cbse mathematics – 2017 · 2020. 6. 8. · cbse mathematics – 2017 Comptt. ... unable to get...

mathematics – 2017 comptt. (delhi) ■ 197

set i

section A

Question numbers 1 to 4 carry one mark each.

Q.1. Let A and B be the matrices of order 3 × 2 and 2 × 4 respectively. Write the order of matrix

(AB).

Q.2. Write the equation of tangent drawn to the curve y = sin x at the point (0, 0).

Q.3. Find : 1

1x x( log )+∫ dx

Q.4. Write the angle between the vectors a b→ →× and b a

→ →× .

section B

Question numbers 5 to 12 carry 2 marks each. Q.5. In the following matrix equation use elementary operation R2 → R2 + R1 and write the

equation thus obtained.

2 31 4

1 02 1

8 39 4

−

=

−−

.

Q.6. Find the value of k for which the function f (x) = x x

xx

k x

2 3 102

2

2

+ −−

≠

=

,

, is continuous at x = 2.

Q.7. The radius r of a right circular cone is decreasing at the rate of 3 cm/minute and the height h is increasing at the rate of 2 cm/minute. When r = 9 cm and h = 6 cm, find the rate of change of its volume.

Q.8. Find : x x2 2−∫ dx.

Time allowed : 3 hours Maximum marks : 100

General InstructIons Same as in Delhi Board 2015.

cbse

mathematics – 2017Comptt. (DELHI)

198 ■ shiv das senior secondary series (Xii)

Q.9. Find the differential equation of the family of curves y2 = 4ax.

Q.10. Find the general solution of the differential equation dydx

+ 2y = e3x.

Q.11. If the points with position vectors 10 3 12 5 11ˆ ˆ, ˆ ˆ ˆ ˆi j i j i j+ − +and λ are collinear, find the value of λ.

Q.12. A firm has to transport atleast 1200 packages daily using large vans which carry 200 packages each and small vans which can take 80 packages each. The cost for engaging each large van is ` 400 and each small van is ` 200. Not more than ` 3,000 is to be spent daily on the job and the number of large vans cannot exceed the number of small vans. Formulate this problem as a LPP given that the objective is to minimize cost.

section c

Question numbers 13 to 23 carry 4 marks each.

Q.13. Prove that : tan–1 1 11 1 4

12

12

1 1+ − −+ + −

= − − ≤ ≤−x x

x xx xπ cos , .

Q.14. If a b y c z

a x b c z

a x b y c

− −− −− −

= 0, then using properties of determinants, find the value of

ax

by

cz

+ + , where x, y , z ≠ 0.

Or

Using elementary operations, find the inverse of the following matrix A

A = 1 22 1−

.

Q.15. If x = a(cos θ + θ sin θ) and y = a(sin θ – θ cos θ), then find d y

dx

2

2 .

Q.16. Find the equation of tangent to the curve y = cos(x + y), – 2π ≤ x ≤ 0, that is parallel to the line x + 2y = 0.

Q.17. Find : x

x x

++ −∫ 5

3 13 102 dx.

Or

Evaluate : 1

42 20

4

cos sin

/

x x+∫π

dx.

Q.18. Find : x dx

x x

2

21 1( )( )− +∫ Q.19. Find the general solution of the following differential equation :

x cos yx

dydx

= y cos yx

+ x


Q.20. If four points A, B, C and D with position vectors 4 3 3 5 7 5 3ˆ ˆ ˆ, ˆ ˆ ˆ, ˆ ˆi j k i x j k i j+ + + + + and 7 6ˆ ˆ ˆi j k+ + respectively are coplanar, then find the value of x.

Q.21. Find the value of p so that the lines

13

7 142

31

− − −= =x yp

z and 7 7

35

111

7− − −= =x

py z

are at right angles.

Or Find the equation of the plane through the line of intersection of the planes x + y + z = 1

and 2x + 3y + 4z = 5 and twice of its y-intercept is equal to three times its z-intercept. Q.22. Solve the following Linear Programming problem graphically : Minimize : z = 6x + 3y

Subject to the constraints :

4 805 115

3 2 1500 0

x y

x y

x y

x y

+ ≥+ ≥+ ≤≥ ≥

,

Q.23. There are three categories of students in a class of 60 students : A : Very hard working students B : Regular but not so hard working C : Careless and irregular 10 students are in category A, 30 in category B and rest in category C. It is found that

probability of students of category A, unable to get good marks in the final year examination is, 0.002, of category B it is 0.02 and of category C, this probability is 0.20. A student selected at random was found to be the one who could not get good marks in the examination. Find the probability that this student is of category C.

section D


Q.24. Let f : IR – − →

43

IR be a function defined as f (x) = 43 4

xx +

. Show that, in

f : IR – − →

43

Range of f, f is one-one and onto. Hence find f –1. Range f → − −

IR 43

.

Or

Let A = IR × IR and be the binary operation on A defined by (a, b) (c, d) = (a + c, b + d). Show that is commutative and associative. Find the identity element for on A, if any. (Not in syllabus)

Q.25. Find matrix A, if 2 11 03 4

1 8 101 2 59 22 15

−

−

=− − −

− −

A

Q.26. Find the intervals in which the function f given by f (x) = sin x + cos x, 0 ≤ x ≤ 2π is strictly increasing or strictly decreasing.

Or


Show that height of the cylinder of greatest volume which can be inscribed in a right circular cone of height h and semi-vertical angle α, is one-third that of the cone. Hence find the greatest volume of the cylinder.

Q.27. Using integration find the area of the region {(x, y) : y2 ≤ 4x, 4x2 + 4y2 ≤ 9}.

Q.28. Find the equation of plane containing the lines x y z− +

−−= =8

319

1610

7 and

x y z− + −−

= =383

298

55 .

Q.29. A fair coin is tossed 8 times, find the probability of (Not in syllabus) (i) exactly 5 heads (ii) at least six heads (iii) at most six heads

Or Three cards are drawn successively with replacement from a well shuffled pack of 52 cards.

Find the mean and variance of the number of red cards.

set ii

Note: Except for the following questions, all the remaining questions have been asked in Set I. Q.10. The position vectors of points A, B and C are λ µˆ ˆ, ˆ ˆi j i j+ +3 12 and 11 3ˆ ˆi j− respectively.

If C divides the line segment joining A and B in the ratio 3 : 1, find the value of λ and μ.

Q.11. Find : 2 2x x−∫ dx.

Q.12. Find the value of p for which the function f (x) = 1 4

2 0

0

−

≠

=

cos ,

,

x

xx

p x is continuous at x = 0.

Q.20. Solve the following Linear Programming problem graphically :

Maximise : z = 8000x + 12000y

Subject to the constraints : 3 4 60

3 300 0

x y

x y

x y

+ ≤+ ≤≥ ≥

,

Q.21. Find : x

x x

++ +∫ 7

3 25 282 dx

Q.22. If x = 3 cos t – 2 cos3 t and y = 3 sin t – 2 sin3 t then find d y

dx

2

2 .

Q.28. Find matrix X if : X1 2 34 5 6

7 8 92 4 6

11 10 9

=

− − −

.

Q.29. Show that the lines x y z+−

− −= =33

11

55

and x y z+−

− −= =11

22

55

are coplanar. Hence find

the equation of the plane containing these lines.


set iii

Note: Except for the following questions, all the remaining questions have been asked in Set I and Set II.

Q.10. Find the value of k for which the function f (x) =

sin cos ,

,

x xx

x

k x

−−

≠

=

4 4

4

ππ

π .

Q.11. Find : 1

42x x−∫ dx

Q.12. Find the general solution of the differential equation dydx x

+ 2 y = x.

Q.20. Maximize z = 105x + 90y graphically under the following constraints :

x + y ≤ 50, 2x + y ≤ 80, x ≥ 0, y ≥ 0

Q.21. Find : x

x x

+

− +∫ 3

5 4 2 dx

Q.22. Prove that : cot–1 1 11 1 2 4

0+ + −+ − −

∈

=sin sin

sin sin, ,x x

x xx

xπ

.

Q.28. Using Integration, find the area of the following region : {(x, y) : y2 ≥ ax, x2 + y2 ≤ 2ax, a > 0} Q.29. Find the equation of the plane passing through the point (–1, 3, 2) and perpendicular to

each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0.

SolutionS

1. The order of matrix (AB) is 3 × 4. 2. y = sin x

Differentiating both sides w.r.t. x,

dydx

= cos x

dydx

at (0, 0) = cos 0 = 1

∴ Slope of tangent, m = 1Equation of the tangent at point (0, 0) and

slope 1 is (y – 0) = 1(x – 0) ⇒ y – x = 0

3. 1

1x x( log )+∫ dx [Let p = 1 + log x, dp = 1x

dx

= dpp∫ = log |p| + c

= log|1 + log x| + c

4. b a a b→ → → →× = − ×( )

∴ The angle between the vector a b→ →× and

b a→ →× = π.5. I A = A

2 31 4

1 02 1−

=

8 39 4

−−

Applying R2 → R2 + R1

2 3

3 7

1 0

2 1−

=

8 3

17 7

−−

6. L.H.L = lim ( )x

f x→ −2

= limx

x xx→ −

+ −−2

2 3 102

= lim( )( )

( )x

x xx→ +

+ −−2

5 22

= lim ( )x

x→ −

+2

5 = 2 + 5 = 7

x x

x x x

x x x

x x

2

2

3 10

5 2 10

5 2 5

5 2

+ −

= + − −= + − += + −

( ) ( )

( )( )


R.H.L = lim ( )x

f x→ +2

= limx

x xx→ +

+ −−2

2 3 102

= lim( )( )

( )x

x xx→ +

+ −−2

5 22

= lim ( )x

x→ +

+2

5 = 2 + 5 = 7

At x = 2, f (x) = k

f (2) = kf (x) is continuous at x = 2. (Given)

∴ LHL = RHL = f (2) 7 = 7 = k ∴ k = 7

7. Given : drdt

= –3 cm/min; and

dhdt

= 2 cm/min; r = 9 cm; h = 6 cm

Volume of the Cone, V = 13

πr2h

Differentiating both sides w.r.t. time,

ddtV =

π3

2 2r h rdhdt

drdt

+

.

= π3

[(81(2) + 6(18) (– 3)]

[ r = 9 cm, h = 6 cm)

= π3

[162 – 324] = π3

(–162)

∴ Change in volume of cone = – 54π cm3/min.

∴ Volume is decreasing at the rate of 54π cm3/minute.

8. x x2 2−∫ dx

= x x2 2 22 1 1− + − dx = ( )x − −∫ 1 12 2 dx

y a dy y y a2 2 2 21

2− =

−∫− + − +a

y y a c2

2 22

log

= 1

21 1 12 2( ) ( )x x− − − −

12 log|( ) ( ) |x x c− + − −

+1 1 12 2

= 1

21 22( )x x x− − −

log|( )x x x c− + −

+1 22

9. y2 = 4ax …[Given equation of family of curves

⇒ yx

2 = 4a

Differentiating both sides w.r.t. x

x y

dydx

y

x

. .2 12

2

− = 0

y 2x ydydx

−

= 0

2x dydx

– y = 0

2x dydx

= y

dydx

yx

=2

or 2x dydx

– y = 0

10. Given differential equation is

dydx

+ 2y = e3x

Here ‘P’ = 2, Q = e3x

I.F = e edx dxP∫ ∫= 2 = e2x

Hence, the solution is y(I.F) = Q I.F.( ) dx∫ y(e2x) = e e dxx x3 2.∫ [am . an = am + n

y(e2x) = e dxx5∫ y(e2x) = e x5

5 + c

y = e

e

c

e

x

x x

5

2 25+

y = e x3

5 + ce–2x

11. Let A( 10 3ˆ î j+ ), B( 12 5ˆ î j− ) and C( λ ˆ î j+ 11 ).

Points A, B, C are collinear.

AB→

= m BC→

(12 – 10) i +(–5 – 3) j = m[(λ – 12) i + (11 + 5) j ] 2 8ˆ î j− = m(λ – 12) i + 16m j

2 = m(λ – 12) – 8 = 16m

2 = − 12

(λ – 12) − 12

= m …(i) [From (i)]

⇒ – λ + 12 = 4 ⇒ – λ = – 8 ⇒ λ = 8

12. Let the firm has x number of large vans and y number of small vans.

From the given information, we have the following corresponding constraint table.


Packages Cost Large van, x 200 ` 400 Small van, y 80 ` 200 1200 ` 3000 (at least) (maximum) Minimum Cost, Z = 400x + 200ySubject to the constraints : 200x + 80y ≥ 1200 ⇒ 5x + 2y ≥ 30 400x + 200y ≤ 3000 ⇒ 2x + y ≤ 15and x ≤ y, x, y ≥ 0Thus, required LPP to minimize cost is Z = 400x + 200ySubject to constraints : 5x + 2y ≥ 30 2x + y ≤ 15 x ≤ y; x, y ≥ 0

13. L.H.S. = tan–1 1 11 1+ − −+ + −

x xx x

= tan–1 1 2 1 21 2 1 2+ − −+ + −

cos coscos cos

θ θθ θ

Let x

x

x

=

⇒ =

⇒ =

−

−

cos

cos

cos

2

2

12

1

1

θ

θ

θ

= tan–1 2 2

2 2

2 2

2 2

cos sin

cos sin

θ θ

θ θ

−

+

1 2 2

1 2 2

2

2

+ =

− =

cos cos

cos sin

θ θ

θ θ;

= tan–1 22

(cos sin )(cos sin )

θ θθ θ−+

Dividing numerator and denominator by cos θ, we have

= tan–1 11−+

tantan

θθ

= tan–1 tan π θ4−

11 4−+

= −

tantan

tanθθ

π θ

= π4

– θ = π4

12

− cos–1 x = R.H.S. [ tan–1 (tan A) = A(Hence proved)

14. a b y c z

a x b c z

a x b y c

− −− −− −

= 0

[Applying R2→ R2 – R1; R3 → R3 – R1

a b y c z

x y

x z

− −−−

00

= 0

Expanding along R1, we have

⇒ a(yz) – (b – y) (–xz) + (c – z) (xy) = 0⇒ ayz + bxz – xyz + cxy – xyz = 0⇒ ayz + bxz + cxy = 2xyz

Dividing both sides by xyz, we have

ayzxyz

bxzxyz

cxyxyz

xyzxyz

+ + = 2

ax

by

cz

+ + = 2Or

Let A = 1 22 1−

A = I A [By Elementary Row Transformation

1 22 1−

= 1 00 1

A

1 20 5−

= 1 02 1−

A [R R R2 2 12→ −

1 20 1

= 1 02

5

1

5−

R R2 21

5→ −

1 0

10

=

1

5

2

52

5

1

5−

A [R R R1 1 22→ −

Now, I = A–1 A

∴ A–1 =

1

5

2

52

5

1

5−

15. x = a(cos t + t sin t)

Differentiating both sides w.r.t. t, dxdt

= a(– sin t + t cos t + sin t . 1) = a(t.cos t) …(i)

Again differentiating both sides w.r.t. t, 2

2d x

dt


= a[cos t . 1 + t (– sin t)] = a[cos t – t sin t] y = a(sin t – t cos t)

Differentiating both sides w.r.t. t, dydt

= a[cos t – (cos t . 1 + t(– sin t))] = a(t . sin t) …(ii)

Differentiating both sides w.r.t. t, 2

2d y

dt

= a[t . cos t + sin t]

Now, dydx

= dy dtdt dx

×

= at sin t × 1cosat t

[From (i) and (ii)]

= tan t

Differentiating w.r.t. x, 2

2d y

dx

= sec2 t . dtdx

= sec2 t × 1cosat t

= cos31

at t

16. y = cos(x + y) …(i)

Differentiating both sides w.r.t. xdydx

= – sin(x + y) . 1 +

dydx

dydx

= – sin(x + y) – sin(x + y) dydx

dydx

+ sin(x + y) dydx

= – sin(x + y)

[1 + sin(x + y)] dydx

= – sin(x + y)

dydx

= − ++ +sin( )sin( )

x yx y1 …(ii)

Since tangent is parallel to the line x + 2y = 0,

∴ Slope of tangent = − 12

…[From (i) and (ii)

∴ − ++ +sin( )sin( )

x yx y1 = − 1

2 2 sin(x + y) = 1 + sin(x + y) sin(x + y) = 1 …(iii)Since sin2 θ + cos2 θ = 1

sin2(x + y) + cos2(x + y) = 1 (1)2 + (y2) = 1 [From (i) and (iii)

y2 = 0 y = 0From (i), y = cos(x + y) 0 = cos(x + 0) (y = 0)

cos x = 0

∴ x = − 32π , [–2π ≤ x ≤ 0 (given)

∴ Point is −

32

0π ,

∴ Equation of tangent at −

32

0π , is

(y – 0) = −

+1

232

x π

2y = –x – 32π

4y = –2x – 3π∴ 2x + 4y + 3π = 0 is the required equation

of tangent.

