Let f(x) = sin (π/6 sin (π/2 sin x)) for all x ∈ R and g(x) = π/2 sin x for all x ∈ R. Let (fog)(x)

Q: Wait, why can't we just say the range of $f(x)$ is $[-1, 1]$ because it's a sine function?

That is a classic trap! While the sine function itself oscillates between $-1$ and $1$, the input to the outermost sine function is not just $x$. It is a scaled value, $\frac{\pi}{6} \sin(\frac{\pi}{2} \sin x)$. Because the input is restricted to a smaller interval, the output of the sine function is also restricted. You must always track how the input interval changes through each nested layer.

Q: In the limit part, why does $g(x)$ go to zero? I thought $x$ was just a variable.

Great question! We are evaluating the limit as $x \to 0$. Since $g(x) = \frac{\pi}{2} \sin x$, we calculate the limit of $g(x)$ as $x$ approaches $0$. Since $\sin(0) = 0$, $g(x)$ also approaches $0$. This is why we can use the substitution $u = g(x)$ and let $u \to 0$.

Q: How do we know for sure that $\sin(f(x))$ can't reach $2/\pi$?

We know the maximum value of $f(x)$ is $1/2$. Since the sine function is strictly increasing on the interval $[0, \pi/2]$, and $1/2 < \pi/6 \approx 0.523$, we know that $\sin(f(x))$ must be less than $\sin(1/2)$. Since $\sin(1/2) < 0.5$ and $2/\pi \approx 0.636$, it is mathematically impossible for $\sin(f(x))$ to reach $2/\pi$.

Question

Let $f(x) = \sin (\frac{\pi}{6} \sin (\frac{\pi}{2} \sin x))$ for all $x \in R$ and $g(x) = \frac{\pi}{2} \sin x$ 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?

Options:

(A) Range of $f$ is $[-1/2, 1/2]$

(B) Range of $f \circ g$ is $[-1/2, 1/2]$

(C) $\lim_{x 	o 0} \frac{f(x)}{g(x)} = \frac{\pi}{6}$

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

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The function $f(x) = \sin\left(\frac{\pi}{6} \sin\left(\frac{\pi}{2} \sin xight)ight)$ appears intimidating, like a fortress with many walls. In mathematics, the secret to breaking down such a structure is to start from the inside.

This is a classic Matryoshka doll problem. We have layers of sine functions, each one wrapping the previous one. Let us peel them back, one by one.

### Phase 1: Peeling the Onion

We begin with the innermost layer, $g(x) = \frac{\pi}{2} \sin x$. We know that for any real number $x$, the sine function $\sin x$ oscillates gracefully between $-1$ and $1$.

When we multiply this by $\frac{\pi}{2}$, we are simply stretching the range. Thus, the range of $g(x)$ becomes $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

Now, we move to the next layer: $\sin(g(x))$. Since $g(x)$ covers the entire interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$, the sine function will cover its full range of $[-1, 1]$.

The next layer is the multiplier $\frac{\pi}{6}$. We take our range $[-1, 1]$ and scale it by $\frac{\pi}{6}$, resulting in $[-\frac{\pi}{6}, \frac{\pi}{6}]$.

Finally, we apply the outermost sine function. We are looking for the range of $\sin(	heta)$ where $	heta \in [-\frac{\pi}{6}, \frac{\pi}{6}]$. Since sine is an increasing function in this interval, the range is simply:

$$[\sin(-\frac{\pi}{6}), \sin(\frac{\pi}{6})] = [-1/2, 1/2]$$

This confirms that the range of $f(x)$ is $[-1/2, 1/2]$.

### Phase 2: The Composite Dance

Now, consider the composite function $(f \circ g)(x) = f(g(x))$. The input to $f$ is now $g(x)$.

We already established that $g(x)$ covers the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$. Since this is the exact interval that allows $f$ to reach its full range, the range of $(f \circ g)(x)$ is identical to the range of $f(x)$, which is $[-1/2, 1/2]$.

