JEE Mathematics 2023 Question with Solution

Given: $f(x) = 4 \sin^{-1} \left( \frac{x^2}{x^2 + 1} \right)$

Find: The range of $f(x)$.

From the given expression,

$$0 \leq \frac{x^2}{x^2+1} < 1$$

because $x^2 \geq 0$ and $x^2+1 > x^2$ for all real $x$.

Applying inverse sine,

$$0 \leq \sin^{-1} \left( \frac{x^2}{x^2+1} \right) < \frac{\pi}{2}$$

Now multiply by $4$:

$$0 \leq 4\sin^{-1} \left( \frac{x^2}{x^2+1} \right) < 2\pi$$

So the range obtained from the working is $[0, 2\pi)$.

However, the provided the solution concludes the range as $[0, 2\pi]$ and marks option D, while the listed options contain $[0, 2\pi]$ as option C. Following the solution's stated conclusion, the defensible listed option is D as marked there, though the option text and working are inconsistent.

Therefore, the correct option according to the solution is D.](streamdown:incomplete-link)

Given: $f(x) = 4 \sin^{-1} \left( \frac{x^2}{x^2 + 1} \right)$

Find: Whether the end points are included.

At $x=0$,

$$\frac{x^2}{x^2+1} = 0$$

so

$$f(0)=4\sin^{-1}(0)=0$$

Hence $0$ is included.

For any real $x$,

$$\frac{x^2}{x^2+1} \neq 1$$

so

$$\sin^{-1} \left( \frac{x^2}{x^2+1} \right) \neq \frac{\pi}{2}$$

Therefore $2\pi$ is not actually attained from the algebraic range.

This shows the natural range from the expression is $[0,2\pi)$, but the source solution states $[0,2\pi]$ and marks D. This mismatch should be noted while selecting from the provided source.](streamdown:incomplete-link)