JEE Advanced · 2011 · 10 Apr 2011 · Paper 2 · Relations And Functions XII

Let

f (x) = x^{2}

and

g (x) = sin x

for all

x \in R

. Then the set of all

x

satisfying

(f \circ g \circ g \circ f) (x) = (g \circ g \circ f) (x)

, where

(f \circ g) (x) = f (g (x))

, is

Answer: (A)

Solution

Computing the compositions:

(f \circ g \circ g \circ f) (x) = f (g (g (f (x)))) = f (g (g (x^{2}))) = f (g (sin (x^{2}))) = f (sin (sin (x^{2}))) = sin^{2} (sin (x^{2}))

(g \circ g \circ f) (x) = g (g (f (x))) = g (g (x^{2})) = g (sin (x^{2})) = sin (sin (x^{2}))

Setting them equal:

sin^{2} (sin (x^{2})) = sin (sin (x^{2}))

sin (sin (x^{2})) [sin (sin (x^{2})) - 1] = 0

So either

sin (sin (x^{2})) = 0

sin (sin (x^{2})) = 1

.

Since

- 1 \leq sin (x^{2}) \leq 1

, we have

sin (sin (x^{2})) = 1

only when

sin (x^{2}) = \frac{π}{2}

, which is impossible.

Thus

sin (sin (x^{2})) = 0

, giving

sin (x^{2}) = nπ

for integer

n

.

Since

- 1 \leq sin (x^{2}) \leq 1

, we need

sin (x^{2}) = 0

, so

x^{2} = nπ

, giving

x = \pm nπ

for

n \in {0, 1, 2, \dots}

JEE Advanced · 2011 · 10 Apr 2011 · Paper 2 · Single correct option