MATHS - KEY ANSWER€¦ · 1. 5*7 = lcm of 5, 7 = 35 2. -1Range of the function y=sec x is 0, 2 3....

II PUC / BOARD EXAMINATION / MATHEMATICS‐KEY ANSWER 1

( ) . ( )

3 1 3

5 5 25

P A P B

x

P(A B)

II PUC MATHEMATICS - KEY ANSWER

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PART-A 1. 5*7 = lcm of 5, 7 = 35

2. Range of the function y=sec-1x is 0,2

3. 1x5, 5x1

4. x2 –36 = 36–36

x2 –36=0

6x36x2

5. y=tan√x sec √x √

6. 2x e dx + ex + C

7. A vector whose magnitude is the same as that of a given vector, but

direction is opposite to that of it, is called negative vector of the given

vector.

8. l = cos900=0, m=cos1350 = √, n cos45

√

9. Any feasible solution which optimizes (maximum or minimum) the

objective function is called its optimal solution.

10.

PART-B

11. f(x) = cosx, g(x) = 3x2

(i) gof(x) = g[f(x)] = g[cosx] = 3cos2x

(ii) fog (x) = f[g(x)], f[3x2] = cos3x2

12. cot-1(–x)=θ

–x = cot θ

X =–cot θ

X = cot( θ

cot-1x = θ cot -1(–x)

cot-1(–x) = cot -1x


1 1

2 2

3 3

1 2 3 11 1

1 3 2 12 2

1 1 8 1

12 ( 2 8 ) 3 (3 1) 1( 2 4 2 )

21 3 0

2 0 1 2 2 2 ] 1 52 2

1 5 .

x y

x y

x y

A rea o f th e tr ia n g le is sq u n its

13 2sin sin sin sin ( )

5 5

2sin

5

2,

5 2 2

1

1sin

13.

14.

15. Sin2x + cos2y = 1

2sinx . cosx – (2siny . cosy)y| = 0

sin2ysin2xy

2siny.cosy2sinxcosxy ||

16. y = xx

logy = logxx

logy = xlogx

1 1. log .1

(1 log ) (1 log)x

dyx x

y dx x

dyy x x

dx

17. f(x) = x2 – 4x = 6

f|(x) = 2x – 4

for f(x) strictly decreasing f| (x)<0

42 4 0

2x x

2 ( , 2)x x


18. I = cot . xlog sinx dx Let log sinx t

I = log sinx cotx. dx . cosxdx dt

I = tdt C cotx dx = dt

= 2 + C

19. I = x sec x dx

I = xtanx – tanx 1 dx

I = xtanx – xlog |secx|+C

20. Order = 2 Degree doesn’t exist

21. ˆ ˆˆ ˆ ˆ ˆ

a i 3j 7k, b=7i- j+8k+

2 2 2

.Projection of

(1) (7) (3) ( 1) (7) (8)

7 ( 1) 8

7 3 56 60

49 1 64 114

ona b

b

a b

22. Let ˆ ˆˆ ˆ ˆ ˆ3 , 2 7a i j k b i j k

ˆˆ ˆ

ˆˆ ˆ1 1 3 ( 1 21) (1 6) ( 7 2)

2 7 1

i j k

a xb i j k

ˆˆ ˆ20 5 5i j k 2 2 220 5 ( 5) 400 25 25A a xb

450

15 2 .A sq units


23. Equation of the plane is

123z4y6x

14z

3y

2xtherefore

4c3,b2,ahere1cz

by

ax

24. 1Σpi i = 1 0.1 + K + 2K + 2K + K = 1 6K = 1–0.1

0.90.15

6K

K = 0.15

PART-C

25. 1 2( , )T T R Triangle T1 is similar to triangle T2.

(i) Triangle T1 is similar to triangle 1 1 2( , )T T T R ∴ R is reflexive.

