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Page 1: jiyg;gpw;F fPNo cs;s ypq;if fpspf; nra;J FOtpy; ,izaTk;! ) · 9/11/2019  · SHRI KRISHNA ACADEMY,NAMAKKAL-9965531727 3 2 2 2 2 2 3 44 (1 ) 4 400 20 p mc p c a m pp §· ¨¸ ©¹

( jiyg;gpw;F fPNo cs;s ypq;if fpspf; nra;J FOtpy; ,izaTk;! )

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Page 2: jiyg;gpw;F fPNo cs;s ypq;if fpspf; nra;J FOtpy; ,izaTk;! ) · 9/11/2019  · SHRI KRISHNA ACADEMY,NAMAKKAL-9965531727 3 2 2 2 2 2 3 44 (1 ) 4 400 20 p mc p c a m pp §· ¨¸ ©¹

SHRI KRISHNA ACADEMY,NAMAKKAL-9965531727 1

SHRI KRISHNA ACADEMY NEET,JEE AND BOARD EXAM COACHING CENTRE

SBM SCHOOL CAMPUS,TRICHY MAIN ROAD,NAMAKKAL CELL:9965531727-9443231727

QUARTERLY EXAMINATION -2019

XI - BUSINESS MATHEMATICS TENTATIVE ANSWER KEY

ANSWER KEY

Q.No PART – I 1.

1 5

30 2 12.

2 47

5 5

a

2. c.4

3.

1.

nb A

4. d. prof.Wassily W.Leontief

5. c.2n-1

6. a.9

7. (all options wrong)correct answer is 18

8. b.20

9. 2.

hc

b

10. d.

2 22 0x ay y

11. a. Latus returm

12. c.Rene Descartes

13. 3 1.

2 2d

14. 1

.3

a

15. b.56/55

16. a.0

17. d.2

18. a.1

19. a.-1

20.

2.5 xC e

x

PART – II

21. 2

2

2

a ab ac

ab b bc

ac bc c

Tacking a,b,c common from c1, c2, c3

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Page 3: jiyg;gpw;F fPNo cs;s ypq;if fpspf; nra;J FOtpy; ,izaTk;! ) · 9/11/2019  · SHRI KRISHNA ACADEMY,NAMAKKAL-9965531727 3 2 2 2 2 2 3 44 (1 ) 4 400 20 p mc p c a m pp §· ¨¸ ©¹

SHRI KRISHNA ACADEMY,NAMAKKAL-9965531727 2

0

0

2

a a a

abc b b b

c c c

a a

abc b b

c c c

Expanding along c1

2

2 2 2

2

2

4

a aabc c

b b

abc ab ab

a b c

22. 0.8 0.2

0.9 0.7

0.2 0.2

0.9 0.3

0.12 0

B

I B

I B

I B is negative Hawkins- simon condition are not satisfied.

23. 2

1

1 11

A B

x xx

Multiplying both side by (x-1)(x+1), we get

1 1 1

11

2

11

2

A x B x

putx A

putx B

24. 2 4 1 3

1 2 0 0

3! 2! 1! 0!

6 4 0 0 1

E X A M

RANK =1

25. Let m1 and m2 be the slope of 2x-y+3=0 and x+y+2+=0

m1=2 , m2= -1

1

2 1tan 3

1 2( 1)

tan (3)

26. The given equation are

2 2 16 3 4 0x y and x y p

The condition for the tangency is c2 = a

2 (1+m

2)

a2 = 16 from the equation 3 4 0x y p

3

4 4

py x

Comparing with y =mx+c

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Page 4: jiyg;gpw;F fPNo cs;s ypq;if fpspf; nra;J FOtpy; ,izaTk;! ) · 9/11/2019  · SHRI KRISHNA ACADEMY,NAMAKKAL-9965531727 3 2 2 2 2 2 3 44 (1 ) 4 400 20 p mc p c a m pp §· ¨¸ ©¹

SHRI KRISHNA ACADEMY,NAMAKKAL-9965531727 3

2

2 2 2

2

3

4 4

(1 )4

400 20

pm c

pc a m

p p

27.

8 12 8 122cos cos

2 2

2cos10 cos 2

2cos10 cos 2

A A A A

A A

A A

28. ( )

0

0

y f x x

x if x

x if x

x -2 -1 0 1 2

y 2 1 0 1 2

Points (-2,-2),(-1,1)(0,0),(1,1),(2,2) and draw a smooth curve.

29. 3 3

3 30

3

330

0

log(1 )lim

sin

log(1 ) 1lim

sinlim

1

x

x

x

x x

x x

x

x x

x

30. Since ABCD are a quadrilateral

0 0

0 0

0 0

0

180 180

180 180

. . sin sin

sin sin 180 180

sin sin 360 ( )

sin sin

0

A c c A

B D D B

L H S A B C D

A B A B

A B A B

A B A B

PART – III

31.

