kvpy sol-20091

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CLASS-XII_STREAM-SB+2_PAGE # 3

6.

P P'

(5, 15) (21, 15)L

A

Mid point of PP = +,

-./

0 "

2

1515,

2

215

L = (13, 15)! Point A will be (13, 0)By property PA + PA = 2a

PA = 22 )150()513( ((

= 22564 "

= 289 = 17 cm

PA = 22 )150()2113( ((

= 22564 "

= 289 = 17 cm

! 2a = PA + PA2a = 17 + 172a = 34 cm

So, length of major axis = 2a = 34 cm.

7.

B C

P(10, 10)

(a, b)(0, 6) 2x + 3y = 18

PB = PC(10 0)2 + (10 6)2 = (a 10)2 + (b 10)2

100 + 16 = a2 + 100 20a + b2 + 100 20ba2 + b2 20a 20b + 84 = 0 ....(i)

Also (a, b) i.e. on 2x + 3y = 182a + 3b = 18

a = 9 2b3

Using equation (i)

2

2

b39 +

,

-./

0( + b2 20 +

,

-./

0(

2

b39 20b + 84 = 0

81 +4

b9 2 27b + b2 180 + 30b 20b + 84 = 0

4

b13 2 17b 15 = 0

13b

2

68b

60 = 0

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sinx siny + cosx cosy = 0

cos(x y) = 0

! x y = 90

By (1) (2) we get

sinx cosx + sinx cosy + siny cosx + siny cosy =25

7

sin(90 + y)cosx + sin(x + y) + sin(x 90) cos y =25

7

cosy cosx + sin(x + y) cosx cosy =25

7

sin(x + y) =25

7.

10.6xy =

0

1

(1, 1)

y = sin x

Clearly, curve meet each other twice in 24 3444 5464 74

84 94

104 114! Total 10 Times.

11. f(x) is differentiable on R.So, it will be contincous on R.Continuity at x = 0LHL

0xlim) x

xsin 2

Put x = 0 h, then h ) 0

0hlim

) h

)h0sin( 2

(

(

0hlim

)

h

h

h

hsin*

(= 0

RHL

") 0xlim x2 + ax + b

Put x = 0 + h, then h ) 0

0hlim

)h2 + ah + b = b

Value of f(x) at x = 0f(0) = b.

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For the critical points

1dy

dz= 0

(y1 2)(y

12 + 2y

1+ 12) = 0

y1

= 2

! y12 = 4x

1

% (2)2 = 4x1% x

1= 1.

2

2

1dy

dz++

,

-

.

.

/

0+ 2z 2

1

2

dy

zd= (y

12 + 2y

1+ 12) + (y

1 2) ( 2y

1+ 2)

= y1

2 + 2y1

+ 12 + 2y12 4y

1+ 2y

1 4

= 3y1

2 + 8.

when y1

= 2 and1dy

dz= 0

21

2

dy

zd> 0

! z is min at (1, 2)

Minimum distance = 22 )32()01( (( = 11" = 2 .

13. We can find the answer through option as the sum of weight of packet taken from trucks is 1022870 gm

and its unit digit is 0. The truck that have heavier bags have unit digit 0. So, the truck have lighter bags in

which the sum of weight of bags must have unit digit 0.

So, according to option D. i.e. truck no. 2, 8

Track 2 have 21 bags and total weight = 21 999 gm = .......8 gmTruck have 27 bags and total weight = 27 999 = 128 999 gm = ......2 gm

So, the unit digit of the weight contain by truck 2, 8 together is 0.

14. dx)]x2cos([)xcos(1

05 44

= dx0cos)xcos(2/1

05 4 + dxcos)xcos(1

2/15 44

=dx)xcos(

2/1

054

dx)xcos(

1

2/154

= 68

9:

;

4

4xsin2/1

0

68

9:

;

4

4xsin1

2/1

= 67

89:

;

4

1 6

7

89:

;

4(

10

=4

2.

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15. IN

= 67

89:

; (

n

x)nxcos( 101

0

+ 51

0

9

n

dxx)nxcos(10

= 0 + 66

7

8

99

:

;

(667

8

99:

;

51

0

81

9

n

dxx)nxsin(

n

9

n

x)nxsin(

n

10

0

= 66

7

8

99

:

;*( 5

1

0

8

2dxx)nxsin(

n

910

= 66

7

8

99

:

;

51

0

10dx)nxsin(

n

!10

= 0 as Denom )

kvpy sol-20091

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