The user wants me to solve a matching-type question from JEE Mathematics on the topic of Application of Derivatives.

The function is given as $f(x) = x + \log_e x - x \log_e x$, for $x \in (0, \infty)$.

The question asks to identify the only correct combination of statements from the three columns provided.

Step 1: Find the first and second derivatives of f(x).

Given function: $f(x) = x + \log_e x - x \log_e x$

First derivative, $f'(x)$: $f'(x) = \frac{d}{dx}(x + \log_e x - x \log_e x)$ $f'(x) = 1 + \frac{1}{x} - \left(1 \cdot \log_e x + x \cdot \frac{1}{x}\right)$ $f'(x) = 1 + \frac{1}{x} - (\log_e x + 1)$ $f'(x) = \frac{1}{x} - \log_e x$

Second derivative, $f''(x)$: $f''(x) = \frac{d}{dx}\left(\frac{1}{x} - \log_e x\right)$ $f''(x) = -\frac{1}{x^2} - \frac{1}{x} = -\left(\frac{1}{x^2} + \frac{1}{x}\right)$

Step 2: Analyze the statements in each column.

Column 1 Analysis (Zeroes)

• (I) $f(x) = 0$ for some $x \in (1, e^2)$ $f(1) = 1 + \log_e 1 - 1 \log_e 1 = 1 + 0 - 0 = 1 > 0$. $f(e^2) = e^2 + \log_e(e^2) - e^2 \log_e(e^2) = e^2 + 2 - e^2(2) = 2 - e^2$. Since $e \approx 2.718$, $e^2 \approx 7.389$, so $f(e^2) < 0$. Since $f(x)$ is continuous and $f(1) > 0$ and $f(e^2) < 0$, by the Intermediate Value Theorem, there exists at least one $x \in (1, e^2)$ such that $f(x) = 0$. Statement (I) is TRUE.

• (II) $f'(x) = 0$ for some $x \in (1, e)$ $f'(1) = \frac{1}{1} - \log_e 1 = 1 - 0 = 1 > 0$. $f'(e) = \frac{1}{e} - \log_e e = \frac{1}{e} - 1 < 0$. Since $f'(x)$ is continuous and $f'(1) > 0$ and $f'(e) < 0$, by the Intermediate Value Theorem, there exists at least one $x \in (1, e)$ such that $f'(x) = 0$. Statement (II) is TRUE.

• (III) $f''(x) = 0$ for some $x \in (0, 1)$ $f''(x) = -\left(\frac{1}{x^2} + \frac{1}{x}\right)$. For $x \in (0, 1)$, both $\frac{1}{x^2}$ and $\frac{1}{x}$ are positive. Thus, $f''(x) < 0$ for all $x \in (0, 1)$. Statement (III) is FALSE.

• (IV) $f''(x) = 0$ for some $x \in (1, e)$ For $x \in (1, e)$, $f''(x) < 0$. Statement (IV) is FALSE.

Column 2 Analysis (Limiting Behaviour)

• (i) $\lim_{x \to \infty} f(x) = 0$ $f(x) = x + \log_e x - x \log_e x = x(1 - \log_e x) + \log_e x$. As $x \to \infty$, $(1 - \log_e x) \to -\infty$, so $x(1 - \log_e x) \to -\infty$. The term $\log_e x$ also goes to $\infty$. To resolve the indeterminate form, we factor out the dominant term $x \log_e x$: $f(x) = x \log_e x \left(\frac{1}{\log_e x} + \frac{1}{x} - 1\right)$ As $x \to \infty$, $\frac{1}{\log_e x} \to 0$ and $\frac{1}{x} \to 0$. The expression in the parenthesis approaches $-1$. Since $x \log_e x \to \infty$, we have $\lim_{x \to \infty} f(x) = \infty \cdot (-1) = -\infty$. Statement (i) is FALSE.

• (ii) $\lim_{x \to \infty} f(x) = -\infty$ From the analysis above, this statement is TRUE.

• (iii) $\lim_{x \to \infty} f'(x) = -\infty$ $f'(x) = \frac{1}{x} - \log_e x$. As $x \to \infty$, $\frac{1}{x} \to 0$ and $\log_e x \to \infty$. Therefore, $\lim_{x \to \infty} f'(x) = 0 - \infty = -\infty$. Statement (iii) is TRUE.

• (iv) $\lim_{x \to \infty} f''(x) = 0$ $f''(x) = -\frac{1}{x^2} - \frac{1}{x}$. As $x \to \infty$, $\frac{1}{x^2} \to 0$ and $\frac{1}{x} \to 0$. Therefore, $\lim_{x \to \infty} f''(x) = -0 - 0 = 0$. Statement (iv) is TRUE.

Column 3 Analysis (Monotonicity)

• (P) f is increasing in (0, 1) The monotonicity of $f(x)$ is determined by the sign of $f'(x)$. For $x \in (0, 1)$, we have $\frac{1}{x} > 1$ and $\log_e x < 0$. So, $f'(x) = \frac{1}{x} - \log_e x > 1 - 0 = 1 > 0$. Since $f'(x) > 0$, $f(x)$ is increasing in $(0, 1)$. Statement (P) is TRUE.

• (Q) f is decreasing in $(e, e^2)$ We found that $f'(x) = 0$ for some $x_0 \in (1, e)$. Since $f''(x) < 0$ for all $x>0$, $f'(x)$ is a strictly decreasing function. For any $x \in (e, e^2)$, we have $x > e > x_0$. Since $f'(x)$ is decreasing, $f'(x) < f'(x_0) = 0$. Since $f'(x) < 0$, $f(x)$ is decreasing in $(e, e^2)$. Statement (Q) is TRUE.

• (R) f' is increasing in (0, 1) The monotonicity of $f'(x)$ is determined by the sign of $f''(x)$. As shown before, $f''(x) = -\left(\frac{1}{x^2} + \frac{1}{x}\right) < 0$ for $x \in (0, 1)$. Since $f''(x) < 0$, $f'(x)$ is decreasing in $(0, 1)$. Statement (R) is FALSE.

• (S) f' is decreasing in $(e, e^2)$ For $x \in (e, e^2)$, $f''(x) = -\left(\frac{1}{x^2} + \frac{1}{x}\right) < 0$. Since $f''(x) < 0$, $f'(x)$ is decreasing in $(e, e^2)$. Statement (S) is TRUE.

Step 3: Evaluate the given options.

• A: (I) (ii) (R): (I) is True, (ii) is True, (R) is False. This combination is incorrect.
• B: (III) (iv) (P): (III) is False. This combination is incorrect.
• C: (II) (iii) (S): (II) is True, (iii) is True, (S) is True. This combination is correct.
• D: (IV) (i) (S): (IV) is False. This combination is incorrect.

The only option where all three statements are correct is C.

Final Answer is C.