(ii) 1 2( , )T T R Triangle T1 is similar to Triangle T2

Triangle T2 is similar to Triangle T1 1 2( , )T T R ∴ R is symmetric and T2

(iii) 1 2( , )T T R Triangle T1 is similar to Triangle T2

2 3( , )T T R Triangle T2 is similar to triangle T3

Triangle T1 is similar to triangle T3 1 3( , )T T R ∴R is transitive

Therefore R is an equivalence relation

26. 1 1 1 1 1 1

4 11 1 4 1 313 72 tan tan tan tan tan tan

4 12 7 3 7 1713 7

27. F(x) F(y)=

cos cos sin sin cos sin sin cos 0

sin cos cos sin sin sin cos cos 0

0 0 1

x y x y x y x y

x y x y x y x y

y)F(x1000y)cos(xy)sin(x0y)sin(xy)cos(x


2

4 3

32

2 4

4

4

4

dxx at at

dtdy

y at atdt

dy atat

dx at

28.

29. f(x) = x2–4x–3 and [1, b] =[1, 4]

Being a polynomial function f(x) is continuous in [1, 4] and hence differentiable in (1, 4) f(b)=f(4)=-3, f(a)=f(1)=-6 and f|(x)=2x-4

| ( ) ( ) 5( ) 2 4 1 (1,4)

2

f b f af c c c

b a

Hence Mean Value Theorem is verified.

30. f(x)=√ where x=36 and ∆ 0.6 f(x + ∆ = f(x)+ f|(x) ∆

1

21 0.1

36.6 6 (0.6 6 6 0.05 6.0512 2

x x x xx

31. Problem must be dx1)(x

e3)(x3

x

dx1)(x

21)(x1xedx

1)(xe3)(x

33x

3

x

2 3 2

1 2

( 1) ( 1) ( 1)

xx e

e dx cx x x

32.

2π

0

2π

0

2

2sin2xx

21dxcos2x)(1

21dxcos | 2

0

1 1

(sin )2 2 2 4


Required equation is (3 2 4) ( 2) 0

.(2,2,1) (1) (6 2 2 4) (2 2 1 2) 2

2 3 0

2

32

(1) , (3 2 4) 9 2) 03

7 5 4 8 0

x y z x y z

Sub in we get

becomes x y z x y z

x y z

37.

38. E1 : Event that six occurs E2 : Event that six does not occur A=Event that man reports that six occurs in the throwing of the die.

83

245

243

243

41

65

43

61

43

61

)P(A/E)P(E)P(A/E)P(E)P(A/E)P(EA)/P(E

43

EAP,

65BP,

61)P(E

2211

111

121

PART-D

39. Given function is f(x)=4x+3, x ∈ R

x4

334x3)(4xf

(f(x))f(x)f).(fL.H.Sx(x)f).(fproveTo:Step(2)

43y(y)f

34xyf(x)yx(y)fLet

(y)ffindTo:Step(1)

1

11

1

1

1

1


43y(y)fwithinvertibleisf

If.fandIf.f:Step(4)

y343-y4

43-yf(y))f(f(y))f. (fL.H.S

y(y))f. (fproveTo:Step(3)

1

11-

11

1

40.

1 2 3 3 1 2 4 1 2

5 0 2 4 2 5 0 3 2

1 1 1 2 0 3 1 2 3

4 1 1 4 1 2 0 0 3

9 2 7 0 3 2 9 1 5

3 1 4 1 2 3 2 1 1

3 1 2 4 1 2 1 2 0

4 2 5 0 3 2 4 1 3

2 0 3 1 2 3 1 2 0

B

A B C

B C

A

1 2 3 1 2 0 0 0 3

( ) 5 0 2 4 1 3 9 1 5

1 1 1 1 2 0 2 1 1

( ) ( )

A B C

A B C A B C

413

729

114

CB)(A


41.

42.

43.

1z;2y;1x

121

zyx

34

5

71151135935

401X

71151135935

Adj.(A);40213

121332

A

(A)adj.A1AwhereD,.AX

34

5D,

zyx

X;213

121332

AwhereD,AX

11

2y)x(12xy)x(1)x(1

1.2(2x)yy)x(1

x2tany)x(1x1

1x)2(tany

x)(tany

12

222

2122

11

2

21

1

21

sec/cm48

1dtdh

2)4(36

12dtdh

dtdh.2h36

dtdv

3h12V

h2r31V,conetheofVolume

cm4handh6rr.61h;sec/3cm12

dtdv


44.

