Mathematics_Today_-_October_2016

Similarly, cosθ=.

GG

bc^ ...(ii)

∵ G

G GG

ab ab⊥∴. = 0

Also, GG

G G G

cpaqbrab=++×()

∴ GG GG G=++ ×

G GGG

ac paa qab ra a b.. ..()

= p + 0 + 0 = p ⇒ cos θ = p ∴ |p| = |cos θ| ≤ 1

Similarly, cosθ = q ⇒ |q| ≤ 1

Also, p = q.

12. (a, c) : ... Roots are complex
    ∴ D < 0 ⇒ (a + b + c)^2 – 4 (ab + bc + ca) < 0
    ⇒ (a + c)^2 + b^2 + 2b(a + c) – 4b(a + c) < 4 ac
    ⇒ (a + c – b)^2 < 4 ac ⇒ − 22 ac<+a c b− < ac
    ⇒ ()ac b acb+>^2 ⇒ +>⇒()a is correct.

⇒ acb+ − > ⇒ + − >

bc ab ca

0 1110

Similarly,

(^11101110)
bc ca ab ca ab bc

• − >+and − >
  ∴ On multiplying,
  111
  111 111 0
  ab bc ca
  bc ca ab ca ab bc
  ⎛ + −
  ⎝
  ⎜
  ⎞
  ⎠
  ⎟
  ⎛ + −
  ⎝
  ⎜
  ⎞
  ⎠
  ⎟ + −
  ⎛
  ⎝
  ⎜
  ⎞
  ⎠
  ⎟>
  ⇒ (c) is correct.

13. (a, c, d) : ... a sinθ + b cosθ = c ...(i)
    & a cosec θ + b sec θ = c ...(ii)
    ∴ On multiplying, abab^22 ++⎛ + c^2
    ⎝⎜

⎞

⎠⎟

sin =

cos

cos

sin

θ

θ

θ

θ

⇒ ++⎛

⎝⎜

⎞

⎠⎟

ab ab^2221 =c^2

sin 2 θ

∴ =

−−

sin2θ 2222 ab

cab

From (i) & (ii), a sin θ + bcosθ = a cosec θ + b secθ

⇒−⎛

⎝⎜

⎞

⎠⎟

+ ⎛ −

⎝⎜

⎞

⎠⎟

ab^11 = 0

sin

sin

cos

cos

θ

θ

θ

θ

⇒abcos +=

sin

sin

cos

22

θ 0

θ

θ

θ

∴abcos^33 θθ+=sin 0

From (i), b cos θ = c – a sinθ ...(iii)

& from (ii), b sec θ = c – a cosec θ ...(iv)

From (iii) & (iv), b^2 = c^2 + a^2 – ac (sinθ + cosecθ)

∴ sinθθ+=cosec acb+ −

ac

222

14. (b,c,d) :
    ... If x → 0 +, 2013x ∈ (1, 2) ⇒ {2013x} = 2013x – 1
    And, x → 0 – , 2013x ∈ (0, 1) ⇒ {2013x} = 2013x

lim ( ) lim ( )

x

x

x

f x x

→→

{}

00 ++=

1

2013

2013 2013

=

→

−

xlim (+ )

x x

0

1

2013 2013 1

=

→

lim ( ) − (∞ )

h

h h

0

1

2013 2013 1 1 form

==→

()−

−

⎡

⎣

⎢⎢

⎤

⎦

⎥⎥

eeh

h

lim 0 2013 1 .()h

1

2013 1

And,

lim ( ) lim

x

x

x

f x

x

→→−−

{}

= ()

00

2013 2013

1

2013

= ()= ()

→ →

−

−

{}−

lim lim

x

x

h

h

x h

0 0

2013 20130

1

2013

1

20130

= ()==

→

−

−

lim

h

h

h

0

2013 11 1

1

2013

... R.H.L ≠ L.H.L

⇒ lim ( )

x

f x

→ 0

2013 does not exist

15. (a, b) : Diagramatically

Clearly, required probability

= ×

⎛

⎝

⎜

⎜

⎞

⎠

⎟

⎟×

⎛ ××

⎝

⎜

⎜

⎞

⎠

⎟

⎟

⎡

⎣

⎢

⎢

⎤

⎦

⎥

⎥

2

3

1

3

2

6

3

1

1

2

1

3

1

6

3

CC

C

CCC

C

= ⋅⋅⋅⋅⋅

⋅

(^233123) ==
20 20
27
100
27%
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