### Phase 3: The Limit's Secret

Next, we tackle the limit $\lim_{x 	o 0} \frac{f(x)}{g(x)}$. Let $u = g(x)$. As $x 	o 0$, $u 	o 0$. The expression becomes:

$$\lim_{u 	o 0} \frac{\sin(\frac{\pi}{6} \sin u)}{u}$$

To solve this, we use the standard limit $\lim_{	heta 	o 0} \frac{\sin 	heta}{	heta} = 1$. We multiply and divide by the inner argument $\frac{\pi}{6} \sin u$:

$$\lim_{u 	o 0} \left( \frac{\sin(\frac{\pi}{6} \sin u)}{\frac{\pi}{6} \sin u} \cdot \frac{\frac{\pi}{6} \sin u}{u} ight)$$

The first part approaches $1$. The second part is $\frac{\pi}{6} \lim_{u 	o 0} \frac{\sin u}{u}$, which is $\frac{\pi}{6} \cdot 1$. Thus, the limit is $\frac{\pi}{6}$.

### Phase 4: The Impossible Quest

Finally, we examine the condition: is there an $x$ such that $(g \circ f)(x) = 1$? This implies:

$$\frac{\pi}{2} \sin(f(x)) = 1 \Rightarrow \sin(f(x)) = \frac{2}{\pi} \approx 0.636$$

We know the maximum value of $f(x)$ is $1/2$. Since $1/2 < \pi/6 \approx 0.523$, the maximum value of $\sin(f(x))$ is $\sin(1/2) < \sin(\pi/6) = 0.5$.

Since $0.5 < 0.636$, it is **impossible** for $\sin(f(x))$ to reach $2/\pi$. You have successfully navigated the layers of this problem; no matter how complex a function looks, it is just a series of simple steps waiting to be unraveled.

Select Answer:

Visualized Solution

Understanding the Functions

Range of $g (x)$

Innermost Layer of $f (x)$

Middle Layer of $f (x)$

Final Range of $f (x)$

Range of $(f \circ g) (x)$

Evaluating the Limit: Setup

Evaluating the Limit: Execution

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

Conclusion for Option D

The Sigma Insight: Composite Functions

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

Let $f (x) = x^{2}$ 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

If $f (x) = {2 + 2 x, 1 - \frac{x}{3}, - 1 \leq x < 0 0 \leq x \leq 3; g (x) = {- x, x, - 3 \leq x \leq 0 0 < x \leq 1$ , then range of $(f \circ g (x))$ is

Let $f (x) = sin^{- 1} x$ and $g (x) = \frac{x ^{2} - x - 2}{2 x ^{2} - x - 6}$ . If $g (2) = lim_{x \to 2} g (x)$ , then the domain of the function $f \circ g$ is :

Let $f (x) = sin x$ and $g (x) = ln ∣ x ∣$ . If the ranges of the composition functions $f \circ g$ and $g \circ f$ are $R_{1}$ and $R_{2}$ respectively, then

If $g (f (x)) = ∣ sin x ∣$ and $f (g (x)) = (sin x)^{2}$ , then

Let $f : R \to R$ and $g : R \to R$ be defined as $f (x) = {lo g_{e} x e^{- x}, x > 0, x \leq 0$ and $g (x) = {x e^{x}, x \geq 0, x < 0$ . Then, $g \circ f : R \to R$ is :

Let $f$ and $g$ be two functions defined by $f (x) = {x + 1, ∣ x - 1∣, x < 0 x \geq 0$ and $g (x) = {x + 1, 1, x < 0 x \geq 0$ . Then $(g \circ f) (x)$ is

For $x \in R$ , two real valued functions $f (x)$ and $g (x)$ are such that, $g (x) = x + 1$ and $f (g (x)) = x + 3 - x$ . Then $f (0)$ is equal to

Let $f, g : (1, \infty) \to R$ be defined as $f (x) = \frac{2 x + 3}{5 x + 2}$ and $g (x) = \frac{2 - 3 x}{1 - x}$ . If the range of the function $f \circ g : [2, 4] \to R$ is $[α, β]$ , then $\frac{1}{β - α}$ is equal to

Let $f : R - {- \frac{1}{2}} \to R$ and $g : R - {- \frac{5}{2}} \to R$ be defined as $f (x) = \frac{2 x + 3}{2 x + 1}$ and $g (x) = \frac{∣ x ∣ + 1}{2 x + 5}$ . Then the domain of the function $f \circ g$ is :

Select Answer:

Visualized Solution

Understanding the Functions

Range of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x)

Innermost Layer of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Middle Layer of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Final Range of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Range of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (f∘g)(x)