Application of Derivatives 2 Question 21
21. The function $y=2 x^{2}-\log |x|$ is monotonically increasing for values of $x(\neq 0)$, satisfying the inequalities… and monotonically decreasing for values of $x$ satisfying the inequalities… .
(1983, 2M)
Match the Columns
Directions (Q.Nos. 22-24) by appropriately matching the information given in the three columns of the following table.
Let $f(x)=x+\log _e x-x \log _e x, x \in(0, \infty)$
Column 1 contains information about zeros of $f(x), f^{\prime}(x)$ and $f^{\prime \prime}(x)$.
Column 2 contains information about the limiting behaviour 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 | ||
|---|---|---|---|---|
| (l) | $f(x)=0$ 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 $\left(e, e^{2}\right)$ |
Show Answer
Answer:
Correct Answer: 21. (a)
Solution:
- Here, $y=\begin{array}{cc}2 x^{2}-\log x, & x>0 \ 2 x^{2}-\log (-x), & x<0\end{array}$
$$ \begin{aligned} \Rightarrow \quad \frac{d y}{d x} & =\begin{array}{c} 4 x-\frac{1}{x}, \quad x>0 \\ 4 x-\frac{1}{x}, \quad x<0 \end{array} \\ & =\frac{4 x^{2}-1}{x}, x \in R-{0}=\frac{(2 x-1)(2 x+1)}{x} \end{aligned} $$
$\therefore \quad$ Increasing when $x \in-\frac{1}{2}, 0 \cup \frac{1}{2}, \infty$
$$ \text { and decreasing when } x \in-\infty,-\frac{1}{2} \cup 0, \frac{1}{2} \text {. } $$
Solutions. $(22-24)$
$$ \begin{aligned} & f(x)=x+\ln x-x \ln x \\ & f(1)=1>0 \\ & f\left(e^{2}\right)=e^{2}+2-2 e^{2}=2-e^{2}<0 \end{aligned} $$
$\Rightarrow \quad f(x)=0$ for some $x \in\left(1, e^{2}\right)$
$\therefore I$ is correct
$f^{\prime}(x)=1+\frac{1}{x}-\ln x-1=\frac{1}{x}-\ln x$
$f^{\prime}(x)>0$ for $(0,1)$
$f^{\prime}(x)<0$ for $(e, \infty)$
$\therefore P$ and $Q$ are correct, II is correct, III is incorrect.
$f^{\prime \prime}(x)=\frac{-1}{x^{2}}-\frac{1}{x}$
$f^{\prime \prime}(x)<0$ for $(0, \infty)$
$\therefore S$, is correct, $R$ is incorrect.
IV is incorrect.
$$ \begin{aligned} \lim _{x \rightarrow \infty} f(x) & =-\infty \\ \lim _{x \rightarrow \infty} f^{\prime}(x) & =-\infty \\ \lim _{x \rightarrow \infty} f^{\prime \prime}(x) & =0 \end{aligned} $$
$\therefore$ ii, iii, iv are correct.