17. x

x x

++ −∫ 5

3 13 102 dx

= 16

6 53 13 102

( )x

x x

++ −∫ dx

= 16

6 30 13 133 13 102

( )x

x x

+ + −+ −∫ dx

= 16

6 13

3 13 10176 3 13 102 2

( )x dx

x x

dx

x x

++ − + −∫ ∫+

Let p x x

dp x dx

= + −= +

3 13 10

6 13

2

( )

= 16

176 3 13

3103

2

dpp

dx

x x∫ ∫+

× + −

= 16

1718 13

3136

136

103

22 2log| |p dx

x x

++ +

−

−

∫

= 16

1718 13

6176

2 2log| |p dx

x

++

−

∫


= 16

1718

12 17

6

log| |p + × × logx

x

+ −

+ +

136

176

136

176

+ c

dxx a a

x ax a

c2 21

2−

= −+

+∫ log c

= 16

16

6 13 176 13 17

log| | logp cxx

+ ++ −+ +

= 16

16

6 46 30

3 13 102log| | logx x cxx

+ − + +−+

= 16

16

2 3 26 5

3 13 102log| | log ( )( )

x x cxx

+ − + +−+

= 1

6

1

6

3 2

3 53 2 13 10log| | ( )

( )x x c

x

x+ − ++ −

+

= 16

16

3 25

3 13 102log| | logx x xx

+ − + −+

− +16

3log c

= 16

16

3 25


+ − + +−+

K

… Where K = c – 16

log 3

Or

142

0

4

cos sin

/

x x+ 2∫π

dx

Dividing numerator and denominator by cos2 x, we have

= sec

tan

/ 2

20

4

1 4

x dx

x+∫π

= 12 1 2

0

2dp

p+∫

Let

When

When

p x

dp x dx

x p

x p

=

=

= =

= =

2

12

2

42

0 0

2

tan

sec

,

,

π

= 12

10

2tan ( )−

p

dxx

x c1 2

1+

= +

−∫ tan

= 1

2[tan–1(2) – tan–1 (0)]

= 1

2tan–1 2 tan− =

1 0 0

18. Let x

x x xx

x

2

2 21 1 1 1( )( )− + −++

= +A B C …(i)

x

x x

x x x

x x

2

2

2

21 11 1

1 1( )( )) ( )( )

( )( )− ++ + + −

− += A( B C

x2 = A(x2 + 1) + (Bx + C) (x – 1)

x2 = A(x2 + 1) + Bx(x – 1) + C(x – 1)Comparing the coefficients of x2, x and

constant term respectively,1 = A + B; 0 = –B + C; 0 = A – C1 = 2C ⇒ B = C ⇒ A = C 12

= C …(ii) …(iii)

[From (ii & (iii)]

∴ A = B = C = 12

Putting the value of A, B, C in (i), we get

x

x x x

x

x

2

2 21 1

12

1

12

121( )( )− + −

+

+= +

x

x x

2

21 1( )( )− +∫ dx

= 12 1

12

112

dxx

x

x−++

+∫ ∫ dx

= 12 1

12 1

12 12 2

dxx

x

x

dx

xdx

− + ++ +∫ ∫ ∫

Let p x

dpx dx

= +

=

2 1

2

= 12 1

14

12 1 2

dxx

dpp

dx

x− ++ +∫ ∫ ∫

= 12

14

12

1 1log| | log| | tanx p x c− + + +−

dxx

x1 2

1+

=

−∫ tan

= 12

14

12

1 12 1log| | log( ) tanx x x c− + + + +−

19. x cos yx

dydx

= y cos

yx

+ x

dydx

yyx

x

xyx

=+cos

cos


⇒ dydx

yx y

x

= +

1

cos Let y vx

yx

v

dydx

v xdvdx

= ⇒ =

∴ = +

. .1

⇒ v + x dvdx v

v= + 1cos

⇒ xdvdx v

= 1cos

⇒ cos v dv = dxx

⇒ cos v dvdx

x=∫ ∫

⇒ sin v = log |x| + log |C|

⇒ sin yx

= log |cx|

20. Let A, B, C, D be the given points.

AB→

= ( 5 7ˆ ˆ î x j k+ + ) – ( 4 3 3ˆ ˆ î j k+ + ) = ˆ ( ) ˆ î x j k+ − +3 4

AC→

= ( 5 3ˆ î j+ ) – ( 4 3 3ˆ ˆ î j k+ + ) = ˆ î k− 3

AD→

= ( 7 6ˆ ˆ î j k+ + ) ( 4 3 3ˆ ˆ î j k+ + ) = 3 3 2ˆ ˆ î j k+ −

The given points are coplanar if AB AC AD→ → →

, , are coplanar

∴ ( , , )AB AC AD→ → →

= 0

1 3 41 0 33 3 2

x −−−

= 0

Expanding along R1⇒ 1(9) – (x – 3) (–2 + 9) + 4(3) = 0⇒ 9 – 7x + 21 + 12 = 0⇒ –7x = –42⇒ x = 6

21. 1

37 14

23

1− − −= =x y

pz

= 0 …(i)

⇒ x y

pz−

−− −= =1

32

27

31

Direction ratios of line (i) are –3, 27p , 1.

7 73

51

117

− − −= =xp

y z …(ii)

⇒ xp

y z−−

−−

−−

= =137

51

117

Direction ratios of line (ii) are − 37

p , –1, –7. Lines are at right anglesUsing a1a2 + b1b2 + c1c2 = 0

− + − + −−

3 1 1 737

27

p p ( ) ( ) = 0

9 2 497

p p− − = 0

⇒ 7p – 49 = 0⇒ 7p = 49⇒ p = 7

OrThe required equation of plane is

(x + y + z – 1) + p(2x + 3y + 4z – 5) = 0⇒ (1 + 2p)x + (1 + 3p)y

+ (1 + 4p)z – 1 – 5p = 0⇒ (1 + 2p)x + (1 + 3p)y + (1 + 4p)z = 1 + 5p …(i)Dividing both sides by (1 + 5p), we have

⇒ ( ) ( ) ( )1 2

1 51 31 5

1 41 5

++

++

++

+ +p xp

p yp

p zp

= 1

⇒ x

pp

ypp

zpp

1 51 2

1 51 3

1 51 4

++

++

++

+ + = 1

2(y-intercept) = 3(z-intercept) …(Given)

2 31 51 3

1 51 4

++

++

=pp

pp p ≠ −

15

3(1 + 3p) = 2(1 + 4p)⇒ 3 + 9p = 2 + 8p⇒ 9p – 8p = 2– 3⇒ p = –1Putting the value of p in (i), we have⇒ (1 – 2)x + (1 – 3)y + (1 – 4)z = 1 – 5⇒ –x – 2y – 3z = –4or x + 2y + 3z = 422. 4x + y ≥ 80; x + 5y ≥ 0, 3x + 2y ≤ 150Let 4x + y = 80; x + 5y = 115; 3x + 2y = 150

x 0 20 15 2y 80 0 20 72

x 0 115 40y 23 0 15

x 0 50 40y 75 0 15


C(40, 1

5)

A

(15, 20)

80

90100

70

60

50

40

30

20

10

0 XX′Y′

Y

B(2, 72)

3x + 2y = 150

x + 5y = 115

Solution set

4x + y = 80

10 20 30 40 50 60 70 80 90 100110 120

Corner points z = 6x + 3y

A(15, 20)B(2, 72)C(40, 15)

90 + 60 = 150 (Minimum)12 + 216 = 228240 + 45 = 285

Minimum value of Z is 150 at A (15, 20)i.e., x = 15, y = 20.23. Let E1 : Very hard working studentsE2 : Regular but not so hard working studentsE3 : Careless and irregular studentsA : Student who could not get good marks

in the examination

P(E1) = 1060

16

= , P(E2) = 3060

36

= ,

P(E3) = 20

6026

= ,

P(A/E1) = 0.002, P(A/E2) = 0.02

P(A/E3) = 0.20

By Bayes’ theorem,

P(E3/A) = P E P A E

P E P A E P E P A EP E P A E

( ) ( / )[ ( ) ( / )] [ ( ) ( / )]

[ ( ) ( /

3 3

1 1 2 2

3 3

×× + ×

+ × ))]

=

26

0 20

16

0 00236

0 0226

0 20

×

×

+ ×

+ ×

.

. . .

= 0 40

0 002 0 06 0 40.

. . .+ + = 0 40

0 462..

= 400462

= 200231

24. See Q. 25 (Or), 2017 Comptt. (I Outside Delhi). [Page 226

OrFor commutativeLet (a, b), (c, d) ∈ A(a, b) (c, d) = (a + c, b + d)(c, d) (a, b) = (c + a, d + b) = (a + c, b + d)∴ (a, b) (c, d) = (c, d) (a, b)∴ is commutative.For associative :Let (a, b), (c, d), (e, f) ∈ A((a, b) (c, d)) (e, f) = (a + c, b + d) (e, f) = (a + c + e, b + d + f)(a, b) ((c, d) (e, f) = (a, b) (c + e, d + f) = (a + c + e, b + d + f)∴ {(a, b) (c, d)} (e, f) = (a, b) {(c, d) (e, f)}∴ is associative.For (a, b) ∈ R × R, if (c, d) ∈ R × R is identity

element, then(a, b) (c, d) = (c, d) (a, b) = (a, b)(a + c, b + d) = (c + a, d + b) = (a, b)∴ a + c = a and b + d = b c = 0 and d = 0⇒ (0, 0) R × R is the identity element.

25.

2 11 03 4 3 2

−

−

×

A =

− − −− −

×

1 8 101 2 59 22 15 3 3

From the above equation, order of A should be 2 × 3.

Let A = a b c

d e f

2 11 03 4

−

−

a b c

d e f

=

− − −− −

1 8 101 2 59 22 15

2 2 2

3 4 3 4 3 4

a d b e c f

a b c

a d b e c f

− − −

− + − + − +

= − − −

− −

1 8 101 2 59 22 15

By equality of matrices, we have


A

B C

D

x

x

h

α

A

B

D

C

FEx

x

y

(h – y)h

r

a = 1, b = –2, c = –52a – d = – 1 2b – e = – 8 2c – f = – 102(1) + 1 = d 2(– 2) + 8 = c 2(– 5) + 10 = f3 = d 4 = e 0 = f

∴ A = 1 2 5

3 4 0

− −

26. f (x) = sin x + cos x, 0 ≤ x ≤ 2π

∴ f ′(x) = cos x – sin x f (x) = 0, ∴ cos x – sin x = 0⇒ – sin x = – cos x ⇒ tan x = 1

∴ x = π π4

54

, as 0 ≤ x ≤ 2π

0 π4

π2

π 2π54π 3

2π

The points x = π4

and x = 54π , divide the

interval [0, 2π] into 3 disjoint intervals —

04

, π

,

π π4

54

,

,

54

2π π,

Intervals Testvalue

Sign of f ′(x) Nature of function

04

, π

x = π6

cos π6

– sin π6

= 32

12

− > 0

f is strictly increasing

π π4

54

,

x = π2

cos π2

– sin π2

= 0 – 1 < 0

f is strictly decreasing

54

2π π,

x = 3

2π cos 3

2π – sin 3

2π

= 0 + 1 > 0

f is strictly increasing

Hence f is strictly increasing in the intervals

04

,π

∪

54

2π π,

and f is strictly decreasing

in the interval π π4

54

,

.

Or1st part :Let cylinder of radius x

and height y is inscribed in a cone of radius r.

ΔABC ~ ΔADF(AA rule)

∴ ABAD

BCDF

= (sides are proportional)

h yh

xr

− =

r h yh

( )− = x …(i)

Vol. of cylinder, V = πx2y

V = π r

h

2

2 .(h – y)2y [From (i)]

ddyV

= πr

h

2

2 [(h – y)2 . 1 + y.2(h – y).(–1)]

= πr

h

2

2 (h – y) [h – y – 2y]

= πr

h

2

2 (h – y) (h – 3y)

For maximum, ddyV

= 0

πr

h

2

2 (h – y) = 0 h – 3y = 0

h – y = 0 ⇒ –3y = –h

⇒ h = y, which ⇒ y = h3

is not possible.

d

dy

r

h

2

2

2

2V = π [(h – y) (–3) + (h – 3y) (–1)]

d

dy at yh

2

2

3

V

=

= πr

h

h hh h2

2 333

3 1− − + − −

( ) ( )

= πr

h

h2

223

3

−( ) < 0 (–ve)

Volume is maximum at y = 13

h.

∴ Height of cylinder = 13

(Height of cone)2nd part :Let x be the radius of the

cylinder which is inscribed in the cone of given height ‘h’ and semi-verticle angle ‘α’.

In rt. ΔABC,

tan α = BCAB

tan α = xAB


OD B

A12

2, −

C12

2, −

4x2 + 4y2 =

9 y2 = 4x

x = 0

x = 12

x = 32

X′

Y′

X

Y

•

• ••

•

⇒ 4x2 + 16x – 9 = 0⇒ 4x2 + 18x – 2x – 9 = 0⇒ 2x(2x + 9) – 1(2x + 9) = 0

⇒ x = − 92

(rejected) or x = 12

Putting x = 12

in (i), we get

y2 = 412

= 2 ∴ y = ± 2

∴ Points of intersection are

12

12

2 2, ,

−and .

From equation (i), y = 4x = 2x1/2

From equation (ii), y = 94

2− x

= 32

22

− x

Area of the bounded region OADCO,

A1 = 2 2 2 1 2

0

1 2

0

1 2

y dx x dx( ) ///

parabola = ∫∫

= 4 4 0

32

0

12

3 2

32

23

12

x

= −./

= 83

12

12

43

12

. .

=

= 4

3 222

4 26

2 23

. = = sq. units

AB = xtanα

= x cot α

Height of cylinder, BD = AD – AB = (h – x cot α) …(i)Volume of cylinder = π(BC)2 (BD) V = πx2(h – x cot α) [From (i)

V = πhx2 – πx3 cot α …(ii)Differentiating both sides w.r.t. x, we have

ddxV = 2πhx – 3πx2 cot α

For maxima or minima, ddxV = 0

πx(2h – 3x cot α) = 0 πx = 0 2h – 3x cot α = 0⇒ x = 0 –3x cot α = –2h

(rejected) x = 23

23

h hcot α

= tan α

d

dx

2

2V = 2πh – 6πx cot α

d

dx xh

2

2 23

V

at

= tanα = 2πh – 6π

23h tanα

cot α

= 2πh – 4πh = 2πh < 0 (–ve)

∴ Volume is maximum at x = 23h tan α.

Now from (ii),Volume of greatest cylinder,

V = π α π αh h h49

827

22

33tan tan

− cot α

= 49

827

32

32π πα αh htan tan−

= 12 827

3 2 3 2π α π αh htan tan−

= 427

πh3 tan2 α cu. units

27. Given curves are y2 = 4x …(i) 4x2 + 4y2 = 9

⇒ x2 + y2 = 94

…(ii)

Equation (ii) is a circle with centre (0, 0) and

radius = 32

.

Solving equations (i) and (ii), we get 4x2 + 4(4x) = 0


Area of the bounded region ABCDA,

A2 = 2 21 2

3 2 22

1 2

3 232

y dx x dx(circle) = −∫ ∫

/

/

/

/

= 22

32

322 3

2

22

2

1

1 2

3 2

x xx

− + −( ) sin

/

/

= 22

94

98

23

2 1

1 2

3 2x xx− + −

sin

/

/

= 2 0 98

23

32

1+ −

sin .

− − +

−122

94

14

98

23

12

1sin .

= 2 198

14

84

98

13

1 1sin ( ) sin− −− −

= 2 298 2

14

98

13

1. sinπ − − −

= 98

22

94

13

1π − − −sin

∴ Total area of the bounded region OABCO = A1 + A2

= 2 23

98

22

94

13

1+ − − −π sin

= 26

98

94

13

1+ − −π sin Sq. units

28. Given lines : x y z− +

−−= =8

319

1610

7, and

x y z− + −−

= =383

298

55

Here x1 = 8, y1 = –19, z1 = 10 a1 = 3 b1 = –16 c1 = 7 a2 = 3 b2 = 8 c2 = –5The equation of plane containing given line is

x x y y z z

a b c

a b c

− − −1 1 1

1 1 1

2 2 2

= 0

x y z− + −

−−

8 19 103 16 73 8 5

= 0

Expanding along R1, we have (x – 8) (80 – 56) – (y + 19) (–15 – 21)

+ (z – 10) (24 + 48) = 0 24(x – 8) + 36(y + 19) + 72(z – 10) = 0Dividing both sides by 12 2(x – 8) + 3(y + 19) + 6(z – 10) = 0 2x – 16 + 3y + 57 + 6z – 60 = 0 2x + 3y + 6z – 19 = 0or 2x + 3y + 6z = 19 is the required

equation of the plane.

29. Here n = 8, p = P(a head) = 12

,

q = 1 12

12

− =

P(r) = nCr.qn – r pr

(i) P(exactly 5 heads) = P(5)

= 8C5q3p5

= 56 12

12

3 5

85

83

8 7 63 2 1

56C C= = =. .. .

