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### IIT-JEE 2011 Paper 2 Offline

Let f(x) = x2 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

A
$$\pm \sqrt {n\pi } ,\,n \in \{ 0,1,2,....\}$$
B
$$\pm \sqrt {n\pi } ,\,n \in \{ 1,2,....\}$$
C
$${\pi \over 2} + 2n\pi ,\,n \in \{ ....., - 2, - 1,0,1,2,....\}$$
D
$$2n\pi ,n \in \{ ....., - 2, - 1,0,1,2,....\}$$

## Explanation

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

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

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

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

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

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

Given, $$(f \circ g \circ g \circ f)(x) = (g \circ g \circ f)(x)$$

$$\Rightarrow {(\sin (\sin {x^2}))^2} = \sin (\sin {x^2})$$

$$\Rightarrow \sin (\sin {x^2})\{ \sin (\sin {x^2}) - 1\} = 0$$

$$\Rightarrow \sin (\sin {x^2}) = 0$$ or $$\sin (\sin {x^2}) = 1$$

$$\Rightarrow \sin {x^2} = 0$$ or $$\sin {x^2} = {\pi \over 2}$$

$$\therefore$$ $${x^2} = n\pi$$ (i.e. not possible as $$- 1 \le \sin \theta \le 1$$)

$$x = \pm \sqrt {n\pi }$$

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