JEE Mathematics 2024 Question with Solution

Given: $f(x) = \frac{1}{2}[g(x) + g(2-x)]$ and $g'\left(\frac{1}{2}\right) = g'\left(\frac{3}{2}\right)$.

Find: Which statement about $f''(x)$ is correct.

Differentiate:

$$f'(x) = \frac{1}{2}\left[g'(x) - g'(2-x)\right]$$

Now evaluate at the symmetric points:

$$f'\left(\frac{1}{2}\right) = \frac{1}{2}\left[g'\left(\frac{1}{2}\right) - g'\left(\frac{3}{2}\right)\right] = 0$$

and

$$f'\left(\frac{3}{2}\right) = \frac{1}{2}\left[g'\left(\frac{3}{2}\right) - g'\left(\frac{1}{2}\right)\right] = 0$$

Differentiate again:

$$f''(x) = \frac{1}{2}\left[g''(x) + g''(2-x)\right]$$

Since $f'\left(\frac{1}{2}\right) = f'\left(\frac{3}{2}\right) = 0$, Rolle's theorem guarantees at least one point in $\left(\frac{1}{2},1\right)$ and at least one point in $\left(1,\frac{3}{2}\right)$ where

$$f''(x) = 0$$

Hence, $f''(x) = 0$ for at least two values of $x$ in $(0,2)$.

Therefore, the correct option is A.

The function

$$f(x) = \frac{1}{2}[g(x) + g(2-x)]$$

is symmetric about $x=1$ because replacing $x$ by $2-x$ gives the same value.

Compute the first derivative carefully:

$$\frac{d}{dx}g(2-x) = -g'(2-x)$$

so

$$f'(x) = \frac{1}{2}\left[g'(x) - g'(2-x)\right]$$

Using $g'\left(\frac{1}{2}\right) = g'\left(\frac{3}{2}\right)$,

$$f'\left(\frac{1}{2}\right) = 0, \qquad f'\left(\frac{3}{2}\right) = 0$$

Apply Rolle's theorem to $f'$ on the interval $\left[\frac{1}{2},1\right]$ and again on $\left[1,\frac{3}{2}\right]$. Since the endpoint values are equal in each symmetric setup, there must be zeros of $f''$ in both subintervals. Thus there are at least two points in $(0,2)$ where $f''(x)=0$.

Note: The provided the solution contains derivative-sign inconsistencies, but it explicitly concludes that option A is correct. The corrected differentiation above supports that conclusion.