Animated Solution for Mathematics - Functions: The values of $f(x)=3\sin \sqrt{\frac{{\pi }^2}{16}-x^2}$ lie in the interval .........

Welcome, fellow traveler on the path to JEE mastery. Today, we stand before a function that, at first glance, might seem like a simple trigonometric expression.

Our goal is to find the range of the function:

To solve this, we must adopt the mindset of a detective. We cannot simply jump to the end; we must peel back the layers one by one, respecting the constraints that the universe of mathematics imposes upon us.

The first layer is the domain. We see a square root, and immediately, our intuition should scream: "The argument must be non-negative!"

We set the radicand to be non-negative:

Rearranging this inequality, we find $x^2\le \frac{{\pi }^2}{16}$, which leads us to the domain $x\in [-\frac{\pi }{4},\frac{\pi }{4}]$. This is the boundary of our world; if we step outside this, the function ceases to exist.

Now, we move to the inner function, $g(x)=\sqrt{\frac{{\pi }^2}{16}-x^2}$. We need to understand how this function behaves within our domain.

At $x=0$, the value is $g(0)=\sqrt{\frac{{\pi }^2}{16}}=\frac{\pi }{4}$. This is the peak of our inner function.

As we move toward the boundaries $x=\pm \frac{\pi }{4}$, the value of $x^2$ increases, causing the expression under the root to decrease toward zero. Thus, the range of our inner function $g(x)$ is $[0,\frac{\pi }{4}]$.

Now, we introduce the outer layer: the sine function. We are evaluating $\sin (g(x))$, where $g(x)$ spans from $0$ to $\frac{\pi }{4}$.

The sine function is strictly increasing in the interval $[0,\frac{\pi }{2}]$. Because it is increasing, the minimum input $0$ yields the minimum output $\sin (0)=0$.

The maximum input $\frac{\pi }{4}$ yields the maximum output $\sin (\frac{\pi }{4})=\frac{1}{\sqrt{2}}$. We have successfully transformed our range through the sine function.

Finally, we apply the scaling factor of $3$. Multiplying our range $[0,\frac{1}{\sqrt{2}}]$ by $3$, we arrive at the final answer.

The range of the function is:

This journey, from the domain constraint to the final scaling, is the essence of solving composite functions. It is about patience, precision, and understanding the behavior of each component.