Sample question paper 2 with solution

Sample Question Paper
MATHEMATICS
Class XII
Time: 3 Hours

Max. Marks: 100

General Instructions
1.

All questions are compulsory.

2. The question paper consists of 29 questions divided into three sections A, B and C. Section
A comprises of 10 questions of one mark each, section B comprises of 12 questions of four
marks each and section C comprises of 07 questions of six marks each.
3. All questions in Section A are to be answered in one word, one sentence or as per the exact
requirement of the question.
4. There is no overall choice. However, internal choice has been provided in 04 questions of
four marks each and 02 questions of six marks each. You have to attempt only one of the
alternatives in all such questions.
5.

Use of calculators is not permitted. You may ask for logarithmic tables, if required.

SECTION-A
1. If f(x) is an invertible function, find the inverse of f(x)=3x-2.

 −1 
2. find the principal value of sin −1  
 2 
x + 3y
3. If 
 7−x

y   4 −1
find the value of x and y.
=
4  0 4 
 


4. A matrix A of order 3X3 has determinant 5. what is the value of 3A .
5 .Find ∫

x + cos 6 x
d ( x)
3 x 2 + sin 6 x

6. Evaluate ∫ xe x dx .

Pratima Nayak,KV Teacher



   


7. If p is a unit vector and x + p . x − p =.
80

)(

(

)

8.write the direction cosine of a line equally inclined the three co-ordinate axes.
9.write the value of the following determinant.
a −b b−c c −a
b−c c−a c−a
c −a a −b b−c

(

) (

)

ˆ
ˆ 0
ˆ
ˆ
10. find the value of p, if 2i + 6 ˆ + 27 k X i + 3 ˆ + pk =
j
j
SECTION-B

5 2 
1
−1  1 
11.Let the value of 2 tan −1   + sec −1 
 7  + 2 tan  8 

5
 


OR

π
 x −1 
−1  x + 1 
Solve for x, tan −1 
,x≤
 + tan 
=1
 x−2
 x+2 2
13.Uasing properties of determinants prove that
2a
2a
a −b−c
2b
2b
b−c−a
=
2c
2c
c −a −b

(a + b + c)

3

14.Show that the function f(x)= x + 2 is continuous at every x ∈ R but fails to be
differentiable at x=-2.
OR
Verify lagrange`smean value theorem for the following function f(x)=
x 2 + 2 x + 3 for [ 4, 6]

15.Differential following function w.r.to x
 1 + x2 − 1
tan −1 

x





OR
Differentiate

( cos x )

sin x

+ ( sin x )

cos x

with respect to x


16.Evalute
π

x sin x

∫ 1 + (cos x)

2

dx

0

17.Solve the following differential equation
( x 2 − y 2 )dx + 2 xydx = 0, y (1) = 1
18.The volume of a spherical balloon is increasing at the rate of 25 cm3 / sec. Find the rate
of change of its surface area at the instant when its radius is 5cm.
19.Form the differential equation representing the family of parabola having vertex at
origin and axis along positive of X-axis.


 

ˆ
ˆ
20.Find the projection of b + c on a when a = i − 2 ˆ + k , b = + 2 ˆ − 2k and
2ˆ
j ˆ
i
j

ˆ
ˆ j
c = 2i − ˆ + 4k .

21.Find the equation of the plan passing through the points (0,-1,0),(1,1,1) and (3,3,0).
22.The probability of A solving a problem is 3/7 and that B solving it is 1/3 what is the
probability
(i) at least one of them solve the problem?
(ii)only one of them will solve the problem?

SECTION-C

(23) Using matrices solve the following system of linear equations2x-y+z=3, -x+2y-z=-4, x-y+2z=1
OR
Using elementary transformations, find the inverse of the following matrix 1 3 − 2
− 3 0 − 5


2 5 0



(24)Determine the points on the curve y =

x2
which are nearest to point (0, 5)
4

OR


Show that the surface area of a closed cuboid with square base and given volume is
maximum when it is a cube.
(25) Find the area of the region included between parabola y 2 =x and the line x +y =2.
a
 a−x

