ö A^{2} + B^{2}+ 2A*B cos q
Here q = 30^{o}
Therefore
R = Ö 14^{2}+ 22^{2} + 14*22 cos 30^{o}
R = 32.181 N
Direction can be calculated by
tana = ( B sinq)/A + B cosq
= 22 sin 30^{o}/ 14 + 22 cos 30^{o}
tana = 0.4119
a = tan^{1} 0.4119
Q 26 Prove that the points with position vectors (  j  k ), ( 4i
+ 5j + k ), ( 3i + 9j + 4k ) and (  4i + 4j + 4k ) are coplanar.
Ans. 26 Given pts. are (  j  k ), ( 4i + 5j + k ), ( 3i + 9j + 4k
) and
(  4i + 4j + 4k )
let the pts. Are A,B,C and D respectively, so
AB = ( 4i + 5j + k )  (  j  k ) = 4i + 6j + 2k
BC = ( 3i + 9j + 4k )  ( 4i + 5j + k ) = i + 4j + 3k
CA = (  j  k )  (  4i + 4j + 4k ) = 4i  5j  5k
Now if they are coplaner then
4 ( 20 + 15)  6 ( 5  12 ) + 2 ( 5  16 ) which is equal to 0. Hence
the pts. Are coplanar

Q 27 Find the equation of plane passing through the origin and perpendicular
to each of the planes, x + 2y  z = 1 and 3x  4y + z = 5. Ans. 27
Given equation
x + 2y  z = 1
3x 4y +z = 5
Any plane through (0,0,0) is
a( x 0) + b(y  0) + c(z  0) = 0,
i.e. ax + by + cz = 0 ……………………….(1)
If (1) is perpendicular to each one of the given planes, we have:
A +2b c = 0 ……………………………..(2)
3a 4b + c = 0 …………………………….(3)
on solving (2) and (3), we get
a/2 = b/4 = c/10 or a/1 = b/2 = c/5 = k.
putting a= k, b= 2k and c = 5k in eq. (1), we obtain the required equation
of the plane as
kx + 2ky + 5kz =0
or
x +2y +5z = 0. ( Ans.)
Q 28 If a car goes from point A to B with a velocity of 40Km/Hr and
returns back with a speed of 60Km/Hr.Find the average speed of the car.
Ans. 28 Let the distance between the two pts. be S Km, then time taken
in first ride will be
T_{1} = S / 40 also time taken in second ride will be
T_{2} = S / 60
Therefore total time will be T = S / 40 + S / 60
Avg. Velocity = total distance travelled / total time taken
= 2S / (S / 40 + S / 60)
= 48 Km / Hr.
Q 29 The ceiling of a long hall is 25m high. What is the maximum horizontal
distance that a ball thrown with a speed of 40m/s can go without hitting
the ceiling of the hall?
Ans. 29 we know that h = u^{2} sin^{2}q / 2g
therefore 25 = 40* 40 sin^{2}q / 2 * 10 ……………………………taking
g = 10 m/s^{2}
sinq = 51/2 / 4
therefore R = u^{2}sin2q / g
= 40 * 40 * 2 sinq cosq / 10
= 40 * 40 * 2* 5^{1/2} * 11^{1/2} / 4*4*10
= 148.32 m
Q 30 Solve the differential equation
(x^{3}  3xy^{2}) dx = (y^{3}  3x^{2}y)
dy.
Ans. 30
(x^{3}  3xy^{2})dx = (y^{3} 3x^{2}y)dy
the given equation can be written as,
dy/dx = x^{3}  3xy^{2}/ y^{3} 3x^{2}y
which is clearly homogeneous.
Putting y= vx and dy/dx = v + x dy/dx in it, we get:
v+ x dv/dx = x^{3}  3x3v2/v^{3}x^{3} 3vx^{3}
= 1  3v^{2}/ v^{3}3v
x dv/dx = [1  3v^{2}/ v^{3}3v]  v = 1  v^{4}/v^{3}
 3v
taking integral both sides,
ò dx/x = ò v^{3}  3v/1  v^{4} dv
by partial fraction
= ò[ 1/ 2(v +1) + ½(v  1)  2v/v^{2}+1]dv
log  x  + log C =1/2 log  v+1 + ½ log v  1  log  v^{2}+1
logCx = log  (Ö v+1) (Öv  1)/ (v^{2}+1)
therefore
Cx = Ö v^{2}  1/( v^{2}+1)
C^{2}x^{2} = (v^{2}  1)/( v^{2}+1)
Putting v = y/x
C^{2}(y^{2}+ x^{2})2 = ( y^{2}  x^{2})
Therefore the solution of the differential equation is
C^{2}(y^{2}+ x^{2})^{2} = ( y^{2}
 x^{2}) ( Ans.)
