IIT-JEE 2011 · Paper 2 · Mathematics Q47

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MathematicsSingle correct+3 / -1Question 47 of 60

Let $f(x)=x^2$ and $g(x)=\sin x$ for all $x\in\mathbb R$ . 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

\pm\sqrt{n\pi}

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

\pm\sqrt{n\pi}

n\in\{1,2,\ldots\}

\dfrac\pi2+2n\pi

n\in\{\ldots,-2,-1,0,1,2,\ldots\}

2n\pi

n\in\{\ldots,-2,-1,0,1,2,\ldots\}

Answer and worked solution

Official answer: A

Write it out. With $u=\sin(\sin x^2)$ , the equation is $u^2=u$ , so $u=0$ or $u=1$ .

$u=1$ is impossible: it would need $\sin x^2=\frac\pi2>1$ .

$u=0$ : $\sin x^2=0$ , so $x^2=n\pi$ and $x=\pm\sqrt{n\pi}$ , $n=0,1,2,\ldots$

(A)

Topics:Composition and inverse functions Trigonometric equations

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IIT-JEE 2011 · Paper 2, Mathematics question 47