Let $f(x)=\sin \left(\frac{\pi}{6} \sin \left(\frac{\pi}{2} \sin x\right)\right)$ for all $x \in \mathbb{R}$ and $g(x)=\frac{\pi}{2} \sin x$ for all $x \in \mathbb{R}$. Let $(f \circ g)(x)$ denote $f(g(x))$ and $(g \circ f)(x)$ denote $g(f(x))$. Then which of the following is (are) true?

(A) Range of $f$ is $\left[-\frac{1}{2}, \frac{1}{2}\right]$; (B) Range of $f \circ g$ is $\left[-\frac{1}{2}, \frac{1}{2}\right]$; (C) $\lim _{x \rightarrow 0} \frac{f(x)}{g(x)}=\frac{\pi}{6}$

(D) There is an $x \in \mathbb{R}$ such that $(g \circ f)(x)=1$

Relations And Functions XII — JEE Advanced · 2015 · 24 May 2015 · Paper 1 Question with Solution

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 \circ g) (x)

denote

f (g (x))

and

(g \circ f) (x)

denote

g (f (x))

. Then which of the following is (are) true?

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

Answer: (A), (B), (C)

Solution

(A) The innermost function is

sin x \in [- 1, 1]

. Then

\frac{π}{2} sin x \in [- \frac{π}{2}, \frac{π}{2}]

, so

sin (\frac{π}{2} sin x) \in [- 1, 1]

. Then

\frac{π}{6} sin (\frac{π}{2} sin x) \in [- \frac{π}{6}, \frac{π}{6}]

, so

f (x) = sin (\frac{π}{6} sin (\frac{π}{2} sin x)) \in [sin (- \frac{π}{6}), sin (\frac{π}{6})] = [- \frac{1}{2}, \frac{1}{2}]

. ✓

(B)

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

. Since

\frac{π}{2} sin x \in [- \frac{π}{2}, \frac{π}{2}]

, the composition has the same range as

f

, which is

[- \frac{1}{2}, \frac{1}{2}]

. ✓

(C)

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

x \to 0

sin x \approx x

sin (\frac{π}{2} x) \approx \frac{π}{2} x

sin (\frac{π}{6} \cdot \frac{π}{2} x) \approx \frac{π ^{2}}{12} x

x \to 0 lim \frac{\frac{π ^{2}}{12} x}{\frac{π}{2} x} = \frac{π ^{2} /12}{π /2} = \frac{π}{6}

✓

(D)

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

since

f (x) \in [- \frac{1}{2}, \frac{1}{2}]

and

sin

is approximately linear near 0... Actually,

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

. The maximum value of

∣ f (x) ∣

\frac{1}{2}

, so the maximum of

∣ sin (f (x)) ∣

sin (\frac{1}{2}) < 1

. Thus

(g \circ f) (x) < \frac{π}{2} sin (\frac{1}{2}) < \frac{π}{2} \cdot 1 = \frac{π}{2}

, and since

\frac{π}{2} > 1

, we need to check if

\frac{π}{2} sin (\frac{1}{2}) \geq 1

. We have

sin (0.5) \approx 0.479

, so

\frac{π}{2} \cdot 0.479 \approx 0.75 < 1

. ✗

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JEE Advanced · 2015 · 24 May 2015 · Paper 1 · Relations And Functions XII

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