Let fx=sin⁡π6sin⁡π2sin⁡x, for all x∈R and gx=π2sin⁡x, for all x∈R. Let fog x denote fgx and gof x denote gfx. Then which of the following is (are) true?

(D) There is an x∈R such that g o fx=1

JEE Advanced · Mathematics · 22. Functions

Let $f (x) = \sin (\frac{π}{6} \sin (\frac{π}{2} \sin x))$ , for all $x \in R$ and $g (x) = \frac{π}{2} \sin x$ , for all $x \in R$ . Let $(f o g) (x)$ denote $f (g (x))$ and $(g o f) (x)$ denote $g (f (x))$ . Then which of the following is (are) true?

A Range of $f$ is $[- \frac{1}{2}, \frac{1}{2}]$
B Range of $f o g$ is $[- \frac{1}{2}, \frac{1}{2}]$
C $\lim_{x \to 0} \frac{f (x)}{g (x)} = \frac{π}{6}$
D There is an $x \in R$ such that $(g o f) (x) = 1$

Verified Solution

Answer & Solution

Correct Answer

(A) Range of $f$ is $[- \frac{1}{2}, \frac{1}{2}]$

Step-by-step Solution

Detailed explanation

f (x) = \sin (\frac{π}{6} \sin (\frac{π}{2} \sin x))

\frac{- π}{2} \leq \frac{π}{2} \sin x \leq \frac{π}{2}

- 1 \leq \sin (\frac{π}{2} \sin x) \leq 1

\sin (\frac{- π}{6}) \leq \sin (\frac{π}{6} (\sin (\frac{π}{2} \sin x))) \leq \sin \frac{π}{6}

\frac{- 1}{2} \leq \sin (\frac{π}{6} (\sin (\frac{π}{2} \sin x))) \leq \frac{1}{2}

f (g (x)) = \sin (\frac{π}{6} (\sin (\frac{π}{2} \sin (\frac{π}{2} \sin x))))

- 1 \leq \sin (\frac{π}{2} \sin x) \leq 1

\frac{- π}{2} \leq \frac{π}{2} \sin (\frac{π}{2} \sin x) \leq \frac{π}{2}

- 1 \leq \sin (\frac{π}{2} (\sin (\frac{π}{2} \sin x))) \leq 1

- \frac{π}{6} \leq \frac{π}{6} \sin (\frac{π}{2} (\sin (\frac{π}{2} \sin x))) \leq \frac{π}{6}

\frac{- 1}{2} \leq \sin (\frac{π}{6} \sin (\frac{π}{2} \sin (\frac{π}{2} \sin x))) \leq \frac{1}{2}

\lim_{x \to 0} \frac{f (x)}{g (x)} = \lim_{x \to 0} \frac{\sin (\frac{π}{6} \sin (\frac{π}{2} \sin x))}{\frac{π}{2} \sin x} \times \frac{\frac{π}{6} \sin (\frac{π}{2} \sin x)}{\frac{π}{6} \sin (\frac{π}{2} \sin x)}

= \lim_{x \to 0} \frac{π}{6} \frac{\sin (\frac{π}{2} \sin x)}{\frac{π}{2} \sin x} = \frac{π}{6}

Range of

g (f (x))

[\frac{π}{2} \sin (\frac{- 1}{2}), \frac{π}{2} \sin \frac{1}{2}]

= [- \frac{π}{2} \sin \frac{1}{2}, \frac{π}{2} \sin \frac{1}{2}]

Hence,

1

, does not belong to this range.

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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