(26) Prove that- ∫ 
 a + x  dx = π a
− a


(27) Mona wants to invest at most Rs.12000 in Saving certificate (SC) and National
saving bonds (NSB).She has to invest at least Rs.2000 in SC and at least Rs.4000 in NSB.
If the rate of interest on SC is 8 pa. and the rate of interest on NSB 10pa, how much
money should she invest to earn maximum yearly income? Also find the maximum
income.
(28)

Find the foot of perpendicular drawn from the point A (1, 0, 3) to the join of the

points B (4, 7, 1) and C (3, 5, 3) and also find the perpendicular distance.
(29) Urn A contains 1 white, 2 black, 3 red balls; Urn B Contains 2 white ,1 black ,1red
ball; Urn C contains 4 white ,5 black and 3 red balls. One urn is chosen at random and
two balls are drawn, these happens to be one white and one red. What is the probability
that they come from urn C.

Solution

SECTION-A
1.

f(x)=3x-2
Let

y =f(x)
y =3x-2
3x+2 =y
f −1 ( x) =3x+2

2.

y = sin −1 (−1/ 2) = −1 (1/ 2) = 6
− sin
−π /

3.

x + 3y
 7−x


4.

3 A = 33 A = 27 × 5 = 135

y   4 −1
7,
−
=
⇒ x + 3 y = y = 1, 7 − x = ⇒ x = y = 1
4,
−
0
4  0 4 
 



5.
I
=

1
x + cos 6 x
log ( 3 x 2 + sin 6 x ) +c
dx
=
2
6
+ sin 6 x

∫ 3x

∫ xe dx = x ∫

e x dx − ∫ 1.e x dx = xe x − e x + c

6.

I=

7.

( x + p ) . ( x − p ) = 80 ⇒

8.

l= m= n= ±

9.

Taking operation C1 → C1 + C2 + C3



x







2 2
2

x − p = 80 ⇒ x = 81 ⇒ x =
9

1
3

a −b b−c c −a
b−c c−a c−a
c −a a −b b−c

10

= 0

2
0
( 2iˆ + 6 ˆj + 27kˆ ) X (iˆ + 3 ˆj + pkˆ ) =⇒ 1 = 6 = 27 ⇒ P = 27
3 P
2

SECTION-B

5 2 
1
−1  1 
11. 2 tan −1   + sec −1 
 7  + 2 tan  8  =

5
 


 1 1 
5 2 
 + 
1
1

2  tan −1 + tan −1  + sec −1 = 2 tan −1  5 8  + tan −1
 7 

1 1
5
8



1 − . 
 5 8

2

5 2 

 7  −1




 2 
 3 1 
+ 
 3 
π
−1  1 
−1  1 
−1
−1  1 
−1  4
−1
7
.
= 2 tan   + tan  = tan 
 =
 + tan   = tan  = tan (1)

3 1
1
4
3
7
7
1 − × 
1 − 
 4 7
 9

Value=

π
4

OR

 x −1 
 x +1  π
tan −1 
+ tan −1 
=
 x − 2

 x + 2 4




x −1 x +1 
+

 π
−1
x−2 x+2  =
⇒ tan 
1 −  x − 1   x + 1   4
  x − 2  x + 2 


 

 ( x − 1)( x + 2 ) + ( x − 2 )( x + 1)  π
⇒ tan −1 
=
 ( x − 2 )( x + 2 ) − ( x − 1)( x + 1)  4

⇒

1
x2 + x − 2 + x2 − x − 2
2x2 − 4
=
1⇒
==
1⇒ x ±
2
2
x − 4 − x +1
−3
2

12. R= {(T1 , T2 ) : T1 ≅ T2

}

(i) R is reflexive:T1 ≅ T1 ⇒ (T1 , T1 ) ∈ R

∴ R is reflexive
(ii)

R is symmetric:Let (T1 , T2 ) ∈ R ⇒ T1 ≅ T2 ⇒ T2 ≅ T1 ⇒ (T2 , T1 ) ∈ R
⇒ R is symmetric.

