JEE Mathematics 2023 Question with Solution

Given: $x \log x \, f'(x) + (\log x) f(x) \geq 1$ for $x \in [2,4]$, with $f(2)=\frac{1}{2}$ and $f(4)=\frac{1}{4}$.

Find: Which of the statements $(A)$ and $(B)$ is true.

From the solution, the conclusion explicitly states that the correct option is D.

The extracted working says:

$$\frac{d}{dx}\left(x \log x \cdot f(x)\right) \geq 0$$

and then discusses monotonic behavior and bounds using $f(2)=\frac{1}{2}$ and $f(4)=\frac{1}{4}$, finally concluding that both statements are true.

Therefore, according to the solution, the correct option is D.

Given: $x \cdot \log x \cdot f'(x) + \log x \cdot f(x) \geq 1$ for $x \in [2,4]$, and $f(2)=\frac{1}{2}, \; f(4)=\frac{1}{4}$.

Find: The correct truth value of statements $(A)$ and $(B)$.

The solution states the following steps:

1. It rewrites the inequality in derivative form and claims

$$\frac{d}{dx} \left( x \cdot \log x \cdot f(x) \right) \geq 0$$

2. It then states

$$\frac{d}{dx} \left( f(x) \cdot \log x \right) \geq 0$$

3. From this, it claims that $f(x)$ is increasing and positive on $[2,4]$.
4. It defines a new function $g(x)=\ln(x)f(x)-x$ and says that $g(x)$ is increasing on $[2,4]$.
5. Using the endpoint values $f(2)=\frac{1}{2}$ and $f(4)=\frac{1}{4}$, it concludes that both statements $(A)$ and $(B)$ are true.

Hence, the solution marks D as the correct option.

Note: The endpoint values themselves appear inconsistent with statements $(A)$ and $(B)$, but under the stated extraction rule, the solution is the primary source for the answer. Therefore, the answer is recorded as D.