= 56 18

132

× × = 732

(ii) P(at least 6 heads) = P(6) + P(7) + P(8) = 8C6q

2p6 + 8C7q1p7 + 8C8q

0p8

= 82

2 68

1

78

0

812

12

12

12

12

C C C

+

+

= 12

8

(8C2 + 8C1 + 8C0)

= 1256

8 72 1

81

1..+ +

= 37256

(iii) P(at most 6 heads) = P(0) + P(1) + … + P(6) = 1 – [P(7) + P(8)]

= 1 – [8C7 q1p7 + 8C8 q0p8]

= 1 – 81

1 78

0

812

12

12

C C

+

= 1 8 1 112

92568− + = − =( ) 247

256Or

p = P(red card) = 2652

12

= ; q = 1 12

12

− =

Also, n = 3

Here random variable X can take values 0,

1, 2, 3.


P(r) = nCr qn – r pr

P(X = 0) = 3C0 q3p0 = 12

3

= 1

8

P(X = 1) = 3C1 q2p1 = 3 12

12

2

= 38

P(X = 2) = 3C2 q1p2 = 3 12

12

2

= 3

8

P(X = 3) = 3C3 q0p3 = 12

3

= 18

Xi Pi PiXi PiXi2

0 18

0 0

1 38

38

38

2 38

68

128

3 18

38

98

ΣPiXi = 12

8 ΣPiXi

2 = 248

∴ Mean = ΣPiXi = 128= 3

2

∴ Variance = ΣPiXi2 – (Mean)2

= 248

32

2−

= 3 94

12 94

− = =− 34

or 0.75

set ii10.

A

3 1

C B

12 ˆ î j+ µ11 3ˆ î j−λ ˆ î j+ 3

Using Section Formula,Position vector of point C = Position vector

of point C

3( 12 ˆ î j+ µ ) + 1(λ ˆ î j+ 3 )

3 + 1 = 11 3ˆ î j−

36 3 34

ˆ ˆ ˆ î j i j+ + +µ λ = 11 3ˆ î j−

(36 + λ) i (3μ +3) j

4 4+ = 11 3ˆ î j−

Equating the corresponding components

364+ λ = 11 3 3

4µ + = – 3

36 + λ = 44 3μ + 3 = –12 ⇒ 3μ = –15

λ = 8 μ = –5

11. 2 2x x−∫ dx

= − −∫ ( )x x2 2 dx

= − − + −∫ [( ) ]x x2 2 22 1 1 dx

= − − −∫ [( ) ]x 1 12 2 dx

= 1 12 2− −∫ ( )x dx

= 1

2

1

11 1 1 12 2 2 1( ) ( ) sinx x c

x− − − +

+− −

… a x dx x a x a c2 2 2 2 212

− = − +( ) +

∫

= ( )sin ( )

xxx x c− −− + +−1

212

12 2 1

12. lim ( )x

f x→0

= lcosim

x

x

x→

−0 2

1 4

= lsinim

x

x

x→0

2

22 2

1 22

2− =

cos sinAA

= lsinim

x

x

x→

×0

2

24 2 2

4

= 8 lsinim

xx

xx→

→

0

2 0

222

= 8 × 1 limsin

A

AA→

=0

1

= 8

At x = 0, f (x) = p ⇒ f (0) = p

Since f (x) is continous at x = 0, (Given)

∴ lim ( )x

f x→0

= f (0)

∴ 8 = p


20. 3x + 4y ≤ 60, x + 3y ≤ 30Let 3x + 4y = 60 Let x + 3y = 30

x 0 20 12 y 15 0 6

x 0 30 12 y 10 0 6

Corner Points

Z = 8000x + 12000y

A(0,0)B(0, 10)C(12, 6)D(20, 0)

01,20,00096,000 + 72,000 = 1,68,000 (Maximum)1,60,000

X′ X

Y

Y′

60

50

40

20

30

10

(0, 0) A 10 20 30 40 50 60

B

D(20, 0)

C(12, 6)(0, 10)

3x + 4y = 60

x + 3y = 30 →

Maximum value of Z is 1,68,000 at C(12, 6)

i.e., x = 12, y = 6.

21. x

x x

++ +∫ 7

3 25 282 dx

= 16

6 73 25 282

( )x

x x

++ +∫ dx

= 16

6 42 25 253 25 282

( )x

x x

+ + −+ +∫ dx

= 16

6 25

3 25 28176 3 25 282 2

( )x dx

x x

dx

x x

++ + + +∫ ∫+

Let p = 3x2 + 25x + 28 dp = (6x + 25) dx

= 16

176 3 25

3283

2

dpp

dx

x x∫ ∫+

× + +

= 16

1718 25

3256

283

256

22 2log| |p dx

x x

++ +

+ −

∫

= 16

1718 25

6176

2 2log| |p dx

x

++

−

∫

= 16

1718

1

2176

256

176

256

176

log| | logp cx

x+ +×

×

+ −

+ +

dxx a a

x ax a

c2 21

2−= −

++

∫ log

= 16

16

6 25 176 25 17

log| | logp cxx

+ ++ −+ +

= 16

16

2 3 46 7

log| | log ( )( )

p cxx

+ +++

= 16

16

3 47


+ + + ++

− +16

3log c

= 16

16

3 47


k+ + + ++

+ ,

where k = c – 16

log 3

22. x = 3 cos t – 2 cos3 t

Differentiating w.r.t. t, we have

dxdt

= –3 sin t + 6 cos2 t sin t

= 3 sin t(2 cos2 t – 1) = 3 sin t . cos 2t [ cos cos 2 1 22 t t− =

…(i) y = 3 sin t – 2 sin3 t

Differentiating w.r.t. t, we have

dydt

= 3 cos t – 6 sin2 t . cos t

= 3 cos t(1 – 2 sin2 t)

= 3 cos t . cos 2t [1 – 2 sin2 t = cos 2t]

…(ii)

From (i) and (ii), dydx

= dydt

dtdx

×

= 3 cos t cos 2t × 13 2sin . cost t

∴ dydx

= cot t

Differentiating both sides w.r.t. x, we have

d y

dx

2

2 = – cosec2 t . dtdx


= −×

1 13 22sin sin cost t t

[From (i)

= − 1

3 23sin . cost t

28. X = 1 2 34 5 6 2 3

×

= − − −

×

7 8 92 4 6

11 10 9 3 3

From the above equation, order of X should be 3 × 2.

Let X =

a b

c d

e f

a b

c d

e f

1 2 34 5 6

=

− − −

7 8 92 4 6

11 10 9

a b a b a b

c d c d c d

e f e f e f

+ + ++ + ++ + +

4 2 5 3 64 2 5 3 64 2 5 3 6

= − − −

7 8 92 4 6

11 10 9

By equating the matrices, we havea + 4b = –7 c + 4 d = 2 e + 4f = 11 …(i) …(iii) …(v)2a + 5b = –8 2c + 5d = 4 2e + 5f = 10 …(ii) (iv) …(vi)From (i) and (ii), From (iii) and (iv), From (v) and (vi),a = 1 and b = –2 c = 2 and d = 0 e = –5 and f = 4

∴ X =

1 2

2 0

5 4

−

−

29. Here x1 = – 3, y1 = 1, z1 = 5 x2 = – 1, y2 = 2, z2 = 5a1 = – 3, b1 = 1, c1 = 5 a2 = – 1, b2 = 2, c2 = 5

Now, 2 1 2 1 2 1

1 1 1

2 2 2

x x y y z z

a b c

a b c

− − −

= 2 1 03 1 51 2 5

−−

= 2(5 – 10) – 1(– 15 + 5) (Expanding along R1)

= –10 + 10 = 0

∴ Given lines are coplanar.Equation of plane containing given lines is

( 3) 1 53 1 51 2 5

x y x− − − −−−

= 0

⇒ (x + 3) (5 – 10) – (y – 1) (– 15 + 5) + (z – 5) (– 6 + 1) = 0

( Expanding along R1)

⇒ – 5(x + 3) + 10(y – 1) – 5(z – 5) = 0⇒ – 5x – 15 + 10y – 10 – 5z + 25 = 0⇒ – 5x + 10y – 5z = 0or x – 2y + z = 0 is the required equation of

the plane.

set iii

10. lim ( )x

f x→π

4

= lim sin cos

x

x xx→

−−π π

44

= limsin cos

p

p p

p→

−

− −

−

−

0

4 4

44

π π

π π

Let

when

π

π

π

4

4

40

− =

− =

→

→

x p

p x

x

p

= limsin cos cos sin cos cos sin sin

p

p p p p

→

−

− +

0

4 4 4 4π π π π

π −− −4p π

= limcos sin cos sin

p

p p p p

p→

− − −

−0

12

12

12

12

4

= limsin

p

p

p→

−

−0

22

4 =

24 0

lim sin

p

pp→

= 24

limsin

p

pp→

=

01

At x = π4

, f (x) = k ⇒ f π4

= k

Since f (x) is continuous at x = π4

, …(Given)

lim ( )x

f x f→

=

π

π

44


∴ 24

= k

11. dx

x x2 4−∫

= dx

x x2 2 24 2 2− + −∫

= dx

x( )− −∫2 22 2

= log |( ) ( ) |x x c− + − − +2 2 22 2

dx

x ax x a c

2 22 2

−= + − +

∫ log | |

= log | |x x x c− + − +2 42

12. dydx x

y+ 2 = x

Here ‘P’ = 2x

, ‘Q’ = x

I.F. = e e e e xdx xdx x xP∫ ∫

= = = =2

2 22log| | log

Here the solution is y(I.F) = Q I.F.( ) dx∫ y(x2) = x x dx. 2∫ y(x2) = x dx3∫ y(x2) = x c

4

4+

y = x

x

c

x

4

2 24+

y = xcx

22

4+ −

20. x + y ≤ 50, 2x + y ≤ 80Let x + y = 50; Let 2x + y = 80

x 0 50 30y 50 0 20

x 0 40 30y 80 0 20

Corner points Z = 105x + 90y

A(0, 50)B(30, 20)C(40, 0)D(0, 0)

0 + 4500 = 45003150 + 1800 = 4950 (Maximum) 4200 + 0 = 42000 + 0 = 0

80

•

70

60

50

40

30

20

10

0 10 20 30 40 50 60 70 80X′ X

Y′

Y

A(0, 50)

B(30,

0)C(4

0, 20

)

D(0, 0)

2x + y = 80

x + y = 50

Maximum value of Z is 4950 at B (30, 20)i.e., x = 30, y = 20.

21. x

x x

+

− +∫ 3

5 4 2 dx

= 12

2 3

4 52

( )x

x x

+

− +∫ dx

= 12

2 6 4 4

4 52

( )x

x x

+ − +

− +∫ dx

= 12

2 4

4 5

102 4 52 2

( )x dx

x x

dx

x x

−

− + − +∫ ∫+

Let p x x

dp x dx

= − += −

2 4 5

2 4( )

= 12 4 2 5 2

52 2 2

dpp

dx

x x+∫ ∫ − + + −

= 12 2 1

1 22 2

5p dp dx

x− +∫ ∫ − +

/

( )

= 12

21

1 2 25 2 2 1× + − + − + +p x x c/ log ( ) ( )

dz

x ax x a c

2 22 2

+= + + +

∫ log

= x x2 4 5 5− + + log |( ) |x x x c− + − + +2 4 52

22. Ist method

L.H.S. = cot–1 1 sin 1 sin1 sin 1 sin

x xx x

+ + − + − −


cot–1 1 sin 1 sin 1 sin 1 sin1 sin 1 sin 1 sin 1 sin

x x x xx x x x

+ + − + + −× + − − + − −

= cot–1 1 sin 1 sin 2 1 sin 1 sin1 sin (1 sin )

x x x xx x

+ + − + + − + − −

= cot–1 2 2 cos2 sin

xx

+

= cot–1 2(1 cos )2 sin

xx

+

21 cos 2 cos2

sin 2 sin cos2 2

xx

x xx

+ = =

= cot–1

22 cos2

2 sin cos2 2

x

x x

= cot–1 2 2

cot x x = = R.H.S.

2nd Method

L.H.S. = cot–1 2 2

2 2

cos sin 2 sin cos2 2 2 2


x x x x

x x x x

+ +

+ +

2 2

2 2



x x x x

x x x x

+ + −

− + −

= cot–1 cos sin cos sin

2 2 2 2

cos sin cos sin2 2 2 2

x x x x

x x x x

+ + − + − −

= cot–1 2 cos

2

2 sin2

x

x

= cot–1 2

cot x

= 2x = R.H.S.

28. See Q. 23, 2016 (I Delhi). [Page 141

Ans. a2π4

23

−

29. Let a, b, c be the direction ratios of the normal to the required plane.

Equation of plane through the point (–1, 3, 2) is a(x + 1) + b(y – 3) + c(z – 2) = 0 …(i) Plane (i) is ⊥ to plane x + 2y + 3z = 5∴ a + 2b + 3c = 0 …(ii) Plane (i) is ⊥ to plane 3x + 3y + z = 0 ∴ 3a + 3b + c = 0 …(iii)

[ a1a2 + b1b2 + c1c2 = 0

Solving (ii) and (iii), we get

a b c2 33 1

1 33 1

1 23 3

= =−

a b c2 9 1 9 3 6−

−− −

= =

⇒ a b c−

−− −

= =7 8 3

or a b c7 8 3= =−

= λ (Say)

∴ a = 7λ, b = –8λ, c = 3λ

Putting the values of a, b and c in (i), we get

7λ(x + 1) + (–8λ) (y – 3) + 3λ(z – 2) = 0Dividing both sides by λ, we have 7(x + 1) – 8(y – 3) + 3(z – 2) = 0 7x + 7 – 8y + 24 + 3z – 6 = 0⇒ 7x – 8y + 3z + 25 = 0 is the required

equation of the plane.


set i

section A


Q.1. If the following function f (x) is continuous at x = 0, then write the value of k.

f (x) = sin

,

,

32 0

0

x

xx

k x

≠

=

Q.2. If A is an invertible matrix of order 2 and det (A) = 4, then write the value of det(A–1).

Q.3. If ( ) ( . )a b a b→ → → →× + =2 2 225 and | |a

→= 5 then write the value of | |b

→.

Q.4. Find 3

3 1x

x −∫ dx.

section B

Question numbers 5 to 12 carry 2 marks each. Q.5. Find the values of x and y from the following matrix equation :

25

7 33 41 2

7 615 14

x

y −

+

−

=

Q.6. If f (x) = sin 2x – cos 2x, find ′ f π

6.

Q.7. A particle moves along the curve 6y = x3 + 2. Find the points on the curve at which y-coordinate is changing 2 time as fast as x-cordinate.

Q.8. Find : x

x32 2−∫ dx.


General InstructIons Same as in Delhi Board 2015.

cbse

mathematics – 2017Comptt. (outside deLHi)

mathematics – 2017 comptt. (outside delhi) ■ 217

Q.9. Find the general solution of the differential equation log dydx

= 3x + 4y.

Q.10. Find the integrating factor of the differential equation dydx

yx

y+ = +1 .

Q.11. If the vectors a i j k b i j k→ →

= + + = − −ˆ ˆ ˆ, ˆ ˆ ˆ3 2 and c i j k→

= + +λ ˆ ˆ ˆ7 3 are coplanar, then find the value of λ.

Q.12. An aeroplane can carry a maximum of 250 passengers. A profit of ` 1,500 is made on each executive class ticket and a profit of ` 1,000 is made on each economy class ticket. The airline reserves at least 25 seats for executive class. However, at least 3 times as many passengers prefer to travel by economy class than by executive class. Frame the Linear Programing Problem to determine how many tickets of each type must be sold in order to maximize the profit for the airline.

section c

Question numbers 13 to 23 carry 4 marks each. Q.13. A school wants to award its students for regularity and hardwork with a a total cash award

of ` 6,000. If three times the award money for hardwork added to that given for regularity amount to ` 11,000, represent the above situation algebraically and find the award money for each value, using matrix method.

Q.14. Find the real solution of the equation tan–1 11

12

−+

=

xx

tan–1 x, (x > 0).

Q.15. If y = (cos x)x + sin–1 3x , find dydx

.

Or

If y = (sec–1 x)2, then show that x2(x2 – 1) d y

dx

2

2 + (2x3 – x) dydx

= 2.

Q.16. Find the intervals in which the function given by f (x) = 2x3 – 3x2 – 36x + 7 is (a) Strictly increasing (b) Strictly decreasing

Q.17. Evaluate : x x

x

sincos1 2

0+∫

π dx.

Q.18. Find : x x

x x

2

21

1 2+ +

+ +∫ ( ) ( ) dx.

Or

Find : ( )x x x− − −∫ 3 3 2 2

dx.

Q.19. Find the general solution of the differential equation 2xy

dydx

= x2 + y2.

Q.20. If a b c→ → →

, and are mutually perpendicular vectors of equal magnitudes, find the angles

which the vector 2 a b c→ → →+ + 2 makes with the vectors a b c

→ → →, and .


Q.21. Find the co-ordinates of the foot of perpendicular drawn from a point A(1, 8, 4) to the line joining the points B(0, –1, 3) and C(2, –3, –1).