(iii)

R is Transitive:Let (T1 , T2 ) ∈ R & (T2 , T3 ) ∈ R ⇒ T1 ≅ T2 , T2 ≅ T3 ⇒ T1 ≅ T3 ⇒ (T1 , T3 ) ∈ R

∴ R is Transitive.
∴ R is equivalence relation.
a −b−c
2a
2a
13.
=
∆
2b
b−c−a
2b
2c
2c
c−a −b

R1 → R1 + R2 + R3
a+b+c a+b+c a+b+c
=
∆
b−c−a
2b
2b
2c
2c
c −a −b

Taking (a + b + c) common
1
1
∆ = (a + b + c) 2b b − c − a
2c

2c

1
2b
c−a−b


C1 → C1 − C 2 , C 2 → C 2 − C 3

0
0
1
∆ = (a + b + c) a + b + c − (a + b + c)
2b
c−a−b
a+b+c
0

1
0 0

1 −1
∆ = (a + b + c) 1 − 1
2b = (a + b + c) 3 0 − 0 + 1

0 1 

0 1 c−b−a
3

∆ = (a + b + c) 3
14. f ( x) = x + 2

f (x) =x+2 if

x>-2

f(x) =0

x=-2

if

f(x)=-(x+2) if x<-2
For continuity
Case(i)
When c>-2
lim x →c = lim x →c ( x + 2) = c + 2 = f (c)

∴ f ( x)is continuous for all c > -2
Case(ii)
When c <- 2 then
lim x →c f ( x) = lim x →c − ( x + 2) = −(c + 2) = f (c)

∴ f(x) is continuous for all c <- 2
Case(iii)
When c=-2
lim x →−2+ f ( x) = lim h→0 f (−2 + h) = lim h→0 (−2 + h + 2) = 0
lim x →−2− f ( x) = lim h→0 f (−2 − h) = lim h→0 − (−2 − h + 2) = 0

∴ lim x →−2 f ( x) = f (−2)

∴ f ( x)is continuous at x=-2
∴ f ( x)is continuous for all x ∈ R


For differentiability at x=-2

Rf ` (−2) = lim h→0
Lf (−2) = lim h→0

f (−2 + h) − f (−2)
= lim h→0
h

(−2 + h − 2) − 0
=1
h

(2 + h − 2) − 0
f (−2 − h) − f (−2)
= −1
= lim h→0
−h
−h

Lf ` (−2) ≠ Rf ` (−2) ⇒ f(x) is not diff. at x=-2
OR
f(x)= x 2 + 2 x + 3 for [ 4, 6]
⇒ f ` ( x) = 2 x + 2

 F(x) is a polynomial function
∴ f(x) is continuous for x ∈ [4,6] and differentiabl for x ∈ (4,6)
Now ∃x = c ∈ (4,6) s.t.
f ` (c ) =

f (b) − f (a )
⇒ 2c + 2 = 12 ⇒ c = 5 ∈ (4,6)
b−a

∴ Lagranges theorem satisfied.
 1 + x 2 − 1
15.y= tan −1 

x




Let x= tan θ
Y=
 1 + tan 2 θ − 1
θ  θ 1 −1
−1  sec θ − 1 
−1 1 − cos θ 
−1 
tan −1 
 = tan 
 = tan  sin θ  = tan  tan 2  = 2 = 2 tan x
tan θ
 tan θ 









y=

1
dy
1
tan −1 x ⇒
=
2
dx 2(1 + x 2 )
OR

Let y=

( cos x )

sin x

+ ( sin x )

cos x

y=u+v


 u = (sin x) cos x ⇒ log u = cos x log(sin x) ⇒

d log u cos x
cos x + log(sin x)(− sin x)
=
sin x
dx

1 du
du
sin x
cos x
]
= cos x. cot x − sin x. log sin x ⇒
= (sin x) cos x [ ( cos x ) + ( sin x )
u dx
dx
v = (cos x) sin x ⇒ log v = sin x log(cos x) ⇒

1 dv
1
(− sin x ) + cos x log(cos x)
= sin x ×
cos x
v dx

dv
= (cos x) sin x [− sin x tan x + cos x log cos x ]
dx
dy du dv
=
+
dx dx dx
⇒

dy
sin x
cos x
] + (cos x) sin x [− sin x tan x + cos x log cos x ]
= (sin x) cos x [ ( cos x ) + ( sin x )
dx
π

π

π

x sin x
(π − x) sin(π − x)
(π − x) sin x
dx ⇒ I = ∫
dx ⇒ I = ∫
dx
2
2
2
0 1 + cos (π − x )
0 1 + cos x
0 1 + cos x

16. I= ∫
π

I= ∫
0

π sin x
1 + cos x
2

π

x sin( x)
dx
1 + cos 2 ( x)
0
π

π sin( x)

∫ ∫ 1 + cos

2I=

π

dx − ∫

2

0

( x)

dx ⇒ I =

1 π sin x
dx
2 ∫ 1 + cos 2 x
0

Let cosx=t ⇒ − sin xdx = dt
−1

[

]

[

]

1
−1
dt ⇒ π tan −1 t −1 = π tan −1 (1) − tan −1 (− 1)
2
1 1+ t

I= π ∫
I=

17.