C1)(xtandx11)(x

1dx22xx

1

11)(x112xx22xx

Caxtan.

a1dx

ax1

θseca

1)θ(tanaaxCθ.a1

θdθsecadxθdθseca.θseca

1

θtanaxPutdxax

1I

1222

2222

122

22

2222

2222

22

45. 2interceptyand3interceptxmakes1

2y

3xlineThe

2b3,114y

9xisEllipse

22


32 2

0

32 22 2 1

0

2 22 2 1

2 2Re , 3 (3 )

3 3

2 33 sin 3

3 2 2 3 2

2 3 3 3 33 3 sin 3(3) 0

3 2 2 3 2

2 9 9 39 3 .

3 2 2 2 2

quired Area A x x dx

x x xx x

sq units

46.

Cx161

4xx)(logyx

Cdx4x.

x1

4x.xlogyx

Cdxxlogx)y(x

CdxQ(IF)y(IF)issolutiongeneralThe

xeeeIF

xlogxQ;x2P

xlogxyx2

dxdy

44

2

442

32

22logxdxx2

Pdx

47.


Let a be the position vector of the given point A with respect to the origin 0 of the rectangular co-ordinate system. Let l be the line which passes through the point A and is parallel to a given vector b. Let r be the position vector of an orbitrary point P on the line.

Then is parallel to the vector b

i. e. AP λb, where λ ∈ R

OP – OA λb

r a λb is the vector form

The cartesian form is as follows:

Let A A(x1, y1, z1) and the direction ratios of the line be a, b, c

Consider P P(x, y, z) Then,

czz

byy

axxisformcartesianThe

λczz;λ

byy;λ

axx

c)b,(a,λ)z,y,xz)y,(x,bλar

ckbjaibandkzjyixazk,yjxir

111

111

111

111

48.

50

0050

050

4949|150

150

xxnx

n

100991

1001

10099C1

0XP11)P(X11)P(X(b)

10099

21

1001

1009950

1001

10099C1)P(X(a)

2,......501,0,x,pqCx)P(X10099p1q,

1001p50,n


PART-E

49. Pf: First we prove that

1

1

4545

a

a

a

a

a

0

a

0

a

0

a

0

a

0

a

a

a

0

a

0

0

a

a

0

a

a

0

a

a

0

a

a

a

0

function)oddanisxcosxsin(0dxxcosxsinNow

0dxf(x)

f(x)x)f(thenfunction,oddisf(x)If:(2)Case

dxf(x)2dxf(x)dxf(x)dxf(x)

f(x)x)f(thenfunction,evenanisf(x)If:(1)Case

dxf(x)dxx)f(dxf(x)

dxf(x)dtt)f(

dxf(x)dt)(t)f(

0t0xatax

dtdx

txputintegral1sttheIndxf(x)dxf(x)dxf(x)

oddisf(x)if0,

evenisf(x)if,dxf(x)2dxf(x)


49. b.

2x)(44)(5xx400

0x402x2x1

4)(5x

4x2x12x4x12x2x1

4)(5x

4x2x45x2x4x45x2x2x45x

4x2x2x2x4x2x2x2x4x

L.H.S.

50.

x+y=50 3x+y=90

X 50 0

Y 0 50

X 30 0

Y 0 90


The feasible region is as shaded in the graph, it is a bounded set, hence the optimal solution exists at the corner points.

Corner points Z = 4x + y

(x, y) = (0, 0) Z = 0

(x, y) = (30, 0) Z = 120

(x, y) = (20, 30) Z = 110

(x, y) = (0, 50) Z = 50

∴ The maximum value of the function is Zmax=120 and it occurs at the point (x, y) = (30, 0)

50. Since f(x) is continuous at x = , we have,

2k

11kπ

)(xx(cosx)Lt

)(xx

1)(kxLt

)(xx

f(x)lim

)(xx

f(x)lim

************** End ***************

MATHS - KEY ANSWER€¦ · 1. 5*7 = lcm of 5, 7 = 35 2. -1Range of the function y=sec x is 0, 2 3....

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Transcript of MATHS - KEY ANSWER€¦ · 1. 5*7 = lcm of 5, 7 = 35 2. -1Range of the function y=sec x is 0, 2 3....