Q.22. Solve the following Linear Programming problem graphically : Minimize : z = 3x + 9y When : x + 3y ≤ 60, x + y ≥ 10, x ≤ y, x ≥ 0, y ≥ 0 Q.23. Bag A contains 3 red and 2 black balls, while bag B contains 2 red and 3 black balls. A ball

drawn at random from bag A is transferred to bag B and then one ball is drawn at random from bag B. If this ball was found to be a red ball, find the probability that the ball drawn from bag A was red.

Or If A and B are two independent events, then prove that the probability of occurrence of

atleast one of A and B is given by 1 – P(A′).P(B ′).

section D


Q.24. If a + b + c ≠ 0 and

a b c

b c a

c a b = 0, then using properties of determinants, prove that

a = b = c.

Q.25. Given a non-empty set X, consider the binary operation : P(X) × P(X) → P(X) given by A B = A ∩ B, ∀ A, B ∈ P(X), where P(X) is the power set of X. Show that is commutative and associative and X is the identity element for this operation and X is the only invertible in P(X) with respect to the operation . (Not in syllabus)

Or

Let f : IR – −

43

→ IR be a function defined as f(x) = 43 4

xx +

. Show that f is a one-

one function. Also check whether f is an onto function or not. Hence f –1 in

(Range of f) → IR – −

43

.

Q.26. Show that the right circular cone of least curved surface and given volume has an altitude equal to 2 times the radius of the base.

Or (i) Find the equation of tangent to the curve x = a cos θ + a θ sin θ, y = a sin θ – a θ cos θ

at any point θ of the curve. (ii) Also show that at any point θ of the curve the normal is at a constant distance from

origin.

Q.27. Find the distance between the line x y z− − −= =53

43

81

and the plane determined by the

points A(2, –2, 1), B(4, 1, 3) and C(–2, –2, 5). Q.28. A coin is biased so that the head is 4 times as likely to occur as tail. If the coin is tossed

thrice, find the probability distribution of number of tails. Hence find the mean and variance of the distribution.


Q.29. Using method of integration find the area of the triangle ABC, co-ordinates of whose vertices are A(1, –2), B(3, 5) and C(5, 2).

Or Evaluate the following definite integral as limit of sums :

( )3 2 12

0

4

x x+ +∫ dx

set ii

Note: Except for the following questions, all the remaining questions have been asked in Set I.

Q.10. Find the general solution of the differential equation xy dydx

= (x + 2) (y + 2).

Q.11. Find : cos

sin

x dx

x8 2−∫ Q.12. The radius r of a right circular cylinder is increasing at the rate of 5 cm/min and its height

h, is decreasing at the rate of 4 cm/min. When r = 8 cm and h = 6 cm, find the rate of change of the volume of cylinder.

Q.21. Evaluate : x xx xtan

sec tan+∫0

π dx

Q.22. Solve the differential equation y e dx xe y dy y

xy

xy. . ( ) , ( )= + ≠2 0 .

Q.23. Find the co-ordinates of the foot of perpendicular drawn from the point (2, 3, –8) to the line 4

2 61

3− −= =x y z

.

Q.28. From a lot of 30 bulbs, which includes 6 defectives, 3 bulbs are drawn one by one at random, with replacement. Find the probability distribution of number of defective bulbs. Hence find the mean and variance of the distribution.

Q.29. Find the distance between the line r i j k i j k→

= + − + + +3 5 2 3 3ˆ ˆ ˆ ( ˆ ˆ ˆ)λ and the plane

determined by the points A(1, 1, 0), B(1, 2, 1) and C(–2, 2, –1).

set iii

Note: Except for the following questions, all the remaining questions have been asked in Set I and Set II.

Q.10. If the vectors ˆ ˆ ˆ, ˆ ˆ ˆ ˆ ˆ ˆi j k i j k i j k− + + + + −3 2 3and λ are coplanar, then find the value of λ.

Q.11. Find the particular solution of the differential equation dydx

y

x= +

+1

1

2

2, given that y(0) = 3 .

Q.12. A balloon, which always remains spherical, has a variable diameter 23

(3x + 1). Find the

rate of change of its volume with respect to x.


Q.21. Solve the following linear programming problem graphically : Find the maximum value of z = 105x + 90y when x + y ≤ 50, 2x + y ≤ 80, x ≥ 20, x ≥ 0, y ≥ 0

Q.22. Find the particular solution of the differential equation 2xy + y2 – 2x2 dydx

= 0; y = 2

when x = 1.

Q.23. Evaluate : x dxx1

0+∫ sin

π.

Q.28. Find the co-ordinates of the point of intersection P of the line r i j k→

= − +2 2ˆ ˆ ˆ +

λ( 3 4 2ˆ ˆ ˆi j k+ + ) and the plane determined by points A(1, –2, 2), B(4, 2, 3) and C(3, 0, 2).

Q.29. A biased die is such that P(4) = 1

10 and other scores are equally likely. The die is tossed

twice. If X is the ‘number of fours obtained’, find the variance of X.

SolutionS

1. 0

lim ( )x

f x→

= 0

3sin

2limx

x

x→

= 0

30

2

33 2lim 32

2x

x

x

x→

→

[at x = 0, f(x) = k, f (0) = k

= 32

(1) = 32

θ→

θ = θ

0

sinlim 1

f (x) is continuous at x = 0. [Given]

⇒ 0

( ) (0)limx

f x f→

= .

∴ 32

= k

2. det (A–1) = 1det (A)

= 14

3. 2 2( ) ( . )a b a b→ → → →× + = 225

But 2 2( ) ( . )a b a b→ → → →× + = 2 2| | | |a b

→ →

∴ 2 2| | | |a b→ →

= 225

(5)2 2| |b→

= 225

2| |b→

= 22525

= 9 ∴ →

| |b = 3

4. 33 1

xx −∫ dx

= [(3 1) 1]3 1x

x− +−∫ dx

= 13 1

1x

−

+∫ dx

= x + 13

log|3x – 1| + c

5. 25 3 4 7 6

7 3 1 2 15 14x

y

− + = −

2 10 3 4 7 614 2 6 1 2 15 14

x

y

− + = −

2 3 6 7 615 2 4 15 14

x

y

+ = −

2x + 3 = 7 2y – 4 = 14 2x = 4 2y = 18∴ x = 2 ∴ y = 96. f (x) = sin 2x – cos 2xDifferentiating both sides w.r.t. x, we get f´ (x) = 2 cos 2x + 2 sin 2x

6

f π ′ = 2 cos

3π + 2 sin

3π

= 1 32 2

2 2

+ = +(1 3)


7. y-coordinate is changing 2 times as fast as x-coordinate.

∴ dydt

= 2 dxdt

…(i)

6y = x3 + 2

y = 3 2

6 6x + …(ii)

Differentiating both sides w.r.t. t

dydt

= 216

. 3 dxdt

x

2 dxdt

= 212

dxdt

x [From (i)]

x2 = 4 ⇒ x = ± 2

From (ii), when x = 2, y = 53

From (ii), when x = –2, y = –1

∴ Required points are

53

2, and (–2, –1).

8. 232

x

x−∫ dx

= – 12

dpp∫

= −

= − = −

2Let 32

2

2

p x

dp x dx

dpx dx

= 121

2p dp−

− ∫ =

121 2

2 1p c− × +

= 232 x− − + c

9. log dydx

= 3x + 4y

dydx

= e3x + 4y ⇒ dy

dx = e3x . e4y

⇒ dy

e y4 = e3x dx

e dy e dxy x−∫ ∫=4 3

e ey x−

−=

4 3

4 3 + c

10. 1dy ydx x

y ++ =

1dy ydx x x

y+ = +

1dy ydx x x

y+ − =

1 11dydx x x

y + − =

Comparing with dydx

+ Py = Q

‘P’ = 1 – 1x

and ‘Q’ = 1x

Integrating Factor (I.F)

= −∫ ∫ =

11 dx

Pdx xe e = −log| |orx

x x ee

x11. Vector , anda b c

→ → →are coplanar

1 3 12 1 1

7 3− −

λ= 0

Expanding along C, we have⇒ 1(–3 + 7) – 2(9 – 7) + λ (–3 + 1) = 0⇒ 4 – 4 – 2λ = 0 ⇒ 2λ = 0 ⇒ λ = 012. Passengers Profit per seat (`)Executive class x 1500 Economy class y 1000 250 (maximum) Let, no. of executive class tickets be x and

economy class tickets be y. Maximise, Z = 1500x + 1000ySubject to the constraints, x + y ≤ 250 x ≥ 25, y ≥ 3x, x, y ≥ 0 13. Let award money for regularity = ` xLet award money for hardwork = ` y x + y = 6000 x + 3y = 11000

1 11 3

x

y

= 6000

11000

A X = B A–1A X = A–1 B X = A–1 B …(i)


|A| = 3 – 1 = 2 ≠ 0

adj A = 3 11 1

− −

As we know, A–1 = 1|A|

. adj A

= 12

3 11 1

− −

From (i), X = A–1 B

x

y

= 1

2

3 1 60001 1 11000

− −

= 1

2

18000 110006000 11000

− − +

= 1

2

70005000

= 35002500

∴ Award money for regularity, x = ` 3500Award money for hardwork, y = ` 2500

14. tan–1 11 1

12

−+

=

xx

, x > 0

⇒ tan–1 (1) – tan–1 (x) = 12

tan–1 x

tan tan tan− − −−+

= −1 1 11A B

ABA B

⇒ π4

12

= tan–1 x + tan–1 x

⇒ π4

32

= tan–1 x ⇒ tan–1 x = π4

23

×

⇒ x = tan π6

⇒ x =

13

15. Let A = (cos x)x …(i)Taking log on both sides log A = x.log (cos x)Differentiating both sides w.r.t. x

1 AA

ddx

= x . 1cos x

(– sin x) + log cos x

Addx

= A(log cos x – x tan x)

Addx

= (cos x)x (log cos x – x tan x)…(ii) [From (i)]

y = (cos x)x + sin–1 3x


dydx

= ddx x

ddx

xA + ×−

1

1 3 23

( )( ) …[From (ii)

= (cos x)x (log cos x – x tan x) +

11 3

12 3

3−

× ×x x

∴ dydx

= (cos x)x (log cos x – x tan x)

+ 32 (1 3 )x x−

Or y = (sec–1 x)2


dydx

= 2(sec–1 x) . 2

1

1x x −

x 2 1x − dydx

= 2 sec–1 x


x 2 1x −2

2 2

1.2 1

2d y dydxdx x

x x−

+ ×

+ 2 1x − dydx

.1 = 2.2

1

1x x −

Multiplying both sides by x 2 1x − , we have

x2 (x2 – 1) 2

32

d y dydxdx

x+ + x(x2 – 1) dydx

= 2

x2 (x2 – 1) 2

3 32 ( )d y dy

dxdxx x x+ + − = 2

x2 (x2 – 1) + −2

32 (2 )d y dy

dxdxx x = 2

(Hence proved)16. f (x) = 2x3 – 3x2 – 36x + 7Differentiating both sides w.r.t. x f´(x) = 6x2 – 6x – 36 = 6(x2 – x – 6) = 6(x2 – 3x + 2x – 6) = 6[x(x – 3) + 2(x – 3)] = 6(x + 2) (x – 3)When f´(x) = 0, x = –2, 3


Intervals Check-ing

points

Value of Sign of

f’(x)

Nature of f (x)

6 (x+2) (x–3)

(–∞, –2) –3 + – – > 0 st. increasing

(–2, 3) 1 + + – < 0 st. decreasing

(3, ∞) 4 + + + > 0 st. increasing

∴ f (x) is strictly increasing in (– ∞, –2) ∪ (3, ∞) f (x) is strictly decreasing in (–2, 3)

17. See Q. 17, 2017 (I Delhi). [Page 165

Ans. I = 2

4π

18. Let x x

x x x x x

2

2 21

1 2 1 1 2+ +

+ + + + += + +

( ) ( ) ( )A B C

…(i)

x x

x x

x x x x

x x

2

2

2

21

1 21 2 2 1

1 2+ +

+ ++ + + + + +

+ +=

( ) ( )( )( ) ( ) ( )

( ) ( )A B C

x2 + x + 1 = A (x2 + 3x + 2) + B (x + 2) + C (x2 + 2x + 1)Comparing the coefficients of x2, x and

constant term1 = A + C, 1 = 3A + B + 2C, 1 = 2A + 2B + C …(ii) …(iii) …(iv)Multiplying (iii) by 2 and (iv) by 1, we get 6A + 2B + 4C = 2 2A + 2B + C = 1 – – – – 4A + 3C = 1 …(v) Multiplying (ii) by 4 and (v) by 1, we get 4A + 4C = 4 4A + 3C = 1 – – – C = 3 …(v)

Putting the value of C in (ii) A + 3 = 1 ⇒ A = –2Putting the value of A and C in (iii) 3 (– 2) + B + 2 (3) = 1 ∴ B = 1Putting the value of A, B and C in (i),

x x

x x x x xdxdx

2

2 21

1 22

11

13

2+ +

+ +−+ +

++

∫ ∫= +( ) ( ) ( ) (

= – 2 log |x + 1| – 11x +

+ 3 log |x + 2| + C

Or

2( 3) 3 2x x x− − −∫ dx

= 21

22( 3) 3 2x x x

−− − − −∫ dx

= 21

2( 2 6 2 2) 3 2x x x

−− + − + − −∫ dx

= 21

2( 2 2) 3 2x x x

−− − − −∫ dx

21

2. 8 3 2x x− − −∫ dx

= − −

= − −

2Let 3 2

( 2 2 )

p x x

dp x dx

= 21

24 ( 2 3)p dp x x

−− − + −∫ dx

= 1

2 2 221

24 [ 2 1 3 1 ]p dp x x

−− − + + − −∫ ∫

dx

= 3/2 21 2

2 3. 4 [( 1) 4]p x

−− − + −∫ dx

= 3/2 2 21

34 2 ( 1)p x

−− − +∫ dx

= 3/2 2 11 1 1

3 2 24 ( 1) 4 ( 1) 4 sin

xp x x −− + − × + − + +

+ c

− − = − + + ∫ 2 2 2 2 2 11sin

2x

a x dx x a x a ca

= −

− − − + − −2 3/2 21

3(3 2 ) 2( 1) 3 2x x x x x

– 8 sin–1 + 12

x + c

19. 2xy dydx

= x2 + y2

dydx

= 22

2 2yx

xy xy+

dydx

= 1 12 2

yxy x+

= ⇒ = = +

Let

. 1 .

yy vx v

xdy dv

v xdx dx

v + x dvdx

= 1 12 2v

v+

x dvdx

= 12 2

vv

v+ −


x dvdx

= 12 2

vv−

x dvdx

= 21

2vv

−

22

1

v

vdv

− = dx

x

22

1

v

vdv

−∫ = dxx

−∫ = −

=

2Let 1

2

p v

dp v dv

dpp∫ = dx

x−∫

log |p | = – log |x| + log |c|

log |p | = log cx

⇒ p = cx

±

x(v2 – 1) = c± ⇒ x2

2 1y

x

− = k

x2 2

2y x

x

−

= k ⇒ y2 – x2 = kx

20. Let | | | | | |a b c→ → →

= = = m (Let) …(i)

Since , ,a b c→ → →

are mutually perpendicular vectors

∴ .a b→ →

= 0, .b c→ →

= 0, .c a→ →

= 0

2|2 2 |a b c→ → →+ +

= (2 2 ) . (2 2 )a b c a b c→ → → → → →+ + + +

→ → →=

2| | .x x x

2|2 2 |a b c→ → →+ + = 4 . 2 . 4 .a a a b a c

→ → → → → →+ +

+ 2 . . 2 .a b b b b c→ → → → → →

+ +

+ 4 . 2 . 4 .c a c b c c→ → → → → →

+ +→ → →

→ → → →

=

= =

2. | |

Given: . . 0

a a a

a b b a

2|2 2 |a b c→ → →+ + = 2 2 24| | | | 4| |a b c

→ → →+ +

2|2 2 |a b c→ → →+ + = 4m2 + m2 + 4m2 [From (i)

2|2 2 |a b c→ → →+ + = 9m2

|2 2 |a b c→ → →+ + = 3m …(ii)

Let 2 2a b c→ → →+ + makes angles α, β, γ with

, anda b c→ → →

respectively, then

cos α = (2 2 )

| ||2 2 |

a a b c

a a b c

→ → → →

→ → → →+ +

+ +

→ →

→ →

θ =

.cos

| || |

a b

a b

= 2 . . 2 .

| |(3 )

a a a b a c

a m

→ → → → → →

→+ + [From (i) and (ii)

cos α = 22| | 0 0

| |(3 )

a

a m

→

→+ +

→ → →

→ → → →

=

= =

2. | |

. . 0

a a a

a b a c

= 22 2

(3 ) 3m

m m=

∴ α = cos–1

23

cos β = .(2 2 )

| ||2 2 |

b a b c

b a b c

→ → → →

→ → → →+ +

+ +

= 2 . . 2 .( ) (3 )

b a b b b cm m

→ → → → → →+ +

= 2

20 | | 0

3

b

m

→+ + =

2

2133

m

m=

∴ β = cos–1

13

cos γ = .(2 2 )

| ||2 2 |

c a b c

c a b c

→ → → →

→ → → →+ +

+ +

= 2 . . 2 .

