Mathematics_Today_-_October_2016

SOLUTION SET-165

1. (c) : Focus is 0 1
   4

⎛⎝⎜, ⎞⎠⎟

P

A(1, 2)

M

S

y=−^14

1

Directrix is y=−^14

4

PS + PA = PM + PA, takes

minimum value when A, P, M are collinear.

∴ Minimum value =+ = 2 1

4

9

4

2. (a) : The number of all 5 digit numbers is 6! – 5! = 600
   Numbers divisible by 11 without digit 1 are
   23045 with 8 permutations
   32450 with 8 permutations
   Likewise there are 16 numbers without 3 and 16
   numbers without 5
   ∴ The number of numbers divisible by 11 are 16 × 3 = 48
   Probability==^48
   600

2

25

3. (b) :^1
   1 21212 3

10

r∑= ()()( )rrr− ++

= − + − +

⎛

⎝⎜

⎞

⎠⎟

(^1) ==
8
1 2
3
1
3
1
21
1
23
40
483
m
n
∴ m + n = 523

4. (d) : Let α, β be the roots. α + β = –a, αβ = 6a
   Eliminating a, we get (α+ 6)(β + 6) = 36
   ∴ αβ=− 66 +=d − +^36
   d

,

The number of pairs (α, β), is the number of divisors

of 36.

i.e., d = ±1, ±2, ±3, ±4, ± 6

∴ The number of values of a is 10.

5. (c) : Let a = cos θ, b = sin θ

z = cos  + i sin  = cosθθsin

22

2

⎛⎝⎜ +i ⎞⎠⎟

(^) =

• −
  = +
  −
  cos sin
  cos sin
  ,
  θθ
  θθ
  22
  22
  i
  i
  ci
  ci
  where,
  c a
  b
  ==cot +cos = +
  sin
  θθ
  2 θ
  11

6. (a, b, c, d) : sin
   cos

cos

sin

3

3

3

3

θ

θ

θ

θ

+ =+ 12 8

sin^32 θ

⇒sin^6  + cos^6 =^3

2

sin^321 θ+

⇒ 1 3

4

− sin^22 θ=^3

2

sin^321 θ+

⇒ sin^2 2  (2sin 2 + 1) = 0

⇒ sin 2 = –^1

2

= sin^7

6

⎛ π

⎝⎜

⎞

⎠⎟

⇒ 2  = nπ+()− 1 n⎛⎝⎜^7 π⎞⎠⎟

6

⇒ (^) θ=+nππ− n θ= ππππ
2
1 7
12
7
12
11
12
19
12
23
12
() ; , , ,

7. (c) : dx
   dy

−x= y ⇒ xe–y = ∫ye–y dy = –(y + 1)e–y + A

x = 0, y = 0 ⇒ A = 1 and x = ey – (y + 1)

At y = ln 3, dx

dy

= 3 – 1 = 2 ⇒ dy

dx

=^1

2

8. (d) : ∫ 01 xdy = ∫()ey dyy−− 1
   0

(^1) = e−^5
2

9. (6) : The minimum number 1 occurs as 2nd, 3rd, ...
   9 th term in the sequence.
   ∴ = ⎛ −
   ⎝⎜

⎞

⎠⎟

=⎛

⎝⎜

⎞

⎠⎟

+⎛

⎝⎜

⎞

⎠⎟

++⎛

⎝⎜

⎞

⎠⎟

= − =

=

N ∑r

r

9

1

9

1

9

2

9

2 8 2 2 510

(^99)
...

10. (P) → 4; (Q) → 1; (R) → 3; (S) → 2
    (P) The desired number is the coefficient of x^10 in
    (x + x^2 + ... + x^6 )^4 or coefficient of x^6 in (1+ x + ... + x^5 )^4
    = (1 – x^6 )^4 (1 – x)–4
    = − ++⎛
    ⎝⎜

⎞

⎠⎟

+⎛

⎝⎜

⎞

⎠⎟

+

⎛

⎝⎜

⎞

⎠⎟

( 14 xxx^62 ...) 114 52 ...

Required coefficient =

9

6 484480

⎛

⎝

⎜

⎞

⎠

⎟− = − =

(Q) a = 2α – 1, b = 2β – 1, c = 2γ – 1, d = 2δ – 1

⇒ α + β + γ + δ = 10

The number of solutions is

9

3

⎛ 84

⎝

⎜

⎞

⎠

⎟=

(R) 1215

1

()++^5 =+^2

⎛

⎝

⎜

⎞

⎠

()xy ⎟()xy++^

5

2 2

5

3 2

⎛ 23

⎝

⎜

⎞

⎠

⎟ ++

⎛

⎝

⎜

⎞

⎠

(xy) ⎟(xy++) ....

The coefficient of xy^2

5

3

=⎛ 32 60

⎝

⎜

⎞

⎠

⎟⋅⋅=

(S)

rC

C

rr

r

r r

rrr r

⋅ = − + = −

==− =

∑∑∑

(^11)
10
1
10
1
(^10101)
()() 11
= +++ +=
1098 1 10 11⋅ =
2
... 55
””