JEE Mathematics 2024 Question with Solution

Given: $g(x) = 3f\left(\frac{x}{3}\right) + f(3-x)$ and $f''(x) > 0$ for all $x \in (0,3)$.

Find: $8\alpha$, where $g$ is decreasing in $(0,\alpha)$ and increasing in $(\alpha,3)$.

Since $f''(x) > 0$ on $(0,3)$, the function $f'(x)$ is increasing on this interval.

Because $g$ decreases before $\alpha$ and increases after $\alpha$, the turning point occurs at $x=\alpha$. Hence,

Differentiate $g(x)$:

so

At $x=\alpha$,

Since $f'(x)$ is increasing, equal values of $f'$ imply equal inputs. Therefore,

Now,

Therefore, the correct option is C.

Given: $f''(x)>0$ on $(0,3)$, so $f$ is convex and $f'$ is strictly increasing.

Find: The value of $8\alpha$.

For $g(x)=3f\left(\frac{x}{3}\right)+f(3-x)$,

The change from decreasing to increasing occurs where $g'(x)$ changes sign, so at the transition point,

Because $f'$ is increasing, it is one-one on the interval. Hence the equality above forces

which gives

and therefore

So the correct answer is $18$.