π2
4

dy − ( x 2 − y 2 )
=
dx
2 xy
Let y=vx ⇒

⇒v+x

dy
dv
=v+x
dx
dx

2v
dv − x 2 (1 − v 2 )
dx
=
⇒
dv = −
2
2
dx
x
2x v
1+ v


⇒∫

dx
2v
dv = − ∫
⇒ log(1 + v 2 ) = − log x + c ⇒ log( x 2 + y 2 ) − log x = c …….(1)
2
x
1+ v

Now put x=1,y=1
C=log2
From (1)
 x2 + y2 
2
2
log 
 = log 2 ⇒ ( x + y ) = 2 x
x 


18.Let any time t,radius=r,volume=V,surface area=S
g.t.

dV
ds
= 25cm 3 / sec, = ?
dt
dt

4
dV
dr
1
 dr 
 V = πr 3 ⇒
= 4πr 2
⇒  =
3
dt
dt
 dt  r =5 4π

S = 4πr 2
⇒

ds
dr
= 8πr
dt
dt

 ds 
⇒   = 10cm 2 / sec
 dt  r =5
19. Equ. Of parabola
y 2 = 4ax ………(i)
d.w.r.to x

dy 4a
y dy
=
⇒a=
dx 2 y
2 dx
From equ. (i)
y = 2x

dy
dx



ˆ 
ˆ
ˆ
ˆ j
20. a = i − 2 ˆ + k , b = + 2 ˆ − 2k , c = 2i − ˆ + 4k .
2ˆ
j ˆ
i
j
 
ˆ
ˆ j
⇒ b + c = 3i + ˆ + 2k
  
 
2
 b + c .a
Projection of b + c on a =
=
Ans.

a
14

(

)

21. points (o,-1,0) ,(1,1,1) ,(3,3,0).


Equ. Of plan passing through point (0,-1,0)

A(x-0) + B(y+1) + C(z-0) = 0 ……….(i)
Point (1, 1, 1)and (3,3,0) satisfy equ.(i) we get
A + 2B + C = 0……….(ii)
3A + 4B + 0.C = 0…….(iii)
Solve eque. (i)@ (ii)
A
B
C
= =
=λ⇒ A
−4 3 −2
A = - 4λ , B = 3λ , C = - 2λ
Putting in (1)
-4 λ x + 3 λ (y + 1) - 2 λ z = 0

-4x + 3 (y + 1) - 2z = 0

4x + 3y + 2z-3 = 0 Ans.

22. Given that probability of A solving the question P(A) =
Probability of B solving the question P (B) =
−

∴ P( A ) =1 -

1
3

3
4
=
7
7

−

and

3
7

P( B ) = 1 -

1
2
=
3
3
−

−

(1) Probability that at least one of them solve the problem = P(A B ) + P( A B) +
P(AB)
=

3 2 4 1 3 1
. + . + .
7 3 7 3 7 3

=

6 + 4 + 3 13
=
21
21


−

−

(2) Probability that only one of them will solve the problem = P(A B ) + P( A B)
=

3 2
4 1
.
+ .
7 3
7 3

=

6+4
21

=

10
Ans.
21

SECTION-C
(23) Given equations can be written as
 2 −1 1   x  3 
− 1 2 − 1  y  = − 4
   

 1 −1 2  z  1 
   


⇒ AX=B → (1)
A =4 ≠ 0

⇒ A −1 exits.
A 11 =(-1) 1+1 (4-1)=33
Similarly other cofactors can be obtained.
 3 1 − 1
adjA 1 
= 1 3 1
Adj. A=

4
A
− 1 1 3 



From (1)
X= A −1 B


 3 1 − 1  3 
X 
 Y  = 1  1 3 1   − 4
⇒ 
 
4 
− 1 1 3   1 
Z 
 

 

