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) $\lim _{x \rightarrow \infty} f^{\prime}(x)=-\infty$ $(\mathrm{R})$ $f^{\prime}$ 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:

  1. 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 $x \in\left(1, e^{2}\right)$

$\therefore \mathrm{I}$ 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.



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