( ) (3 )c a c b c c

m m

→ → → → → →+ +

= 2

20 0 2| |

3

c

m

→+ + =

2

22 2

33m

m=

∴ γ = cos–1 23


21.

•

•

•

A(1, 8, 4)

B D C

(0, –1, 3) (2, –3, –1)

Direction ratios of BC are 2 – 0, –3 + 1, – 1 – 3 2, –2, –4 or 1, –1, –2Equation of line BC is

11 1y yx x z za b c

−− −= =

10 31 1 2

yx z+− −− −

= = = p (let)

Let D(p, –p – 1, –2p + 3) …(i) be the general point of line BC.

Direction ratios of AD are p – 1, –p – 1 – 8, –2p + 3 – 4 p – 1, –p – 9, –2p – 1Given that BC ⊥ AD,Using a1a2 + b1b2 + c1c2 = 0

∴ (1(p – 1) – 1(–p – 9) – 2(–2p – 1) = 0 p – 1 + p + 9 + 4p + 2 = 0 6p + 10 = 0

p = 10 56 3

− −=

From (i), D = 5 5 103 3 3

, 1, 3−

− +

= 5 2 193 3 3

, ,−

22. x + 3y ≤ 60, x + y ≥ 10, x ≤ y Let x + 3y = 60; Let x + y = 10; Let x = y

x 0 60 15y 20 0 15

x 0 10 5y 10 0 5

x 0 20 5y 0 20 5

Corner points Z = 3x + 9y

A(0, 10)B(0, 20)C(15, 15)D(5, 5)

0 + 90 = 900 + 180 = 18045 + 135 = 18015 + 45 = 60 (Minimum)

Y

XX′

60

50

40

30

20

10

10 20 30 40 50 600

D(5, 5)

B(0, 2

0)

C(15, 1

5)

x + 3y = 60

x + y = 10

x = y

Y′

∴ Minimum value of Z is 60 at (5, 5) i.e. x = 5, y = 5 of the feasible region.

23. Let E1 : Red ball is transferred from bag A to bag B.

E2 = Black ball is transferred from bag A to bag B.

A = A red ball is drawn from bag B Bag A Bag B

Red Black 3 2

Red Black 2 3

P(E1) = 35

; P(A/E1) = 36

P(E2) = 25

; P(A/E2) = 26

By Bayes’ theorem,

P(E1/A) = ×× + ×

1 1

1 1 2 2

P(E ) P(A / E )[P(E ) P(A)E ] [P(E ) P(A)E ]

=

3 35 6

3 3 2 25 6 5 6

×

× + ×

= 99 4+

= 913

OrA and B are Independent events (Given)

∴ P(A ∩ B) = P(A) × P(B) …(i)P (at least one of A and B) = P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = P(A) + P(B) – P(A) × P(B) [From (i)


= P(A) + P(B)[1 – P(A)] = 1 – P(A′) + P(B).P(A′)

+ =′⇒ = −′

P(A) P(A ) 1

P(A ) 1 P(A)

= 1 – P(A′) [1 – P(B)] = 1 – P(A′)P(B′) [Hence proved]

[ 1 P(B) P(B )− = ′

24.

a b c

b c a

c a b

= 0 [C1 → C1 + C2 + C3

a b c b c

a b c c a

a b c a b

+ ++ ++ +

= 0

Taking common (a + b + c) from C1

(a + b + c) 111

b c

c a

a b

= 0

(a + b + c) 100

b c

c b a c

a b b c

− −− −

= 0

2 2 1

3 3 1

R R R

R R R

→ − → −

(Expanding along C1)

⇒ (a + b + c) [(c – b) (b – c) – (a – b) (a – c)] = 0

⇒ (a + b + c) [bc – c2 – b2 + bc – a2 + ac + ba – bc) = 0

⇒ (a + b + c) (–a2 – b2 – c2 + ab + bc + ca) = 0

⇒ – (a + b + c) (a2 + b2 + c2 – ab – bc – ca) = 0

⇒ – 12

(a + b + c) (2a2 + 2b2 + 2c2 – 2ab – 2bc – 2ca) = 0

⇒ – 12

(a + b + c) (a2 + a2 + b2 + b2 + c2

+ c2 – 2ab – 2bc – 2ca) = 0⇒ (a + b + c) [(a2 + b2 – 2ab) + (b2 + c2

– 2bc) + (c2 + a2 – 2ca)] = 0⇒ (a + b + c) [(a – b)2 + (b – c)2

+ (c – a)2] = 0Either a + b + c = 0

or (a – b)2 + (b – c)2 + (c – a)2 = 0But a + b + c ≠ 0 …(Given)

(a – b)2 = 0 (b – c)2 = 0 (c – a)2 = 0 a – b = 0 b – c = 0 c – a = 0 a = b b = c c = a ∴ a = b = c Hence proved25. (i) CommutativeFor any two elements A, B ∈ P(X), we have A ∩ B = B ∩ A⇒ A B = B A for A, B ∈ P(X)Thus, is commutative on P(X).(ii) AssociativeFor any three elements A, B, C ∈ P(X), we

have (A B) C = (A ∩ B) C = (A ∩ B) ∩ Cand A (B C) = A (B ∩ C) = A ∩ (B ∩ C)Since (A ∩ B) ∩ C = A(B ∩ C)Thus, is associative on P(X).(iii) Identity elementLet E be the identity element in P(X) w.r.t.

. Then, A E = A = E A for all A ∈ P(X) A ∩ E = A = E ∩ A for all A ⊂ X E = XThus, X is the identity element with respect

to on P(X).(iv) Invertible elementLet A be an invertible element in P(X) andLet S be its inverse. Then A S = X = S A A ∩ S = X = S ∩ A A = S = XThus, X is the only invertible element in

P(X) with respect to and it is the inverse of itself.

Or

f (x) = 43 4

xx +

For f is one-one—Let x1, x2 ∈ R F(x1) = F(x2)

+ +

=1 2

1 2

4 43 4 3 4

x xx x

⇒ x1(3x2 + 4) = x2(3x1 + 4)⇒ 3x1x2 + 4x1 = 3x1x2 + 4x2⇒ 4x1 = 4x2⇒ x1 = x2∴ f is one-one.


For f is onto —1st methodLet y be any element of R.

Let y = f (x)

y = 43 4

xx +

⇒ 3xy + 4y = 4x⇒ 3xy – 4x = – 4y⇒ x(3y – 4) = – 4y

⇒ x = 43 4

yy−−

or 44 3

yy−

We have, f (x) = 43 4

xx +

f 44 3

yy

−

=

44

4 3

43 4

4 3

yy

yy

−

+ −

= 1612 16 12

yy y+ −

= 1616

y

f (x) = y ∴ f is onto

f –1 y = 44 3

yy−

26. Let radius of the base of cone = xLet height of the cone = yLet volume of the cone

= 13

π k (a constant)

13

πx2y = 13

πk π

1 2Volume of cone = 3

r

x2y = k …(i)

y = 2k

x …(ii)

Curved Surface Area of cone, S = πrl

S = πx 2 2x y+

S = πx 2

24

k

xx + [From (ii)

S2 = π2x2 2

24

k

xx

+ [Squaring both sides

Let S2 = z ∴ z = π2x4 + π2k2x–2

Differentiating both sides w.r.t. x, we get

dzdx

= 4π2x3 – 2π2k2x–3 …(iii)

For stationary points, dzdx

= 0

4π2x3 – 2 2

32 k

x

π = 0

2 6 2 2

34 2x k

x

π − π = 0

4π2x6 – 2π2k2 = 0 4π2x6 = 2π2k2

2x6 = k2

⇒ 2x6 = (x2y)2 [From (i) 2x6 = x4y2

2x2 = y2

2 x y=Differentiating (iii) both sides w.r.t. x, we

get

2

2d z

dx = 12π2x2 + 6π2k2x–4

22 2 22

2 4at

2

62

4

12 vey

x

yd z k

dx y=

π

= π + = +

∴ z is minimum at x = y/ 2

and S2 is minimum at x = y/ 2

∴ S is minimum at x = 2

y or y = 2 x

∴ Curved Surface Area of cone is least

when y = 2 x.

or Altitude = 2 (radius of the base)Or

Part I : x = a cos θ + aθ sin θDifferentiating both sides w.r.t. θ, we getdxdθ

= – a sin θ + aθ cos θ + a sin θ = aθ cos θ

y = a sin θ – aθ cos θDifferentiating both sides w.r.t. θ, we get

θdyd

= a cos θ – (aθ sin θ + a cos θ) = aθ sin θ

y

x

2nd method

For y = 43

There is no X

such that

f (x) = 43

∴ f is not

invertible

But f : R – 4

4 3y

y−Range of f is onto

so invertible and

f–1(y) = 4

4 3y

y−


Slope of tangent, dydx

= dyd

ddxθθ×

= aa

θ θθ θ

θθ

sincos

sincos

× =1

Equation of tangent is (y – y1) = slope of tangent (x – x1) y – (a sin θ – aθ cos θ)

= sincos

θθ

[x – (a cos θ + aθ sin θ)]

⇒ sin θ[x – a cos θ – a θ sin θ] = cos θ[y – a sin θ + aθ cos θ]⇒ x sin θ – a sin θ cos θ – aθ sin2 θ = y cos θ – a sin θ cos θ + a cos2 θ⇒ x sin θ – y cos θ = aθ(sin2 θ + cos2 θ)⇒ x sin θ – y cos θ = aθ [ sin2 θ + cos2 θ = 1]

Part II :

Slope of normal = − 1tan θ

= –cot θ = – cossin

θθ

Equation of normal is y – y1 = Slope of normal (x – x1) y – (a sin θ – aθ cos θ)

= – cossin

θθ

[x – (a cos θ + aθ sin θ)]

y – sin θ – a sin2 θ + aθ cos θ sin θ = –x cos θ + a cos2 θ + aθ cos θ sin θ x cos θ + y sin θ – a(sin2 θ + cos2 θ) = 0 x cos θ + y sin θ – a = 0Perpendicular distance from the origin, (0, 0)

= 0 02 2

(cos ) (sin )

cos sin

θ θ

θ θ

+ −

+

a

⊥ = + +

+

distanceax by c

a b1 1

2 2

= − a

1 = a, which is a constant

27. Given : Points A(2, –2, 1), B(4, 1, 3) and C(–2, –2, 5)

Using equation of plane

1 1 1

2 1 2 1 2 1

3 1 3 1 3 1

x x y y z z

x x y y z z

x x y y z z

− − −− − −− − −

= 0

2 2 14 2 1 2 3 12 2 2 2 5 1

x y z− + −− + −

− − − + − = 0

2 2 1

2 3 24 0 4

x y z− + −

− = 0

(expanding along R1)

⇒ (x – 2) (12 – 0) – (y + 2) (8 + 8) + (z – 1) (0 + 12) = 0

⇒ 12x – 24 – 16y – 32 + 12z – 12 = 0⇒ 12x –16y + 12z – 68 = 0⇒ 3x – 4y + 3z – 17 = 0 …(i)

(Dividing both sides by 4)

D.r’s of normal to the plane (i) are 3, –4, 3.D.r’s of the given line are 3, 3, 1.Now, a1a2 + b1b2 + c1c2

= 3(3) – 4(3) + 3(1) = 9 – 12 + 3 = 0∴ Normal to the plane ⊥ to the given line So given line is parallel to the plane (i).Required distance = Distance between point

(5, 4, 8) and plane (i) Given line (5, 4, 8)

Plane (i)

= 1 1 12 2 2

ax by cz d

a b c

+ + +

+ +

= 3(5) 4(4) 3(8) 179 16 9

− + −+ +

= = = = = − = = −

1 1 1Here 5, 4, 8

3, 4, 3, 17

x y z

a b c d

= 15 16 24 1734

− + −

= 6 34 6 343434 34

× = = 3 3417

units

28. P(a head) = 4P(a tail) P(H) = 4P(T) …(i) P(H) + P(T) = 1 4P(T) + P(T) = 1 [From (i)

5P(T) = 1 P(T) = 15

Let p = 15

, q = 1 – 15

= 45

, n = 3


Here random variable X (no. of tails) can take values 0, 1, 2, 3.

Using P(r) = nCr.qn– rpr

P(X = 0) = 3C0q3p0 = q3 =

34 645 125

=

P(X = 1) = 3C1q2p1 = 3q2p = 3

24 1 485 5 125

=

P(X = 2) = 3C2qp2 = 3qp2 = 3 4 1 125 5 125

=

P(X = 3) = 3C3q0p3 = p3 =

31 15 125

=

Probability distribution is

Xi 0 1 2 3P(Xi) 64

12548

12512125

1125

Xi Pi PiXi PiXi2

0 64125

0 0

1 48

125

48125

48125

2 12125

24125

48125

3 1125

3125

9125

ΣPiXi = 75

125 ΣPiXi

2 = 105125

∴ Mean = ΣPiXi = 35

=75125

Variance = ΣPiXi2 – (ΣPiXi)2

= 2105 3

125 5 −

= 21 925 25

− = 1225

29. Equation of AB is 1

1 2 13 5 1

x y

− = 0

⇒ x(– 2 – 5) – y(1 – 3) + 1(5 + 6) = 0

⇒ – 7x + 2y + 11 = 0 ⇒ 2y + 11 = 7x

⇒ x = 2 117

y + …(i)

Y

X

Y′

X′

5

5

3

2

1

0 1 2 3 4 5–1

–2

C(5, 2)

B(3, 5)

B(3, 5)

A(1, –2)

Equation of BC is 1

3 5 15 2 1

x y

= 0

⇒ x(5 – 2) – y(3 – 5) + 1(6 – 25) = 0

⇒ 3x + 2y – 19 = 0

⇒ x = 19 23

y− … (ii)

Equation of AC is 1

1 2 15 2 1

x y

− = 0

⇒ x(– 2 – 2) – y(1 – 5) + 1(2 + 10) = 0⇒ 4y + 12 = 4x⇒ x = y + 3 …(iii)From (i), (ii) and (iii),Area of the shaded region

=

2 5

2 2

19 23

( 3) yy dy dy−

−

+ +∫ ∫

– 5

2

2 117

y dy−

+ ∫


= 2 5 52 2 2

2 2 2

2 21 12 3 2 7 2

3 19 11y y yy y y− −

+ + − − +

= 4 4 12 2 3

6 6 − + − + [(95 – 25) – (38 – 4)]

– 17

[(25 + 55) – (4 – 22)]

= 8 + 4 + 13

(70 – 34) – 17

(80 + 18)

= 12 + 12 – 14 = 10 sq. unitsOr

f (x) = 3x2 + 2x + 1, b = 4, a = 0, nh = b – a = 4 – 0 = 4

f (a) = f (0) = 0 + 0 + 1 = 1f (a + h) = f (0 + h) = 3h2 + 2h + 1f (a + 2h) = f (0 + 2h) = 3(2h)2 + 2(2h) + 1

= 12h2 + 4h + 1: : :: : :: : :f (a + (n – 1)h) = f (0 + (n – 1)h)

= 3[(n – 1)h]2 + 2(n – 1)h + 1

( )b

a

f x dx∫ = 0

limh

h→

[f (a) + f (a + h) + f (a + 2h) +

… + f (a + (n – 1)h]4

2

0

(3 2 1)x x dx+ +∫ = 0

limh

h→

[1 + 3h2

+ 2h + 1) + (12h2 + 4h + 1) + …

+ 3(n – 1)2h2 + 2(n – 1)h + 1]

= 0

limh

h→

[n + 3h2(12 + 22 + … + (n – 1)2)

+ 2h(1 + 2 + … + (n – 1)]

= 0

limh

h→

2 2 . ( 1)( 1)(2 1)6 2

3 h n nn n nn h −− −

+ +

= 0

limh

h→ 2

( )(2 ) ( )hnnh nh h nh h nh nh h

+ − − + −

= 0

limh

h→

42

4 (4 )(8 ) 4(4 )h h h

+ − − + −

(nh = 4)

= 4 + 2(4 – 0) (8 – 0) + 4(4 – 0)= 4 + 64 + 16 = 84

set ii

10. xy dydx

= (x + 2) (y + 2)

2y

y + dy = ( 2)xx+ dx

( 2) 2

2y

y

+ − + ∫ dy =

2xx x

+∫ dx

2

21

y

+

−∫ dy = 21x

+∫ dx

∴ y – 2 log |y + 2| = x + 2 log |x| + c

11. 2

cos

8 sin

x

x−∫ dx

= 2 2(2 2)

dp

p−∫ =

=

Let sin

cos

p x

dp x dx

= sin–12 2

p + c −

= + −∫ 1

2 2sin

dx xc

aa x

= sin–1

sin2 2

x + c

12. drdt

= 5 cm/min, dhdt

= – 4 cm/min

r = 8 cm, h = 6 cm

Volume of cylinder, v = πr2h Differentiating w.r.t. t, we get

dvdt

= 2. . 2dh drdt dt

r h r

π +

= π [64 (–4) + 6(16) (5)]

= π [–256 + 480]

= 224π cm3/min

∴ Volume is increasing at the rate of

224π cm3/min.

