⇒ (x – 2014) (x – 2015) + (x – 2015) (x – 2013)

+ (x – 2013)(x – 2014) = 0 ⇒ g(x) = 0 [say]

∴ g(2013) = (–1)(–2) = 2 > 0, g(2014) = (– 1)(1) = –1 < 0

and g(2015) = (2)(1) = 2 > 0

As g(x) changes sign between 2013 & 2014, 2014 &

2015, the equation g(x) = 0 has roots between them.

∴ Both roots are real and different.

5. (d) : Let sin
   cos

sin

cos

sin

cos

x

A

y

B

z

C

===k

∴ sin x = kcos A ...(i)

∴ −

−

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟= −−+

−

⎛

⎝

∑ sin cos ⎜

sin cos

cos sin

sin cos

(^22) Ax 112 2
xA
Ax
⎜⎜ xA
⎞
⎠
⎟⎟
= −
−
⎛
⎝
⎜⎜
⎞
⎠
⎟⎟
∑
∑
sin cos
sin cos
(^22) xA
xA
= ∑(sinx + cos A) = ∑sinx + ∑cos A

()+ ()−
−
∑∑ ∑∑
∑∑
sin cos sin cos
sin cos
xAxA
xA

()−()
−
∑∑
∑
sin cos
(sin cos )
xA
xA
22

6. (d) : xxxr px
   r

r r

r

() 1 2

00

++ =

=

∞

=

∞

∑∑

⇒ (1 + x + x^2 )(1 + x + x^2 + x^3 +... to ∞)

= p 0 + p 1 x + p 2 x^2 + p 3 x^3 + ... to ∞

⇒ ++

−

⎛

⎝⎜

⎞

⎠⎟

( )^1 =+ + ++...1 ∞

1

2

01 2

xx^2

x

ppxpx to

⇒ ppxpx px01 2++ + +^233 ... to∞

= (1–x)–2 (1 – x^3 )

= (1 + 2x + 3x^2 + 4x^3 + ... to ∞) (1 – x^3 )

∴ On comparing, we get

p 0 = 1, p 1 = 2, p 2 = 3 = p 3 = p 4 = p 5 = ........

∴ =+++++

=

∑pppppr p

r

0123 671

0

671

...

= 1 + 2 + (3 + 3 + 3 + ... to 670 terms)

= 3 × 671 = 2013

7. (b) : Let I dx
   xx

=

∫(tan) 1 + 2

=

+

= −−

∫ ∫ +

cot

(cot )

()

(cot )

2

2

2

2

xdx 1

xx

xdx

xx

cosec

=− +

+

=

+

∫dx x x + ⇒ =

xxxx

(cot) cfx x

(cot) cot

2 1 () cot

8. (d) : For non-trivial solution, we must have
   sin
   sin
   sin

2

2

2

11

11

11

0

11

11

11

0

α

β

γ

= ⇒ =

p

q

r

(where = sinp^2 αβγ, qrsin , sin )

p

pq

pr

RR

==

⇒− −

−−

=

→−

22

22

11

110

101

0

RR

RRR

1

331

,

→−

⎡

⎣

⎢

⎤

⎦

⎥

⇒ p(q – 1)(r – 1) – (1 – p)(r – 1) + 1{0 – (1 – p)

(q – 1)} = 0

⇒ − +

−

+

−

+

−

()p =

pqr

11

1

1

1

1

1

0

⇒

−

+

−

+

−

(^1) =
1
1
1
1
1
2221
sin αβγsin sin
⇒ ++=⇒ +=
⇒ =− <
sec sec sec ∑( tan )
tan
222 2
2
11 1
20
αβγ α
∑ α , which is impoossible

9. (c) : ... 256 a 1 a 2 a 3 a 4 ≥ ar
   r=

∑

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟

1

4 4

∵ 44 aa a a123 4≥a123 4+++a a a

⇒

+++

≤⇒≤

aa a a

123 4 (^4) aa a a123 4
4
A.M. G.M.
But, A.M ≥ G.M ⇒ A.M = G.M
⇒ a 1 = a 2 = a 3 = a 4 ...(i)
Also, 502a 1 + 503a 2 + 504a 3 + 505 a 4 = 2014
∴ 2014 a 1 = 2014 ⇒ a 1 = 1 = a 2 = a 3 = a 4
Now, aaaaarr
r
= + + + =+++=

∑ 122 33 44
1
4
1111 4^

10. (d) : ∵ xyt
    t

xyt

t

22 4 42

2

+=−^11 and +=+

⇒ xy xyt++ =+−

t

44 222 2 1 2

2

and xyt

t

442

2

+=+^1

⇒ x^2 y^2 = –1, which is not possible for x, y ∈ R.

∴ (a), (b) and (c) are not possible.

11. (a,b,c): ∵ G

G G

abc=== 1

Let θ be the angle between GG

G G

ac& and & bc

∴ cos ==

.

||||

θ.

GG

GG

ac GG

ac

ac ...(i)