Identify Correct Column Match for Function Behavior Analysis II

To determine the correct combination, we analyze the given function $f(x)$ and its derivatives on the domain $x \in (0, \infty)$.

Step 1: Definition and Derivatives of $f(x)$

The given function is: $f(x) = x + \log_e x - x \log_e x$

Taking the first derivative with respect to $x$: $f'(x) = \frac{d}{dx}\left(x + \log_e x - x \log_e x\right) = 1 + \frac{1}{x} - \left(1 \cdot \log_e x + x \cdot \frac{1}{x}\right)$ $f'(x) = 1 + \frac{1}{x} - \log_e x - 1 = \frac{1}{x} - \log_e x$

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

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Step 2: Evaluation of Statements in Column 1

1. Statement (II): $f'(x) = 0$ for some $x \in (1, e)$ Evaluate $f'(x)$ at the endpoints $x = 1$ and $x = e$: $f'(1) = \frac{1}{1} - \log_e 1 = 1 > 0$ $f'(e) = \frac{1}{e} - \log_e e = \frac{1}{e} - 1 < 0 \quad (\text{since } e \approx 2.718 > 1)$ Since $f'(x)$ is continuous on $[1, e]$ and changes sign from positive to negative, by the Intermediate Value Theorem, there exists a root $x_0 \in (1, e)$ such that $f'(x_0) = 0$. Hence, Statement (II) is TRUE.

2. Statement (III): $f'(x) = 0$ for some $x \in (0, 1)$ For $x \in (0, 1)$, we have $\frac{1}{x} > 1$ and $\log_e x < 0$. Therefore: $f'(x) = \frac{1}{x} - \log_e x > 1 - 0 > 0$ Thus, $f'(x) \neq 0$ for any $x \in (0, 1)$. Hence, Statement (III) is FALSE.

3. Statement (IV): $f''(x) = 0$ for some $x \in (1, e)$ For $x \in (0, \infty)$, $1 + x > 0$ and $x^2 > 0$, so: $f''(x) = -\frac{1+x}{x^2} < 0$ Thus, $f''(x)$ has no real roots in $(0, \infty)$. Hence, Statement (IV) is FALSE.

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Step 3: Evaluation of Statements in Column 2

1. Statement (iii): $\lim_{x \to \infty} f'(x) = -\infty$ $\lim_{x \to \infty} f'(x) = \lim_{x \to \infty} \left(\frac{1}{x} - \log_e x\right) = 0 - \infty = -\infty$ Hence, Statement (iii) is TRUE.

2. Statement (iv): $\lim_{x \to \infty} f''(x) = 0$ $\lim_{x \to \infty} f''(x) = \lim_{x \to \infty} \left(-\frac{1+x}{x^2}\right) = \lim_{x \to \infty} \left(-\frac{1}{x^2} - \frac{1}{x}\right) = 0$ Hence, Statement (iv) is TRUE.

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Step 4: Evaluation of Statements in Column 3

1. Statement (R): $f'$ is increasing in $(0, 1)$ Since $f''(x) = -\frac{1+x}{x^2} < 0$ for all $x \in (0, \infty)$, the function $f'(x)$ is strictly decreasing throughout its domain $(0, \infty)$. Thus, $f'(x)$ cannot be increasing in $(0, 1)$. Hence, Statement (R) is FALSE.

2. Statement (S): $f'$ is decreasing in $(e, e^2)$ Since $f''(x) < 0$ for all $x \in (0, \infty)$, $f'$ is strictly decreasing on $(0, \infty)$, and therefore $f'$ is strictly decreasing on the sub-interval $(e, e^2)$. Hence, Statement (S) is TRUE.

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Conclusion

Evaluating the options given in Question 53:

• (A) (I) (ii) (R): Incorrect because (R) is FALSE.
• (B) (II) (iii) (S): Correct because (II), (iii), and (S) are all TRUE.
• (C) (III) (iv) (P): Incorrect because (III) is FALSE.
• (D) (IV) (i) (S): Incorrect because (IV) and (i) are FALSE.

Thus, the only CORRECT combination is (B).