21. Let I = 0

tansec tan

x xx x

π

+∫ dx …(i)

I = 0

( ) tan( )sec( ) tan( )

x xx x

ππ − π −π − + π −∫ dx

= −∫ ∫

0 0

( ) ( )a a

f x dx f a x dx

I = 0

( )( tan )( sec ) ( tan )

x xx x

ππ − −

− + −∫ dx


I = 0

( ) tansec tan

x xx x

ππ −

+∫ dx …(ii)

Adding (i) and (ii), we have

2I = 0 0

tan ( ) tansec tan sec tan

x x x xx x x x

dx dxπ π

π −+ +

+∫ ∫

2I = 0

tansec tan

x dxx x

ππ

+∫

∴ I = 0

tan1 sec tan2 sec tan sec tan

x x xx x x x

π−×

+ −π∫ dx

= 2

2 20

(sec tan tan )2 sec tan

x x x

x x

ππ −

−∫ dx

…(sec2 x – tan2 x = 1)

= 2

02

[sec tan (sec 1)]x x xπ

π − −∫ dx

= π π

2 0sec tanx x−[ ]

= 2

[(sec tan ) (sec 0 tan 0 0)]π π − π + π − − +

= 2π [(–1 – 0 + π) – (1 – 0)] = π

2( π – 2)

22. y . .xye dx = 2( )

xyxe y dy+

dxdy

= 2

.

xy

xyxe y

y e

+

v + y dvdy

= 2v

vv ye y

ye

+

= ⇒ = = +

Let

.1 .

xx vy v

y

dx dvv y

dy dy

y dvdy

= ( )v

vy ve y

ye

+ – v

y dvdy

= v v

vve y ve

e

+ −

ev dv = dy

vdy e dv=∫ ∫

y = ev + c

y = e

xy + c

23.

A(2, 3, – 8)

Given line

B CD

4 12 6 3

yx z− −= =

4 12 6 3

yx z− −− −

= = = p (Let) …(i)

Let D(–2p + 4, 6p, –3p + 1) …(ii) be the general point of given line BC.

Direction ratios of AD are –2p + 4 – 2, 6p – 3, –3p + 1 + 8 –2p + 2, 6p – 3, –3p + 9Direction ratios of given line BC are –2, 6, –3.AD ⊥ BC, using a1a2 + b1b2 + c1c2 = 0⇒ –2(–2p + 2) + 6(6p – 3) – 3(–3p + 9) = 0⇒ 4p – 4 + 36p – 18 + 9p – 27 = 0⇒ 49p = 49 ⇒ p = 1

From (ii), Required foot of ⊥ D

= (–2 + 4, 6, –3 + 1) or D(2, 6, –2)

28. p = P(defective bulb) = 6 130 5

=

q = 1 – p = 1 – 1 45 5= , n = 3

Here random variable X can take values 0, 1, 2, 3.

Using P(r) = nCr qh – r pr

P(X = 0) = 3C0q3p0 = q3 =

34 645 125

=

P(X = 1) = 3C1q2p1 = 3

24 1 485 5 125

=

P(X = 2) = 3C2q1p2 = 3

24 1 125 5 125

=


⇒ (x – 1) (–1 – 1) – (y – 1) (3) + z(3) = 0 (Expanding along R1)

⇒ –2x + 2 – 3y + 3 + 3z = 0⇒ –2x – 3y + 3z + 5 = 0or 2x + 3y – 3z – 5 = 0 …(i)Direction ratios of normal to the plane (i)

are 2, 3, –3.Direction ratios of given line are 3, 1, 3Now, a1a2 + b1b2 + c1c2 = 2(3) + 3(1) + (–3) (3) = 6 + 3 – 9 = 0∴ Given line is parallel to plane (i)Required distance = Distance between point

(3, 5, –2) and plane (i)

= 1 1 12 2 2

ax by cz d

a b c

+ + +

+ +

= = = − = = = − = −

1 1 13, 5, 2

2, 3, 3, 5

x y z

a b c d

= 2(3) 3(5) 3( 2) 54 9 9

+ − − −+ +

= + + −6 15 6 522

= 2222

= 22 units

set iii10. Given vectors are coplanar

1 1 13 1 21 3

−

λ − = 0

⇒ 1(–3 – 2λ) + 1(–9 – 2) + 1(3λ – 1) = 0(Expanding along R1)

⇒ –3 – 2λ – 11 + 3λ – 1 = 0

⇒ λ = 15

11. 2

21

1

dy ydx x

++

=

2 21 1

dy dx

y x+ +=

2 21 1

dy dx

y x+ +=∫ ∫ …(By integrating)

⇒ tan–1 y = tan–1 x + c …(i)

⇒ tan–1 3 = tan–1 0 + c =∴ = =

(0) 3

0, 3

y

x y

P(X = 3) = 3C3q0p3 =

31 15 125

=

Probability distribution is

Xi 0 1 2 3P(Xi) 64

12548

12512125

1125

Xi Pi PiXi PiXi2

0 64125

0 0

1 48

125

48125

48125

2 12125

24125

48125

3 1125

3125

9125

ΣPiXi = 75

125 ΣPiXi

2 = 105125

Mean = ΣPiXi = 35

=75125


= 2105 3

125 5 −

= 21 925 25

− = 1225

29. Equation of plane through points A, B, C is

1 1 1

2 1 2 1 2 1

3 1 3 1 3 1

x x y y z z

x x y y z z

x x y y z z

− − −− − −− − −

= 0

1 1 0

1 1 2 1 1 02 1 2 1 1 0

x y z− − −− − −

− − − − − = 0

1 1

0 1 13 1 1

x y z− −

− −= 0


3π = 0 + c

Putting the value of c in (i), we get

tan–1 y = tan–1 x + 3π

12. radius, r = 13

(3x + 1)

Volume of Sphere, V = 43

πr3

V = 43

π × 127

(3x + 1)3

Differentiating both the sides w.r.t. x,Vd

dx = 24

813(3 1) 3xπ × + × = π4

9(3x + 1)2

21. x + y ≤ 50, 2x + y ≤ 80, x ≥ 20 Let x + y = 50, Let 2x + y = 80, Let x = 20

x 0 50 20y 50 0 30

x 0 40 30y 80 0 20

X

2x + y = 80

X′

Y′

80

70

60

50

40

30

20

10

0 10 20 30 40 50 60

x =

20

x + y = 50

C(30, 20)

B(20, 30)

A(20, 0) D(40, 0)

Y

Corner points

Z = 105x + 90y

A(20, 0)B(20, 30)C(30, 20)D(40, 0)

2100 + 0 = 21002100 + 2700 = 48003150 + 1800 = 4950 (Maximum)4200 + 0 = 4200

Maximum value of Z is 4950 at C(30, 20) i.e., x = 30, y = 20.

22. 2xy + y2 – 2x2 dydx

= 0

–2x2 dydx

= – (2xy + y2)

⇒ 2

22

2

dy xy ydx x

+=

⇒ 2 2

22 ( )

2dv x vx v xdx x

v x ++ = = ⇒ = = +

Let

.1 .

yy vx v

xdy dv

v xdx dx

⇒ 2 2

2(2 )

2dv x v vdx x

x v+= −

22 22

dv v v vdx

x + −=

22dv dx

xv=

− =∫ ∫ 122

xv dv dx

⇒ 12

1v−

− = log|x| + c

⇒ 2v− = log|x| + c

⇒ 2xy

− = log|x| + c

∴ –2x = y log|x| + cy …(i)

⇒ –2(1) = 2 log 1 + c(2) [Putting y = 2, x = 1]

c = –1 [ log 1 = 0]

Putting the value of c in (i), we get –2x = y log|x| – y –2x = –y(1 – log|x|)

y = 2

1 log | |x

x− , x ≠ 0, x ≠ e

23. Let I = 0

1 sinx

xdx

π

+∫ …(i)

I = 0

( )1 sin( )

xx

dxπ

π −+ π −∫ ( ) ( )

0 0

a af x dx f a x dx

= −∫ ∫

I = 0

1 sinx

xdx

ππ −+∫

…(ii)


Adding (i) and (ii), we get

2I = 0

1 sin xdx

ππ

+∫

I = 0

1 sin2 1 sin 1 sin

xx x

dxπ

π π −+ −

×∫

= 20

1 sin2 cos

x

xdx

ππ −∫

= 2

0

sec tan )2

(sec x x xπ

π −∫

= [ ]02tan secx x ππ −

= [ ]2

(tan sec ) (tan 0 sec 0)π π − π − −

= 2

(0 ( 1)) (0 1)π − − − −

= 2

(2)π = π

28. Equation of plane through points A, B, C is

1 1 1

2 1 2 1 2 1

3 1 3 1 3 1

x x y y z z

x x y y z z

x x y y z z

− − −− − −− − −

= 0

⇒ 1 2 2

4 1 2 2 3 23 1 0 2 2 2

x y z− + −− + −− + −

= 0

1 2 2

3 4 12 2 0

x y z− + −= 0

⇒ (x – 1) (– 2) – (y + 2) (– 2) + (z – 2) (6 – 8) = 0

⇒ –2(x – 1) + 2(y + 2) – 2(z – 2) = 0

Dividing both sides by –2

⇒ (x – 1) – (y + 2) + (z – 2) = 0⇒ x – 1 – y – 2 + z – 2 = 0⇒ x – y + z = 5 …(i)

Point (2, –1, 2) lies on the given line, andDirection ratio of given line are 3, 4, 2.Cartesian equation of given line is

+− −= =12 23 4 2

yx z = q (Let)

Let P(3q + 2, 4q – 1, 2q + 2) …(ii) be the required point

Then point P lies on the plane (i) 3q + 2 – (4q – 1) + (2q + 2) = 5⇒ 3q + 2 – 4q + 1 + 2q + 2 = 5⇒ q = 0Putting the value of q in (ii), we have∴ Point P(0 + 2, 0 – 1, 0 + 2)∴ Point (2, –1, 2) is the required point.

29. p = P(4) = 110

, q = 1 – p = 1 – 1 910 10

=

Here random variable X can take values 0, 1, 2.

Using P(r) = nCr qn – r pr

P(X = 0) = 2C0q2p0 = q2 =

29 8110 100

=

P(X = 1) = 2C1q1p1 = 2qp = 2

9 1 1810 10 100

=

P(X = 2) = 2C2q0p2 = p2 =

21 110 100

=

Xi Pi PiXi PiXi2

0 81100

0 0

1 18100

18100

18100

2 1

100 2

100 4

100

ΣPiXi = 20

100 ΣPiXi

2 = 22100


= 222 20

100 100 − = 22 4

100 100− = 18

100 or 0.18

250 ■ SHIV DAS SENIOR SECONDARY SERIES (XII)

CBSE

MATHEMATICS – 2018 (COMPTT.)


GENERAL INSTRUCTIONS

(i) All questions are compulsory.

(ii) The question paper consists of 29 questions divided into four sections A, B, C and D. Section Acomprises of 4 questions of one mark each, Section B comprises of 8 questions of two marks each,Section C comprises of 11 questions of four marks each and Section D comprises of 6 questions ofsix marks each.

(iii) All questions in Sections A are to be answered in one word, one sentence or as per the exactrequirement of the question.

(iv) There is no overall choice. However, internal choice has been provided in 3 questions of four markseach and 3 questions of six marks each. You have to attempt only one of the alternatives in all suchquestions.

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

SECTION A


Q.1. Find the value of tan–1 3 – sec–1(– 2).

Q.2. If A =

1 2 22 12 2 1

x

− −

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ is a matrix satisfying AA′ = 9I, find x.

Q.3. Find the value of [ˆ, ˆ, ˆ]i k j .

Q.4. Find the identity element in the set Q+ of all positive rational numbers for the operation ✻

defined by a ✻ b =

32ab for all a, b ∈ Q+. (Not in syllabus)

SECTION B


Q.5. Prove that : 3 cos–1 x = cos–1 (4x3 – 3x), x ∈

12

1,⎡⎣⎢

⎤⎦⎥.

(Common for Delhi & Outside Delhi)

MATHEMATICS – 2018 (COMPTT.) ■ 251

Q.6. If A =

2 35 2−

⎡

⎣⎢

⎤

⎦⎥ be such that A–1 = kA, then find the value of k.

Q.7. Differentiate tan–1

cos sincos sin

x xx x

−+

⎛⎝⎜

⎞⎠⎟ with respect to x.

Q.8. The total revenue received from the sale of x units of a product is given by R(x) = 3x2 + 36x + 5in rupees. Find the marginal revenue when x = 5, where by marginal revenue we mean therate of change of total revenue with respect to the number of items sold at an instant.

Q.9. Find :

3 52

−∫ sincos

xx

dx.

Q.10. Solve the differential equation cos

dydx

⎛⎝⎜

⎞⎠⎟

= a, (a ∈ IR).

Q.11. If a b c→ → → →

+ + = 0 and | |a→

= 5, | |b→

= 6 and | |c→

= 9, then find the angle between a→

and b→

.

Q.12. Evaluate P(A ∪ B), if 2P(A) = P(B) =

513

and P(A/B) =

25

.

SECTION C

Question numbers 13 to 23 carry 4 marks each.Q.13. Using properties of determinants, prove that

5 2 22 5 22 2 5

a a b a c

b a b b c

c a c b c

− + − +− + − +− + − +

= 12(a + b + c) (ab + bc + ca)

Q.14. If sin y = x cos(a + y), then show that

dydx

a ya

= +cos ( )cos

2.

Also, show that

dydx

= cos a, when x = 0.

Q.15. If x = a sec3 θ and y = a tan3 θ, find

d y

dx

2

2 at θ =

π3

.

Or

If y = extan−1, prove that

( ) ( )1 2 12

2

2+ + −x xd ydx

dydx

= 0.

Q.16. Find the angle of intersection of the curves x2 + y2 = 4 and (x – 2)2 + y2 = 4, at the point inthe first quadrant.

OrFind the intervals in which the function f (x) = – 2x3 – 9x2 – 12x + 1 is

(i) Strictly increasing (ii) Strictly decreasingQ.17. A window is in the form of a rectangle surmounted by a semicircular opening. The total

perimeter of the window is 10 metres. Find the dimensions of the window to admitmaximum light through the whole opening.


Q.18. Find :

42 42( )( )x x− +∫ dx

Q.19. Solve the differential equation (x2 – y2) dx + 2xydy = 0.Or

Find the particular solution of the differential equation (1 + x2)

dydx

+ 2xy =

11 2+ x

, given that

y = 0 when x = 1.Q.20. Find x such that the four points A(4, 4, 4), B(5, x, 8), C(5, 4, 1) and D(7, 7, 2) are coplanar.

Q.21. Find the shortest distance between the lines

x y z− − −= =12

23

34

and

x y z− − −= =23

44

55

.

Q.22. Two groups are competing for the positions of the Board of Directors of a corporation. Theprobabilities that the first and second groups will win are 0.6 and 0.4 respectively. Further,if the first group wins, the probability of introducing a new product is 0.7 and thecorresponding probability is 0.3 if the second group wins. Find the probability that the newproduct introduced was by the second group.

Q.23. From a lot of 20 bulbs which include 5 defectives, a sample of 3 bulbs is drawn by random,one by one with replacement. Find the probability distribution of the number of defectivebulbs. Also, find the mean of the distribution.

SECTION D


Q.24. Show that the relation R on the set Z of all integers defined by (x, y) ∈ R ⇔ (x - y) is divisibleby 3 is an equivalence relation.

Or

A binary operation ✻ on the set A = {0, 1, 2, 3, 4, 5} is defined as a ✻ b =

a b a b

a b a b

+ + <+ − + ≥

⎧⎨⎩

,,

ifif

66 6

.

Write the operation table for a ✻ b in A.Show that zero is the identity for this operation ✻ and each element ‘a’ ≠ 0 of the set isinvertible with 6 – a, being the inverse of ‘a’. (Not in syllabus)

Q.25. Given A =

5 0 42 3 21 2 1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

, B–1 =

1 3 31 4 31 3 4

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

, compute (AB)–1.

Or

Find the inverse of the matrix A =

1 2 21 3 00 2 1

−−

−

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

by using elementary row transformations.


Q.26. Using integration, find the area of the region : {(x, y) : 0 ≤ 2y ≤ x2, 0 ≤ y ≤ x, 0 ≤ x ≤ 3}

Q.27. Evaluate:

x x x

x x

sin cossin cos

/

4 40

2

+∫π

dx

Or

Evaluate:

( )3 2 12

1

3

x x dx+ +∫ as the limit of a sum.

