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