sol for 2nd paper-1

8/12/2019 SOL FOR 2nd PAPER-1

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25. (A)

26. (B) Cyclisation

27. (B)

28. (A) IE of 2ndperiod > 3rd

29. (B) cis-trans

30. (B, C) one forming

31. (D) 5 unpaired e-

32. (A) Ksp = S S

moles/lit

33. (C)N ch arg of e

31

34. (B) no. of =80 82

42

=

35. (A)

36. (D)

37. (C)

38. (B)

39. (B) Fe 14

wVN (KMno )

E=

40. (B) 1w

1000.804

41. (A)

42. (C)

43. (A) CN COOH

44. (B)

O

3 3| |

CH C Cl CH Mg BR+

Mathematics:

45. (A, B)

Let 1 2 3 c c i c j c k = + +

2

2 2 2

1 2 3 c c c i j 2+ + = + =

Also,( ) ( ) ( ) ( )1 2 3 1 2 3 c i c j c k . i j c i c j c k . j k 1

2 2 2

+ + + + + += =

c1+ c2= c2+ c3= 1 c1= c3and c2= 1 c3

Hence from (1), 23 3

3c 2c 1 0 = c3=1

3 or 1 when c3=

1

3

( )

1 c i 4j k 3

= +

when c3= 1, c i k= +

46. (B, C)


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Let the line bex y z

,m n

= =

lthen

3 3 0

2 1 1 0

m m

=

l

l m n = 0

Again,( )2 2 22 m n

cos 606 m n

+ += + +

l

l5l

2

m2

+ 4mn + 8ln + 8lm = 0

Eliminating l, we get 2m2+ 5mn + 2n

2= 0 m = -2n or m =

n

2

So, m = -2n, l= -n or m = -n, l= -m. The desired direction ratios are 1, 2, -1 and 1, -1, 2

47. (A, D)

Let f(x) = x3 3x + a f(x) = 3x2 3 = 3(x 1) (x + 1)Now, f(1) = a 2, f(-1) = a + 2

All the roots would be real and distinct if,

f(1) f(-1) < 0 (a 2) (a + 2) < 0 -2 < a < 2Thus the given equation would have real and distinct roots if a (-2 , 2).48. (A, C)

Let f(x) = ( )33 x+ 2

3

3xf (x) 0 x [1, 3]

2 3 x = >

+

f(x) increases on the interval [1, 3] the least value of the function

( )2m f (1) 3 1 2= = + = and the greatest value of the function mf (3) = 23 3 30+ =

Therefore (3 1) 2

3

2

1

(1 x ) dx (3 1) 30+

Hence,3

3

1

4 (1 x ) dx 2 30 +

49. (A, C)

f(x) = [tan x] + tan x [tan x] [t] t [t] = + , where t = tan x

Clearly 0 t , at 0 x b = e.2b

2b,e

= if a < b

If co-ordinates of p be (x, y) then PS + PS+ SS= 2a + 2ae = 2a (1 + e)S = a (1 + e). semi perimeter of the and S - SS= a(1 e)

51. (A, B, C)

Let (xi, yi) =2

2ti,ti

i = 1, 2, 3, 4

The equation of the normal at 22t ,t

of, xy = 2 is ( )22y t x 2tt

=

xx2

x1 O

yf(x1) > 0, f(2) < 0


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[equation of the tangent at 22t ,t

to xy = 2 is2

2 ty x 4t

+ = slope of the tangent

is2

1

t ]

ty = t3x + 42 2 t . If it passes through (3, 4) then

4t = 3t

3

+

4

2 2 t

or

4 3

2 t 3t 4t 2 0 + =

If the roots of this equation are t1, t2, t3, t4then t1+ t2+ t3+ t4=

3

2x1+ x2+ x3+ x4= 3

t1t2= 0 and t1t2t3=4

2 and t1t2t3 t4= -1 y1y2y3y4= x1x2x3x4= ( )

4

2 = -4

And y1+ y2+ y3+ y4=1 2 3 4

1 1 1 12

t t t t

+ + +

=

1 2 3

1 2 3 4

2 t t t4

t t t t=

52. (A, B, C)

Centre of the first circle is (-3, 0) and the radius is 3 and radius of the second circle is (1, 0)

and the radius is 1. Since the distance between the centrs is equal to the sum of the radii,the two circle touch each other externally at the origin, the common tangent at the origin is

y = axis.

