Mathematics_Today_-_October_2016

SECTION-II

Multiple Correct Answer Type

11. If G

GG

abc,, be unit vectors such that G

G

abc⊥ , is

inclined at the same angle to both G

G

aband and

GGcpaqbrab=++×G ()G G then

(a) p = q (b) |p| ≤ 1

(c) |q| ≤ 1 (d) pq > 1

12. If a, b, c be three positive numbers and (ab + bc
    
    
    • ca)x^2 + (a + b + c)x + 1 = 0 has complex roots,
      then
      (a) acb+>
      (b) abc++> 4
      (c) 111 111
    
    
    
    

(^1110)
ab bc ca bc ca ab
ca ab bc
⎛ + −
⎝
⎜
⎞
⎠
⎟ + −
⎛
⎝
⎜
⎞
⎠
⎟
×+⎛ −
⎝⎜
⎞
⎠⎟

(d) none of these

13. If a sin θ + b cos θ = c = a cosec θ + b sec θ,then

(a) sin 2θ= 2222

−−

ab

cab

(b)

tan^3 θ=a

b

(c) a cos^3 θ + b sin^3 θ = 0

(d)

sinθθ+=cosec acb+ −

ac

222

14. Let fx x()= {}x

1

(where {·} denotes the fractional

part of x) then

(a)

lim ( )

x

f x

→+

=

0

2013 2013

(b)

lim ( )

x

f x

→−

=

0

2013 1

(c)

lim ( )

x

fex

→+

=

0

2013

(d)

lim ( )

x

f x

→ 0

2013 does not exist

15. A person draws 3 balls randomly from a bag
    containing 3 white and 3 black balls and then he
    put 3 red balls into the bag and draws 3 balls again
    randomly. The probability that now he has all 3
    balls of different colour is
    (a) > 20 % (b) > 25 %
    (c) > 30 % (d) > 33 %

SOLUTIONS

1. (c) :

Axxx

r

r

r

= ⎛−

⎝⎜

⎞

⎠⎟

⎧

⎨

⎪

⎩⎪

⎫

⎬

⎪

⎭⎪

=− + −∞

=

∞

∑

1

2

1 1

2

1

4

2

0

sin sin^24 sin ... to

=

−−⎛

⎝⎜

⎞

⎠⎟

=

+

1

1 1

2

2

sin^2 x^2 sin^2 x

B rxxx

r

==+++∞

=

∞

∑sin^2 sin sin ...

0

1to^24 =^1

cos^2 x

∴ =+⇒

+

AB x x x=

x

x

x

: sin :( cos ) cos

sin

sin

cos

4122

2

4

2

2 2

2

2

2

⇒ cos^4 x = 2sin^2 x + sin^4 x ⇒sinx=±^1

2

Hence, there will be 8 solutions in [–2π, 2π]

2. (b) :

lim

( ) ( ) .... ( )

x

xx x

→∞ x

{}++++++

+

1 2 2013

2013

2013 2013 2013

2013 20133

=

⎛ +

⎝⎜

⎞

⎠⎟

++⎛

⎝⎜

⎞

⎠⎟

+

+

→∞

lim

...........

x

x^2013 xx

1 2013 2013

1 2 1

2013

xx

x

x

⎛ +

⎝⎜

⎞

⎠⎟

⎧

⎨

⎪

⎪

⎩

⎪

⎪

⎫

⎬

⎪

⎪

⎭

⎪

⎪

+⎛

⎝⎜

⎞

⎠⎟

⎛

⎝

⎜

⎜

⎞

⎠

⎟

⎟

1

1 2013

2013

2013

2013

= ++++++

+

( 01 ) ( 01 ) .... ( 01 ) =

10

2013

2013 2013 2013

[As x → ∞, 1/x → 0]

3. (d) : Let x = z^3 ⇒ dx = 3z^2 dz
   3 2
   2

zdz

∫zz−

=

−

= − +

∫∫−

3

1

3 11

1

zdz

z

z

z

()dz

=+ (^313) []zzcx x clog | − |+=⎡⎣^33 +log | − 1 |⎤⎦+
⇒ α = β = γ = 3 ⇒ α, β, γ are in A.P. and G.P. both.

4. (c) : ... f(x) is an identity function ∴ f(x) = x
   ∴ Given equation becomes

xr

r

{}− + − =

=

∑ ()^20120

1

1

3

⇒

−

+

−

+

−

(^1) =
2013
1
2014
1
2015
0
xxx