Question

Let f(x) = sin sin 6 2             sin x for all x   and g(x) = 2  sin x for all x  . 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 1 1 , 2 2      
• B. Range of f o g is 1 1 , 2 2      
• C.   0   lim  6   x f x g x
• D. There is an x   such that (g o f)(x) = 1

Answer 1

Answer: (A), (B), (D)

Options (A), (B), and (D) are true.

Answer 2

Solution:

We are given:

$$f(x)=\sin\Biggl(\frac{\pi}{6}\sin\Bigl(\frac{\pi}{2}\sin x\Bigr)\Biggr) \quad \text{for all } x\in \mathbb{R},$$ $$g(x)=2\pi\sin x \quad \text{for all } x\in \mathbb{R}.$$

(A) Range of $f$:

For any $x$, $\sin x\in [-1,1]$. Thus:

• $\frac{\pi}{2}\sin x\in \left[-\frac{\pi}{2},\frac{\pi}{2}\right]$,

• Then $\sin\Bigl(\frac{\pi}{2}\sin x\Bigr)\in [-1,1]$,

• So $\frac{\pi}{6}\sin\Bigl(\frac{\pi}{2}\sin x\Bigr)\in \left[-\frac{\pi}{6},\frac{\pi}{6}\right]$,

• Finally, taking sine we get

$$f(x)=\sin\Bigl(y\Bigr) \quad \text{with } y\in\left[-\frac{\pi}{6},\frac{\pi}{6}\right].$$

The sine function in this restricted interval is increasing. Hence, the range of $f$ is:

$$\Bigl[\sin\Bigl(-\frac{\pi}{6}\Bigr), \sin\Bigl(\frac{\pi}{6}\Bigr)\Bigr]=\left[-\frac{1}{2},\frac{1}{2}\right].$$

So, (A) is true.

(B) Range of $f\circ g$:

We have

$$(f\circ g)(x)= f(g(x)) = \sin\Biggl(\frac{\pi}{6}\sin\Bigl(\frac{\pi}{2}\sin(2\pi\sin x)\Bigr)\Biggr).$$

Notice that regardless of the actual form of the argument, the definition of $f$ guarantees that its output always lies in the interval $\left[-\frac{1}{2},\frac{1}{2}\right]$ (because the inner sine always produces a number in $[-1,1]$). Thus, the range of $f\circ g$ is also

$$\left[-\frac{1}{2},\frac{1}{2}\right].$$

So, (B) is true.

(C) Evaluate $\displaystyle \lim_{x\to0}\frac{f(x)}{g(x)}$:

For small $x$, use the linear approximations:

• $\sin x\approx x$,

• Thus, $\sin\left(\frac{\pi}{2}\sin x\right)\approx \sin\left(\frac{\pi}{2} x\right)\approx \frac{\pi}{2}x$.

• Then, $\frac{\pi}{6}\sin\left(\frac{\pi}{2}\sin x\right)\approx \frac{\pi}{6}\cdot\frac{\pi}{2}x = \frac{\pi^2}{12}x$,

• So, $f(x)= \sin\Bigl(\frac{\pi^2}{12}x\Bigr)\approx \frac{\pi^2}{12}x.$

Similarly, $g(x)=2\pi\sin x\approx2\pi x.$

Therefore,

$$\frac{f(x)}{g(x)}\approx \frac{\frac{\pi^2}{12}x}{2\pi x}=\frac{\pi^2}{12}\cdot\frac{1}{2\pi}=\frac{\pi}{24}.$$

Thus,

$$\lim_{x\to0}\frac{f(x)}{g(x)}=\frac{\pi}{24}\neq\frac{\pi}{6}.$$

So, (C) is false.

(D) Existence of $x$ such that $(g\circ f)(x)=1$:

We have

$$(g\circ f)(x)= g(f(x))=2\pi\sin\bigl(f(x)\bigr).$$

Since $f(x)$ takes all values in $\left[-\frac{1}{2},\frac{1}{2}\right]$ and the sine function is continuous and increasing on that interval, $\sin(f(x))$ will vary continuously between

$$\sin\left(-\frac{1}{2}\right)\quad \text{and}\quad \sin\left(\frac{1}{2}\right).$$

Numerically, $\sin\left(\frac{1}{2}\right)\approx 0.4794$ and $\sin\left(-\frac{1}{2}\right)\approx -0.4794$. Therefore, $2\pi\sin(f(x))$ will sweep the interval approximately

$$\left[-2\pi(0.4794),\,2\pi(0.4794)\right]\approx [-3.012,\,3.012].$$

Since $1$ lies in this interval, by the Intermediate Value Theorem there exists some $x$ for which

$$(g\circ f)(x)=1.$$

So, (D) is true.

Final Answer: The correct statements are (A), (B), and (D).