Q.28. Find the vector equation of the line passing through (1, 2, 3) and parallel to each of the planes

r i j k→

− +.(ˆ ˆ ˆ)2 = 5 and r i j k→

+ +. ( ˆ ˆ ˆ)3 = 6. Also find the point of intersection of the line thus

obtained with the plane r i j k→

+ +.( ˆ ˆ ˆ)2 = 4.

Q.29. A company produces two types of goods, A and B, that require gold and silver. Each unitof type A requires 3 g of silver and 1 g of gold while that of B requires 1 g of silver and2 g of gold. The company can use atmost 9 g of silver and 8 g of gold.If each unit of typeA brings a profit of � 40 and that of type B � 50, find the number of units of each type thatthe company should produce to maximize the profit. Formulate and solve graphically theLPP and find the maximum profit.

SOLUTIONS

3. 1st method

[ˆ ˆ ˆ]i k j =

ˆ ( ˆ ˆ) ˆ ( ˆ)i k j i i• •× = − = – 12nd method

[ˆ ˆ ˆ]i k j =1 0 00 0 10 1 0

Expanding along C1, 1(0 – 1) = – 1

4. a ✻ b =

32ab …[Given]

a ✻ e =

32ae …(i)

Identity element, a ✻ e = a

⇒

32ae = a …[From (i)

⇒

32e = 1

⇒ e =

23

5. R.H.S. = cos–1(4x3 – 3x)= cos–1(4 cos3 θ – 3 cos θ) Let x = cos θθθθθ= cos–1 (cos 3θ) ∴ cos–1 x = θθθθθ= 3θ= 3 cos–1 x = L.H.S. (Hence proved)

1. tan–1 ( )3 – sec–1 (–2)

=

π3

– [π – sec–1 (2)]

Q sec ( ) sec ,| |−− −−−− == −− ≥≥[[ ]]1 1 1x x xππ

=

π3

– π +

π3

=

−− ππ3

2. AA′ = 9I

1 2 22 12 2 1

1 2 22 1 22 1

91 0 00 1 00 0 1

x

x− −

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

−

−

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

=⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

1 4 4 2 2 2 2 4 22 2 2 4 1 4 2

2 4 2 4 2 4 4 1

2+ + + + − + −

+ + + + − + −− + − − + − + +

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

x

x x x

x

=

9 0 00 9 00 0 9

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

Equating the corresponding elements4 + 2x = 0 5 + x2 = 92x = – 4 ⇒ x = – 2 x2 = 4 ⇒ x = ± 2.

Therefore, x = – 2 is the only common solution.

⎡

⎣⎢


6. A =

2 35 2−

⎡

⎣⎢

⎤

⎦⎥

|A| = – 4 – 15 = – 19 ≠ 0, ∴ A–1 exists.

adj A =

− −−

⎡

⎣⎢

⎤

⎦⎥

2 35 2

As we know, A–1 =

1| |A

. adj A

=

− − −−

⎡

⎣⎢

⎤

⎦⎥

1

19

2 35 2

=

219

319

519

219−

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

Now,A–1 = kA …(Given)

219

319

519

219−

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥ = k

2 35 2−

⎡

⎣⎢

⎤

⎦⎥

219

319

519

219−

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥ =

2 35 2

k k

k k−⎡

⎣⎢

⎤

⎦⎥

⇒ 2k =

2

19(Equating the corresponding elements)

∴ k =

119

7. Let y = tan–1

cos sincos sin

x xx x

−+

⎛⎝⎜

⎞⎠⎟

Dividing Numerator and Denominator by cos x,we have

y = tan–1

11

−+

⎛⎝⎜

⎞⎠⎟

tantan

xx

y = tan–1

tan π4

−⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

x

y =

π4

– x


dydx

= – 1

8. R(x) = 3x2 + 36x + 5Differentiating both sides w.r.t. x, we have

Marginal Revenue, R′(x) = 6x + 36When x = 5, R′(5) = 6(5) + 36

= 30 + 36 = � 66∴ Marginal Revenue is � 66.

9.

3 52

−∫ sincos

x

x dx

=

3 52 2cos

sincosx

x

x−⎛

⎝⎜⎞⎠⎟∫ dx

=

( sec sec tan )3 52 x x x−∫ dx

= 3 tan x – 5 sec x + C

10. cos

dydx

⎛⎝⎜

⎞⎠⎟

= a

dydx

= cos–1 a

dy = cos–1 a . dx

dy∫ = cos–1 a

dx∫ (Integrating both sides)

y = x . cos–1 a + Cy – C = x cos–1 a

yx− C = cos–1 a

cos

yx−−⎛⎛

⎝⎝⎜⎜⎞⎞⎠⎠⎟⎟

C = a is the solution of the

differential equation.

11. a b c→ → →

+ + = 0 …(Given)

⇒ a b c→ → →

+ = −Squaring both sides, we get

( ) ( ) ( ) ( )a b a b c c→ → → → → →

+ + = − −• •

[ . | |Q a a a a→→ →→ →→ →→

== ==2

2

a a a b b a b b c c→ → → → → → → → → →

• • • • •+ + + =

| | | | | |a a b b c→ → → → →

+ + =•2 2 22

| | | || |cos | | | |a a b b c→ → → → →

+ + =2 2 22 θ52 + 2(5) (6) cos θ + 62 = 92

… [ | | , | | , | |Q a b c→→ →→ →→

== == ==5 6 9

60 cos θ = 81 – 25 – 36

⇒ cos θ =

2060

13

=

∴ Required angle, θθθθθ = cos–1

13

⎛⎝⎜

⎞⎠⎟

12. Given: 2P(A) = P(B) =

513

⇒ P(A) =

526

and P(A/B) =

25


As we know, P(A/B) =

P A BP B

( )( )∩

25 5

13

= ∩P A B( )

P(A ∩ B) =

25

513

213

× =

∴ P(A ∪∪∪∪∪ B) = P(A) + P(B) – P(A ∩ B)

=

526

513

213

+ −

=

5 10 426

+ − = 1126

13. L.H.S. =

5 2 22 5 22 2 5

a a b a c

b a b b c

c a c b c

− + − +− + − +− + − +

Applying C1 → C1 + C2 + C3

=

5 2 2 2 22 5 2 5 22 2 5 2 5

a a b a c a b a c

b a b b c b b c

c a c b c c b c

− + − + − + − +− + + − + − +− + − + + − +

=

a b c a b a c

a b c b b c

a b c c b c

+ + − + − ++ + − ++ + − +

2 25 2

2 5

Taking common (a + b + c) from C1, we get

= (a + b + c)

1 2 21 5 21 2 5

− + − +− +

− +

a b a c

b b c

c b c

Applying R2 → R2 – R1 and R3 → R3 – R1

= (a + b + c)

1 2 20 2 4 2 20 2 2 2 4

− + − ++ −− +

a b a c

a b a b

a c a c

Expanding along C1, we get= (a + b + c) [(2a + 4b) (2a + 4c) –

(2a – 2b) (2a – 2c)]= 4(a + b + c) [(a + 2b) (a + 2c) –

(a – b) (a – c)]= 4(a + b + c) [a2 + 2ac + 2ab + 4bc –

a2 + ac + ab – bc]= 4(a + b + c) (3ab + 3bc + 3ca)= 12 (a + b + c) (ab + bc + ca) = R.H.S.

(Hence proved)

14. sin y = x cos(a + y) …(i)

sincos( )

ya y+

= x


cos( ). cos sin ( sin( ))

cos ( )

a y ydydx

y a ydydx

a y

+ − − +

+2 = 1

cos(a + y – y)

dydx

= cos2(a + y)

…[ Q cos A cos B + sin A sin B = cos(A – B)]

dydx

=

cos ( )cos

2 a ya++ (Hence proved)

Putting the value of x = 0 in (i), we getsin y = 0.cos(a + y)sin y = 0 ⇒ y = 0

∴ Point A(x, y) = A(0, 0)

Now,

dydx

⎤⎤⎦⎦⎥⎥atA( , )0 0

=

cos ( )cos

2 0aa+ =

coscos

2 aa

= cos a (Hence proved)15. x = a sec3 θ y = a tan3 θDifferentiating Differentiatingboth sides w.r.t. θ both sides w.r.t. θ

dxdθ

= a.3 sec2 θ.sec θ tan θ

dydθ

= 3a tan2 θ . sec2 θ

dydx

=

dyddxd

θ

θ

=

33

2 2

2a

a

tan secsec sec tan

θ θθ θ θ

dydx

=

tansec

sincos

cosθθ

θθ

θ= ×1

dydx

= sin θ

Again differentiating both sides w.r.t. x,

d y

dx

2

2 = cos θ .

ddxθ

= cos θ .

13 3a sec . tanθ θ

=

13

4

acostan

θθ

∴

d y

dx

2

2

3

⎤⎤

⎦⎦⎥⎥⎥⎥ ==at θθ ππ

=

13

3

3

4

a

cos

tan

π

π

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

=

13

12

3

4

a

⎛⎝⎜

⎞⎠⎟

( )=

13

116

13

33a

× × × =

3144 a


x m x m

2x m

y m

Or

y = extan−1

…(i)Differentiating both sides w.r.t. x, we have

dydx

= extan−1.

11 2+ x

(1 + x2).

dydx

= y [From (i)


(1 + x2).

d y

dx

dydx

dydx

x2

2 2+ =.

(1 + x2).

d y

dx

dydx

dydx

x2

2 2+ −. = 0

(1 + x2).

d y

dx

dydx

x2

2 2 1++ −−( ) = 0 (Hence Proved)

16. Point of intersectionx2 + y2 = 4 (x – 2)2 + y2 = 4y2 = 4 – x2 …(i) (x – 2)2 + 4 – x2 = 4

[From (i)x2 – 4x + 4 – x2 = 0– 4x = – 4

⇒ x = 1

Putting the value of x in (i), y = ± 3

∴ y = 3 , but y ≠ – 3 ( Q Pt . lies in Ist Quadrant)

∴ Point of intersection is A(1, 3 )x2 + y2 = 4 (x – 2)2 + y2 = 4

Differentiating both Differentiating bothsides w.r.t. x, sides w.r.t. x,

2x + 2y

dydx

= 0 2(x – 2) + 2y

dydx

= 0

2y

dydx

= – 2x 2y

dydx

= – 2(x – 2)

dydx

xy

= − = m1

dydx

xy

= − −( )2 = m2

m1 at (1, 3 ) =

− 13

m2 at (1, 3 ) =

13

Let θ be the required angle.

tan θ =

m mm m

2 1

2 11−

+ .

tan θ =

13

13

113

13

− −⎛⎝⎜

⎞⎠⎟

+⎛⎝⎜

⎞⎠⎟

−⎛⎝⎜

⎞⎠⎟

=

23

3 13−

tan θ =

23

32

3× =

θ = 60° or

π3

∴ Required angle of intersection of the curves

is

ππ3

.

Orf (x) = – 2x3 – 9x2 – 12x + 1

Differentiating both sides w.r.t. x, we havef ′(x) = – 6x2 – 18x – 12

= – 6(x2 + 3x + 2)= – 6(x2 + 2x + x + 2)= – 6[x(x + 2) + 1(x + 2)]= – 6(x + 2) (x + 1)

When f ′(x) = 0, x = – 2, – 1

Intervals Checking Value of Sign of Nature ofPoint – 6 (x + 2) (x + 1) f ′′′′′(x) f(x)

(– ∞, – 2) – 3 – – – < 0 strictly decreasing(– 2, – 1) – 1.5 – + – > 0 strictly increasing(– 1, ∞) 1 – + + > 0 strictly decreasing

(i) f is strictly increasing in (– 2, – 1).(ii) f is strictly decreasing in (– ∞∞∞∞∞, – 2) ∪∪∪∪∪ (– 1, ∞∞∞∞∞).

17.

Let length of rectangular window = 2x mLet breadth of rectangular window = y m∴ Radius of rectangle window = x mPerimeter of window = 10 m∴ y + 2x + y + πx = 10 [ Q Perimeter of semi-circle = πππππr

2y = 10 – 2x – πx

y =

10 22

− −x xπ …(i)

y =

52

− −⎛⎝⎜

⎞⎠⎟

x xπ

Area of window = Area of rectangle+ Area of semi-circle

A = (2x)y +

12

πx2

A = 2x

10 22

12

− −⎛⎝⎜

⎞⎠⎟

+x xπ πx2 [From (i)]

A = 10x – 2x2 – πx2 +

12

πx2

A = 10x – 2x2 –

12

πx2


ddxA = 10 – 4x – πx

For Stationary points, 10 – 4x – πx = 0

⇒ – (4 + π)x = – 10 ⇒ x =

104 + π

m

ddx

2

2A = – 4 – π

ddx

x

2

2 104

A

at

⎤

⎦⎥⎥ =

+π

= – (4 + π) < 0 (–ve)

∴∴∴∴∴ Area is maximum at x =

104 ++ ππ

m.

∴ Maximum light will enter the window

when radius =

104 ++ ππ

m.

Length of rectangle window = 2x =

204 ++ ππ

m,

Breadth of rectangle window, y

= 5 –

104 2

104+ +

⎛⎝⎜

⎞⎠⎟

−π

ππ

=

20 5 10 54

+ − −+

π ππ

=

104 ++ ππ

m

18. Let

42 42( )( )x x− +

=

A B Cx

x

x−+ +

+2 42 …(i)

42 42( )( )x x− +

=

A B C( ) ( )( )( )( )

x x x

x x

2

24 2

2 4+ + + −

− +4 = A(x2 + 4) + Bx(x – 2) + C(x – 2)

Comparing the coefficient of x2, coefficent of xand constant terms respectively,

Coefficient of x2 Coeficient of x Constant term 0 = A + B 0 = – 2B + C 4 = 4A – 2C – B = A ...(ii) 2B = C ...(iii) 4 = 4(– B) – 2(2B)Putting the Putting the (From (ii) and (iii))

value of B in value of B in 4 = – 4B – 4B (ii), (iii), 4 = – 8B

–

−⎛⎝⎜

⎞⎠⎟

12

= A 2

−⎛⎝⎜

⎞⎠⎟

12

= C ⇒ B =

− 12

⇒

12

= A ⇒ C = – 1

Putting the value of A, B, and C in (i),

42 42( ) ( )x x

dx− +∫ =

12

2

12

1

42x

x

xdx

−+

− −

+

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟

∫

=

12

12 2

22 2 2log| |x dxx

x

dx

x− −

× +−

+∫ ∫4 22

=

12

14

12 2

2 42 1log| | log( ) tanx x x−− −− ++ −− ++−− C

Q

Q

f xf x

dx f x

dx

x a axa

’( )( )

log| ( )|

tan

== ++

++== ++

⎡⎡

⎣⎣

⎢⎢⎢⎢⎢⎢⎢⎢

∫∫

∫∫ −−

C

C2 211

19. (x2 – y2)dx + 2xy dy = 0 [Given]2xy dy = – (x2 – y2)dx

⇒

dydx

x yxy

= − +2 2

2 ⇒

dydx

xy

yx

= +−2 2

v + x

dvdx

x v xx vx

= − +×

2 2 2

2

let y vxyx

v

dydx

v xdvdx

== ⇒⇒ ==

== ++

⎡⎡

⎣⎣

⎢⎢⎢⎢⎢⎢⎢⎢ . .1

⇒ x v xdv

dxx v

x v

dvdx

vv

= − ⇒ =− −2 2

2

212

12

( ) – v

⇒ x

dvdx

v vv

dvdx

vv

x= ⇒ =− − − −2 2 21 22

12

⇒

21 2

vdv

v

dxx( )+

−=

21 2

v

v

dxx

dv+∫ ∫= − …[Integrating both sides]

dpp

dxx

=∫ ∫− …

Let p v

dp vdv

== ++==

⎡⎡

⎣⎣⎢⎢⎢⎢

1

2

2

log|p| = –log|x|+ log c

⇒ log|p| = log

cx

⇒ p =

cx

⇒ 1 + v2 =

cx

⇒ 1 +

yx

cx

2

2 = ⇒

x yx

cx

2 2

2+ =

⇒

x yx

2 2+ = c ⇒ x2 + y2 = cx …(i)

∴ 12 + 12 = c(1) [ Q x = y = 1 (given)]⇒ 2 = cPutting the value of c in (i),

x2 + y2 = 2xOr

Given differential equation is

(1 + x2)

dydx

+ 2xy =

11 2( )+ x

Dividing both sides by (1 + x2), we get

dydx

xyx x

+ =+ +2

11

12 2 2( )

On comparing with

dydx

+ Py = Q,

Here ‘P’ =

21 2

xx+

, ‘Q’ =

11 2 2( )+ x

I.F. = e e edxxx

dx xP∫ ∫= =+ +2

1 12 2log| | = 1 + x2

Q

f xf x

dx f x c′′

== ++⎡⎡

⎣⎣⎢⎢⎢⎢

⎤⎤

⎦⎦⎥⎥⎥⎥∫∫ ( )

( )log| ( )|


Hence the solution is y(I.F.) =

Q(I.F.)∫ dx

y (1 + x2) =

11 2 2

21( )

( )+

+∫ xx dx

y (1 + x2) =

11 2+∫ x

dx

y (1 + x2) = tan–1 x + c …(i)0.(1 + 1) = tan–1 (1) + c [ Qy = 0, x = 1]

0 =

π4

+ c ⇒ c =

− π4

Putting the value of c in (i), we get

y.(1 + x2) = tan–1 x –

ππ4

is the particular

solution.

or y =

tan( )

–1

2 21 4 1x

x x++ ++−− ππ .

