Question

Let $f(x)=x+\log _{\mathrm{e}} x-x \log _{\mathrm{e}} x, x \in(0, \infty)$.
- Column 1 contains information about zeros of $f(x), f^{\prime}(x)$ and $f^{\prime \prime}(x)$.
- Column 2 contains information about the limiting behavior of $f(x), f^{\prime}(x)$ and $f^{\prime \prime}(x)$ at infinity.
- Column 3 contains information about increasing/decreasing nature of $f(x)$ and $f^{\prime}(x)$.

| Column 1 | Column 2 | Column 3 |
| --- | --- | --- |
| (I) $f(x)=0$ for some $x \in\left(1, e^{2}\right)$ | (i) $\lim _{x \rightarrow \infty} f(x)=0$ | (P) $f$ is increasing in $(0,1)$ |
| (II) $f^{\prime}(x)=0$ for some $x \in(1, e)$ | (ii) $\lim _{x \rightarrow \infty} f(x)=-\infty$ | (Q) $f$ is decreasing in $\left(e, e^{2}\right)$ |
| (III) $f^{\prime}(x)=0$ for some $x \in(0,1)$ | (iii) $\lim _{x \rightarrow \infty} f^{\prime}(x)=-\infty$ | (R) $f^{\prime}$ is increasing in $(0,1)$ |
| (IV) $f^{\prime \prime}(x)=0$ for some $x \in(1, e)$ | (iv) $\lim _{x \rightarrow \infty} f^{\prime \prime}(x)=0$ | (S) $f^{\prime}$ is decreasing in $\left(e, e^{2}\right)$ |

Which of the following options is the only INCORRECT combination?

Accepted Answer

(C) (III) (i) (R)

Answer

(A) (I) (iii) (P)

Answer

(B) (II) (iv) (Q)

Answer

(D) (II) (iii) (P)

Column 1	Column 2	Column 3
(I) $f (x) = 0$ for some $x \in (1, e^{2})$	(i) $lim_{x \to \infty} f (x) = 0$	(P) $f$ is increasing in $(0, 1)$
(II) $f^{'} (x) = 0$ for some $x \in (1, e)$	(ii) $lim_{x \to \infty} f (x) = - \infty$	(Q) $f$ is decreasing in $(e, e^{2})$
(III) $f^{'} (x) = 0$ for some $x \in (0, 1)$	(iii) $lim_{x \to \infty} f^{'} (x) = - \infty$	(R) $f^{'}$ is increasing in $(0, 1)$
(IV) $f^{''} (x) = 0$ for some $x \in (1, e)$	(iv) $lim_{x \to \infty} f^{''} (x) = 0$	(S) $f^{'}$ is decreasing in $(e, e^{2})$

JEE Advanced · 2017 · 21 May 2017 · Paper 1 · Application of Derivatives

Solution

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