1 1

1 1

1

2 1

4 2 2 1

3 1 1 0

4 2 2 11

3 1 1 02

5 311

4 22

2

5 3 5 312

4 2 4 22

A B

adjA adjB

A B

A B

AB

adjAB AB

From 1 and 2 A-1

B-1

=(AB)-1

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Page 5: jiyg;gpw;F fPNo cs;s ypq;if fpspf; nra;J FOtpy; ,izaTk;! ) · 9/11/2019  · SHRI KRISHNA ACADEMY,NAMAKKAL-9965531727 3 2 2 2 2 2 3 44 (1 ) 4 400 20 p mc p c a m pp §· ¨¸ ©¹

SHRI KRISHNA ACADEMY,NAMAKKAL-9965531727 4

32.

1

1

2 0 1 0

5 1 0 4

0 1 3 5

1 0

3 1 1

15 6 5

5 2 2

3 1 1

15 6 5

5 2 2

x

y

z

AX B

A A exists

adjA

A

33.

34.

35.

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SHRI KRISHNA ACADEMY,NAMAKKAL-9965531727 5

36. Given x+y=6 and x+2y=4 are two diameters of the circle.

Solve the two equation y = -2 , x = 8

The poin of intersection of two diameters = (8,-2) and center = C(8,-2)

Equation of the circle is (x-8)2+(y+2)

2=r

2 at passing through the point (2,6)

r2=100

Required equation of the circle x2+y

2-16x+4y-32=0.

37.

38. Given tanA – tanB = x and cotB – cotA = y

1 1 1 1

tan tan cot cot

1 1

1 1tan tan

tan tan

1 tan tan

tan tan tan tan

1 tan tan

tan tan

1

tan

cot

x y A B B A

A B

B A

A B

A B A B

A B

A B

A B

A B

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SHRI KRISHNA ACADEMY,NAMAKKAL-9965531727 6

39.

40. f(x) is continuous at x=1

Lf(x)=Rf(x)=1

1 1

1

1

lim lim ( ) 15

lim ( ) 15

2 5 1

lim ( ) 15

2 15 2

x x

x

x

f x

f x

a b

f x

a b

From 1 & 2 a = 5 and b = -5

PART – IV

41.(a) 4 11 5

35 35 351 3 71 6 25

4 2 335 35 35

1 2 16 1 10

35 35 35

35 0 01

0 35 035

0 0 35

1 0 0

0 1 0

0 0 1

,ly

AB

I

lll BA I

AB BA I

A and B are inverce of each other.

(b) 0.4 0.7

0.1 0.7

0.6 0.7

0.1 0.3

0.11

B

I B

I B

The system is viable.

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SHRI KRISHNA ACADEMY,NAMAKKAL-9965531727 7

1

1

0.3 0.71

0.1 0.60.11

0.3 0.7 6.81

0.1 0.6 10.20.11

83.45

61.82

I B

X I B D

A

B

Out put A=83.45 tonnes & B=61.82 tonnes.

42.(a)

(b)

11 415 15 1365C C

10 414 14 1001C C

11 314 14 364C C

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SHRI KRISHNA ACADEMY,NAMAKKAL-9965531727 8

43.(a) Let y represent the costand x represent the no.of TV sets.

1 1

2 1 2 1

220000 80

287500 220000 125 80

220000 80

67500 45

1500 100000

y y x x

y y x x

y x

y x

y x

The cost of 100 TV set.

=1500(100)+100000

=RS.250000.

(b)

44.(a)

1 1

1

2 21 1

2 2

. . tan tan

tan

1

tan tan (1)4

m m nL H S

n m n

m m n

n m n

m m n

n m n

m n

m n

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SHRI KRISHNA ACADEMY,NAMAKKAL-9965531727 9

(b)

2

2

2 2

2 2

cos 4 cos 2(2 )

1 2sin (2 )

1 2(2sin cos )

1 2(4)sin cos

1 8sin cos . .

x x

x

x x

x x

x x R H S

45.(a)

1

3 2

2

3 4

5 6 3 2

x xY

x x x

Tacking log on both sides

3 21log log 3 log 4 log 5 6 3 log 2

2y x x x x x

2

3 2

3 2

3 22

1 1 1 3 10 6 1

2 3 4 5 6 3 2

3 41 1 3 10 6 1

2 3 4 5 6 3 25 6 3 2

dy x x

y dx x x x x x

x xdy x x

dx x x x x xx x x

(b)

Given

21m

y x x

Differentiating both sides,with respect to x,

112 2 2

2

12

2

1

11 1 1 2

2

1

1

1

m

m

dym x x x x

dx

m x xy

x

y x my

Differentiating both sides,with respect to x

2 2 2

1 1 2 12 1 2 2xy x y y m yy

Dived each term by 2y1

2 2

11 0x xy m y

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SHRI KRISHNA ACADEMY,NAMAKKAL-9965531727 10

46.(a)

(b)

47.(a)

2

2

2

2

2

6 4 21 0

6 4 21

6 9 9 4 21

3 4 3

4

x x y

x x y

x x y

x y

X aY

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SHRI KRISHNA ACADEMY,NAMAKKAL-9965531727 11

3

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14 1

4

X x

Y y

a a

Axis: x = -3

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Directrix: y=2

Length of latus rectum: 4(1)=4

(b)

22

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1 11 1

3 5 1 1

x A Bx C

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x A x Bx C x

Sub 1x

4A

Equating coefficient of x2

A + B = 0

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Equating coefficient of x -B + C = 3

C= -1

22

3 5 4 4 1

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