20. AB→

= ( ) ˆ ( ) ˆ ( ) ˆ5 4 4 8 4− + − + −i x j k

= ˆ ( ) ˆ î x j k+ − +4 4

AC→

= ( ) ˆ ( ) ˆ ( ) ˆ5 4 4 4 1 4− + − + −i j k

= ˆ ˆ î j k+ −0 3

AD→

= ( ) ˆ ( ) ˆ ( ) ˆ7 4 7 4 2 4− + − + −i j k= 3 3 2ˆ ˆ î j k+ −

Points A, B, C and D are coplanar …(Given)

[ ]AB AC AD→ → →

= 0

1 4 41 0 33 3 2

x −−−

= 0

Expanding along R1⇒ 1(9) – (x – 4) (– 2 + 9) + 4(3) =0⇒ 9 – (x – 4)7 + 12 = 0⇒ 9 – 7x + 28 + 12 = 0⇒ – 7x = – 49 ⇒ x = 7

21. Herex1 = 1, y1 = 2, z1 = 3; x2 = 2, y2 = 3, z2 = 5,a1 = 2, b1 = 3, c1 = 4; a2 = 3, b2 = 4, c2 = 5Shortest distance between the lines, d

x x y y z z

a b c

a b c

b c b c c a c a a b a b

2 1 2 1 2 1

1 1 1

2 2 2

1 2 2 12

1 2 2 12

1 2 2 12

− − −

− + − + −( ) ( ) ( )

d =− + − + −

1 1 22 3 43 4 5

3 5 4 4 4 3 5 2 2 4 3 32 2 2[ ( ) ( )( )] [( ( ) ( )] [ ( ) ( )]

d =

− − + −

+ +

1 15 16 1 10 12 2 8 9

1 4 1

( – ) ( ) ( )

d =

− +1 2 2

6

– =

− 1

6

∴ d == 1

6 or

d == 6

6 units.

22. Let E1, E2, A be the following events.E1 : Ist group will winE2 : 2nd group will winA : Introducing a new productP(E1) = 0.6, P(A/E1) = 0.7P(E2) = 0.4, P(A/E2) = 0.3

Using Bayes’ theorem,

P(E2/A) =

P E P A EP(E )P(A / E ) + P(E )P(A / E

2

1 1 2 2

( ) ( / ))

2

=

0 4 0 30 6 0 7 0 4 0 3

. .( . . ) ( . . )

×× + ×

=

0 120 42 0 12

.. .+

=

0 120 54

1254

.

.= =

29

23. p = P(defective bulb) =

520

14

=

q = 1 – p = 1 –

14

34

= , n = 3

Here random variable X can take values 0, 1, 2, 3.Using P(r) = nCr.qn – r (p)r

P(X = 0) = 3C0.q3p0 = q3 =

34

2764

3⎛⎝⎜

⎞⎠⎟

=

P(X = 1) = 3C1.q2p1 = 3q2p = 3

34

14

2764

2⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

=

P(X = 2) = 3C2.q1p2 = 3qp2 = 3

34

14

964

2⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

=

P(X = 3) = 3C3.q0p3 = p3 =

14

164

3⎛⎝⎜

⎞⎠⎟

=

Probability distribution isXi 0 1 2 3

P(Xi)

2764

2764

964

164

∴ Mean = ΣPiXi

=

0 1 2 32764

2764

964

164

× + × + × + ×⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

=

0 27 18 364

+ + + =

4864

= 34


1 ✻ 5 = 1 + 5 – 6 = 02 ✻ 5 = 2 + 5 – 6 = 1 …5 ✻ 5 = 5 + 5 – 6 = 44 ✻ 5 = 4 + 5 – 6 = 3

Identity :Let a be an arbitrary element ofset A = {0, 1, 2, 3, 4, 5}

a ✻ 0 = a + 0 = a …(i)0 ✻ a = 0 + a = a …(ii)

From (i) and (ii),a ✻ 0 = 0 ✻ a = a ∀ a ∈{0, 1, 2, 3, 4, 5}

Hence, 0 is the identity for binary operation ✻.Inverse :

1 ✻ 5 = 1 + 5 – 6 = 02 ✻ 4 = 2 + 4 – 6 = 03 ✻ 3 = 3 + 3 – 6 = 04 ✻ 2 = 4 + 2 – 6 = 05 ✻ 1 = 5 + 1 – 6 = 0

Therefore, Inverse of 1 is 5 or (6 – 1)Inverse of 2 is 4 or (6 – 2)Inverse of 3 is 3 or (6 – 3)Inverse of 4 is 2 or (6 – 2)Inverse of 5 is 4 or (6 – 1)Inverse of a is (6 – a)

Here, for each element a ≠≠≠≠≠ 0, (6 – a) is its inverse.

25. A =

5 0 42 3 21 2 1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

|A| = 5(3 – 4) + 4(4 – 3) (Expanding along R1) = – 5 + 4 = –1 ≠ 0

∴ A–1 does exist.a11 = 3 – 4 a12 = – (2 – 2) a13 = 4 – 3

= – 1 = 0 = 1a21 = – (0 – 8) a22 = 5 – 4 a23 = – (10 – 0)

= 8 = 1 = – 10a31 = 0 – 12 a32 = – (10 – 8) a33 = 15 – 0

= – 12 = – 2 = 15

adj A =

− −−

−

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

1 8 120 1 21 10 15

Now, A–1 =

1| |A

. adj A

=

1

1

1 8 120 1 21 10 15

−

− −−

−

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

=

1 8 120 1 21 10 15

−−

− −

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

✻ 0 1 2 3 4 5

0 0 1 2 3 4 5

1 1 2 3 4 5 0

2 2 3 4 5 0 1

3 3 4 5 0 1 2

4 4 5 0 1 2 3

5 5 0 1 2 3 4

a b+ <⎧

⎨⎪

⎩⎪

0

24. R = {(x, y) : x, y ∈ Z and (x – y) is divisibleby 3 }

Reflexive : R is reflexive, for x ∈ Zx – x = 0, which is divisible by 3(x, x) ∈ R ∴ R is reflexive.

Symmetric :R is symmetric, (x, y) ∈ R where x, y ∈ Zx – y is divisible by 3.

Let (x – y) = 3m, where m ∈ Z– (y – x) = 3m(y – x) = – 3m

∴ (y – x) is divisible by 3.(y, x) ∈ R ∴ R is symmetric.

Transitive :R is transitive, (x, y) ∈ R where x, y ∈ Zx – y is divisible by 3x – y = 3m, where m ∈ Z …(i)

For (y, z) ∈ Z, where y, z ∈ Zy – z is divisible by 3.y – z = 3n, where n ∈ Z …(ii)

Adding (i) and (ii),(x – y) + (y – z) = 3m + 3nx – z = 3(m + n)

⇒ (x – z) is divisible by 3.(x, y) ∈ R, (y, z) ∈ R

∴ (x, z) ∈ R where x, y, z ∈ Z∴ R is transitive.Since R is reflexive, symmetric and transitive.∴ R is an equivalence relation.

OrOperation table for ✻ is :

0 ✻ 0 = 0 + 0 = 00 ✻ 1 = 0 + 1 = 10 ✻ 2 = 0 + 2 = 2

…

0 ✻ 5 = 0 + 5 = 5

a b+ ≥⎧

⎨⎪

⎩⎪

6


Applying R3 → 5R3

=

1 0 1

0 1

0 0 1

1 0 1

0

2 2 5

2

5

1

5

1

5

−

−

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

=

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

A


=

1 0 0

0 1

0 0 1

3 2 6

0

2 2 5

2

5

1

5

1

5−

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

=

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

A

Applying R2 → R2 +

25

R3

=

1 0 00 1 00 0 1

3 2 61 1 22 2 5

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

=⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥A

I = A–1A

∴ A–1 =

3 2 6

1 1 2

2 2 5

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

26. 0 ≤ 2y ≤ x2; 0 ≤ y ≤ x; 0 ≤ x ≤ 3Let 2y = 0, 2y = x2 Let y = x Let x = 0⇒ y = 0 x = 3For Point of intersection,We have x2 = 2y y = x …(i)∴ x2 = 2x …[from (i)

x2 – 2x = 0x(x – 2) = 0x = 0 or x = 2

Now, 2y = x2

y =

12

x2

As we know, (AB)–1 = B–1A–1

(AB)–1 =

1 3 31 4 31 3 4

1 8 120 1 21 10 15

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

−−

− −

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

=

1 0 3 8 3 30 12 6 451 0 3 8 4 30 12 8 451 0 4 8 3 40 12 6 60

+ − − − + + −+ − − − + + −+ − − − + + −

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

=

− −− −− −

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

2 19 272 18 253 29 42

Or

Let A =

1 2 21 3 0

0 2 1

−−

−

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

We know that A = IA

=

1 2 21 3 0

0 2 1

1 0 00 1 00 0 1

−−

−

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

=⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥ A


=

1 2 20 5 20 2 1

1 0 01 1 00 0 1

−−

−

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

=⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

A

Applying R2 →

15

R2

=

1 2 2

0 1

0 2 1

1 0 0

0

0 0 1

2

5

1

5

1

5

−

−

−

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

=

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

A


=

1 0 1

0 1

0 2 1

1 0 1

0

0 0 1

2

5

1

5

1

5

−

−

−

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

=

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

A

Applying R3 → R3 + 2R2

=

1 0 1

0 1

0 0

1 0 1

0

1

2

51

5

1

5

1

52

5

2

5

−

−

⎡

⎣

⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥

=

⎡

⎣

⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥

A X′ X

Y

8

7

6

5

4

3

2

1

–4 –3 –2 –1 0 1 2 3 4

Y′ x = 3

x 0 3 2y 0 3 2

2y =

x2

y = x

x 0 ±2 ±4y 0 2 8


The area of the shaded region

=

12

2

0

2

2

3

x dx x∫ ∫+

=

12

13

12

30

2 22

3× +[ ] [ ]x x dx

=

16

12

2 0 3 23 2 2( ) ) ( )− + −

=

86

12

+ (9 – 4) =

8 156

+ = 236

sq. units

27. Let I =

x x xx x

sin cossin cos

/

4 40

2

+∫π

dx …(i)

=

π π π

π π

π2 2 2

2 24 4

0

2 −⎛⎝⎜

⎞⎠⎟

−⎛⎝⎜

⎞⎠⎟

−⎛⎝⎜

⎞⎠⎟

−⎛⎝⎜

⎞⎠⎟

+ −⎛⎝⎜

⎞⎠⎟

∫x x x

x x

sin . cos

sin cos

/

dx

Q f x dx f a x dxa a

( ) ( )0 0∫∫ ∫∫

⎡⎡

⎣⎣

⎢⎢⎢⎢

== −−

I =

ππ

24 4

0

2 −⎛⎝⎜

⎞⎠⎟

+∫x x x

x x

cos sin

sin cos

/

dx …(ii)

Adding (i) and (ii), we get

2I =

x x x x x x

x x

sin cos cos sin

sin cos

/ + −⎛⎝⎜

⎞⎠⎟

+∫π

π2

4 40

2

dx

2I =

ππ2

4 40

2 sin cos

sin cos

/ x x

x x+∫ dx

Dividing num. and deno. by cos4 x, we get

2I =

ππ

2 1

2

40

2tan . sectan

/x x

x +∫ dx

=

π2

12 12

0

×+

∞

∫dp

p

=

π4

10

tan ( )− ∞[ ]p

2I =

π4

[tan–1 (∞) – tan–1 (0)]

∴ I =

π π8 2

0−⎡⎣⎢

⎤⎦⎥ =

ππ2

16

OrLet f (x) = 3x2 + 2x + 1, b = 3, a = 1,

nh = b – a = 3 – 1 = 2

f (a) = 3(1)2 + 2(1) + 1 = 6f (a + h) = 3(1 + h)2 + 2(1 + h) + 1

= 3 + 3h2 + 6h + 2 + 2h + 1= 3h2 + 8h + 6

f (a + 2h) = 3(1 + 2h)2 + 2(1 + 2h) + 1= 3 + 12h2 + 12h + 2 + 4h + 1= 12h2 + 16h + 6

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .

. . . . . .f (a + (n – 1)h) = 3[1 + (n – 1)h]2

+ 2[1 + (n – 1) h] + 1= 3 + 3(n – 1)2h2 + 6(n – 1)h + 2 +

2(n – 1)h + 1= 3(n – 1)2h2 + 8(n – 1)h + 6

f x dx

a

b

( )∫ = limh

h→0

[f (a) + f (a + h) + f (a + 2h)

+ ... + f(a + (n –1)h)]

( )3 2 12

1

3

x x dx+ +∫=

limh

h→0

[6 + (3h2 + 8h + 6) + (12h2 + 16h +

6) + … + 3(n – 1)2h2 + 8(n – 1)h + 6]=

limh

h→0

[6n + 3h2(12 + 22 + … + (n – 1)2]

+ 8h[1 + 2 + … + (n – 1)]

= limh→0

h

6 3 2 1 2 1

6

8 1

2n h

n n n hn n+ +⎡⎣⎢

⎤⎦⎥

− − −( )( ) ( )

= limh→0

6 21

2nh nh nh h nh h+ − − +⎡

⎣⎢( )( )

4nh nh h+ − ⎤⎦⎥

( )

=

6 21

22 2 4 4 2 2( ) ( )( )( ) ( )( )+⎡

⎣⎢⎤⎦⎥

− − + −h h h

( Q nh = 2)= 12 + 1(2 – 0) (4 – 0) + 8(2 – 0)= 12 + 8 + 16= 36 sq. units

28. Part I : Pt. (1, 2, 3) ∴ A→

= ˆ ˆ ˆi j k+ +2 3

Let D.rs of the required line are a, b, cRequired line is || to the given two planes∴ Required line is ⊥ to the normal to the

given two planes

Let

when

when

p xdp x x dxdp

x xdx

x p

x p

====

==

== == ∞∞

== ==

⎡⎡

⎣⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

tantan sec

tan sec

,

,

2

2

2

2

2

20 0

ππ


P

Y

X′ X

Y′

9

8

7

6

5

4

3

2

1

0 1 2 3 4 5 6 7 8

A(0, 4)

D(0, 0)

B(2, 3)

3x + y = 9

x + 2y = 8

�

C(3, 0)•

∴ a – b + 2c = 0and 3a + b + c = 0

a b c− −

−− +

= =1 2 1 6 1 3

= p (Let)

a = – 3p, b = 5p, c = 4p⇒ a = – 3, b = 5, c = 4

∴ B→

= − + +3 5 4ˆ ˆ ˆi j kVector equation of a line is

r→ =

→+

→A Bλ

∴ r i j k i j k→→ == ++ ++ ++ −− ++ ++( ˆ ˆ ˆ) ( ˆ ˆ ˆ)2 3 3 5 4λλPart II :

Now we have,

r i j k i j k→ = + + + − + +(ˆ ˆ ˆ) ( ˆ ˆ ˆ)2 3 3 5 4λ

r i j k→ = − + + + +( ) ˆ ( ) ˆ ( ) ˆ1 3 2 5 3 4λ λ λ

Let P(1 – 3λ, 2 + 5λ, 3 + 4λ) …(i) be the requiredpoint of intersection, must satisfy the givenequation of plane i.e.,

r i j k→ + +.( ˆ ˆ ˆ)2 = 4

⇒ [( ) ˆ ( ) ˆ ( ) ˆ]1 3 2 5 3 4− + + + +λ λ λi j k .

[ˆ ˆ ˆ]2 i j k+ + = 4

⇒ 2(1 – 3λ) + (2 + 5λ) + (3 + 4λ) = 4⇒ 2 – 6λ + 2 + 5λ + 3 + 4λ = 4⇒ 3λ = 4 – 7⇒ 3λ = – 3 ⇒ λλλλλ = – 1Putting the value of λ in (i), we get

P(1 + 3, 2 – 5, 3 – 4)Therefore, P(4, – 3, – 1) is the required point

of intersection.

29. Silver Gold ProfitType A, x 3g 1g � 40Type B, y 1g 2g � 50

9g 8g(at most) (at most)

Maximise, Z = � (40x + 50y)Subject to the constraints, 3x + 1y ≤ 9

1x + 2y ≤ 8x, y ≥ 0

Let 3x + y = 9 Let x + 2y = 8

x 0 3 x 0 8

y 9 0 y 4 0

Corner Points Z = �(40x + 50y)

A (0, 4) 0 + 200 = � 200

B (2, 3) 80 + 150 = � 230 →Maximum

C (3, 0) 120 + 0 = � 120

D (0, 0) 0 + 0 = � 0

Maximum value of Z is �� 230 at B(2, 3) i.e.

The company should produce 2 units ofType A and 3 units of Type B goods tomaxize the Profit.

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