1 
= − 2 ⇒ x=1, y=-2, z=-1
 
 − 1
 

OR

A =AI
R 1 → R 1 +3R 1
1 3 − 2  1 0 0
0 9 − 11 = 3 1 0 A
 


 2 5 0  0 0 1 
 



R 1 → R 1 +3R 3 , R 2 →

Applying R 3 → R 3 -2R 1

1
9
11
R 2 R 3 → R 3 +R 2 R 3 →
R 3R 2 →R 2 + R3
9
25
9

R 1 → R 1 -10 R 3

1 0 0
⇒ 0 1 0  =


0 0 1 



−2
5
4
25
1
25


 1
− 2

 5
−3
 5


− 3
5 
11 

25 
9 
25 


⇒ I=BA ⇒ B is inverse of A

(24)let (x,y) be the foot of perpendicular on the given curve which is nearest to (0,5)

[

D= ( x − 0) + ( y − 5)
2

2

]

1
2


If D is minimum then D 2 = x 2 +(y-5) 2 =E is also minimum
dE
= 2y-6
dY
For minimum distance
At y=3

dE
= 0 ⇒ y=3
dY

d 2E
=2 〉 0
dy 2

⇒ for y=3 ,distance is minimum
⇒ x= ± 2 3
Reqd. point is ( ± 2 3 ,3 )
OR

Let x be the side of the square base & y be the height of cuboid

V=x 2 y ⇒ y =

V
x2

Surface area (S)=2x 2 +4 xv / x 2

=ds/dx=4x-4v/x 2
1

For minimum ds/dx=0 => x 3 =v

,x=v 3

1
3

3

S’’ =4+8v/x when x =v =4+8 >0 therefore S is ,minimum
3

When x 3 =v=>x =x 2 y =>x =y
There when S is mini mum cuboid is cube.
Q25 given curves are y 2 =x and x+y =2 solving points of intersection aare (1,1),(4,-2)
1

The area of shaded region

∫

( x 2 − x1 )dy

−2
1

= ∫ (2 − y − y 2 )dy
−2

=9/2 sq units.


(26)
a

Given integral =

∫

−a

a

a
(a − x )
2

2

dx −

∫

−a

xdx
(a 2 − x )
2

Putting x=asin θ ,dx=acos θdθ ,and changing limits in second integral

Π
Π
Π
Given integral= a  − (−  +a [cos θ ] 2 =a Π
2
2
−Π
2

(27)
Let she invests Rs. x in saving certificates and Rs. y in National saving bonds
Then LPPis
To maximize Z=0.08x+0.1y
Subject to constraints
x ≥ 2000, y ≥ 4000, x + y ≤ 12000
corner points of feasible region ABC are A(2000,4000), B(8000,4000),c(2000,10000)

at A, Z=160+400=560

at B, Z=640+400=1040
at C,Z=160+1000=1160
thus Rs. 2000 should be invested in saving certificates and Rs.10000 in National saving
bonds. Maximum yearly income is Rs. 1160.


(28)Let A(1,0,3),B(4,7,1),C(3,5,3) be the given points . Let P be the foot of perpendicular
from A on BC. If P divides BC in k:1 then coordinates of P are (

3k + 4 5k + 7 3k + 1
)
,
,
k +1 k +1 k +1

d.r.’s of BC are 1,2,-2
d.r.’s of AP are

2 k + 3 5k + 7 − 2
,
,
k +1 k +1 k +1

since AP ⊥ BC
therefore k=

−7
4

5 7 17
thus coordinates of P are ( , , )
3 3 3
reqd. ⊥ dis tan ce = AP =

117
7
17
 5
2
2
2
( 3 − 1) + ( 3 − 0) + ( 3 − 3)  = 3



(29)
Let E1,E2 ,E3 be the events that the balls are drawn from urn A, urn B, urn C
respectively and let E be the event that balls drawn are one white and one black
Then P(E1)=P(E2)=P(E3)=
P(E/E1)=

1
3

C (1,1) * C (3,1) 1
=
C (6,2)
3

Similarly P(E/E2)=

1
2
, P(E/E3)=
11
3

Using Bayes’ Theorem reqd. probability= P(E3/E)

2
15
33
= (
)=
1 1 2
59
+ +
15 9 33


Sample question paper 2 with solution

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Sample question paper 2 with solution