Let y = mx + c be a direct common tangent to the two circles, then

2 2

3m c m c3 and 1

1 m 1 m

+ += =

+ +

-6cm + c2= 9 and 2 cm + c2= 1

cm = -1 and c2= 3 c = 3 and1

m3

=

equation of the common tangents are1 1

y x 3, y x 3, x 03 3= + = =

Since the lines1

y x 33

= + and1

y x 33

= make angle of 60 with x = 0,

P

Q

R

(-3, 0) (1, 0)

The triangle PQR formed by these tangents is equilateral so that the centroid, circumcentreand orthocenter of the triangle coin side with its incentre (1, 0), the centre of the circle ofsmaller radius inscribed in the triangle PQR.

53. (B, D)

p. q. 4a + 2. 2a. 2a. 2- p. 4a2 4a . 42= 042 4a+ {(p + q)a pq} = 0 (a 0)y R, 16a2 4.4 {(p + q) a pq} 0 or (a p(a q) 0a p or a q

54. (A, B, C)

Let BP = n, CQ = n + 1, AR = n + 2Then BP = BR = nCQ = CP = n + 1 and AR = AQ = n + 2

BC = 2n + 1, CA = 2n + 3, AB = 2n + 2 and s =1

2[2n + 1 + 2n + 3 + 2n + 2] = 3n + 3

(3n 3) (n 2) (n) (n 1) + + + and in radius = 4s

=

n(n 2)4

3

+ = n2+ 2n 48 = 0 n = 6

So the sides are 13, 14, 15 and perimeter = 2s = 42 unit

31 8 6 7 = 7 3 4 = 84 unit

A

CB

R Q

I

P


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radius of circumcircle13 14 15 65

R cm4 84 8

= =

55. (A, B)

For simplicity we just consider a fourth order matrix, then

0 1 8 15

1 0 5 12A8 5 0 7

15 12 7 0

=

Which is a skew symmetric. Also,

A = (1 5 8 12 + 7 15)2= perfect square

56. (B, D)

We haven n n n

0 1 2 nc c c c

.......n n 1 n 2 2n

+ + + ++ +

1 1 1 1

n n 1 n n n n 1 n 2n 1

0 1 2 n

0 0 0 0

c x dx c x dx c x dx ....... c x dx + = + + + +

{ }1 1 2

n 1 n n n 2 n n n 1 n n n 1

0 1 2 n

0 0 1

x c c x c x ....... c x dx x (1 x) dx x (x 1) dx = + + + + = + =

Assertion-Reasoning:

57. (A)If AB = B, BA = A then A, B are idempotent matrix A2= A, B2= BA2+ B2= A + BBoth are true and it is correct reason.

58. (A)A + C = 180; B + 0 = 180

cos A = -cos C cos B + cos D = 0cos A + cos D = 0

= cos A + cos B + cos C + cos D = 0

but A + C = 180 sin A = sin CB + D = 180 sin B = sin DOnly II is correct

59. (C)

In =n 2 n 2

n n 2 2 n 2 2tan (x) tantan x dx tan x sec x tan x dx Inn 1 n 1

= = =

Put n = 6, 5(I6+ I4) = tan5x

A is true R is false

60. (D)

1y 9m

= and 1y m 4= 2

1y 36= y1= 6 product of sub tangent and sub normal is

y12. A is false R is true.

COMPREHENSION-1:

Consider a system of n chairs in circle. Now let us think of a frog jumping analogy. The frog at

the 1stseat can jump either to that seat itself or can jump to say 2, 3, 5 and back to one and so on.

The total number of ways the frog can jump back to original seat is equal to the number of waysthe experiment can be carried out. Rest of the problem B simple with the above analogy.

COMPREHENSION-2:


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sol for 2nd paper-1

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