SATHEE: Application of Derivatives 2 Question 21

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 $f (x), f^{'} (x)$ and $f^{''} (x)$ .

Column 2 contains information about the limiting behaviour of $f (x), f^{'} (x)$ and $f^{''} (x)$ at infinity.

Column 3 contains information about increasing/decreasing nature of $f (x)$ and $f^{'} (x)$ .

Column-1		Column-2	Column-3
(l)	$f (x) = 0$ 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 $(e, e^{2})$

Show Answer

Answer:

Correct Answer: 21. (a)

Solution:

Here, $y = \begin{array}{cc} 2 x^{2} - \log x, & x > 0 \\ 2 x^{2} - \log (- x), & x < 0 \end{array}$

$\begin{aligned} \Rightarrow \frac{d y}{d x} & = \begin{array}{c} 4 x - \frac{1}{x}, x > 0 \\ 4 x - \frac{1}{x}, x < 0 \end{array} \\ = \frac{4 x^{2} - 1}{x}, x \in R - 0 = \frac{(2 x - 1) (2 x + 1)}{x} \end{aligned}$

$∴$ Increasing when $x \in - \frac{1}{2}, 0 \cup \frac{1}{2}, \infty$

$and decreasing when x \in - \infty, - \frac{1}{2} \cup 0, \frac{1}{2} .$

Solutions. $(22 - 24)$

$\begin{aligned} f (x) = x + \ln x - x \ln x \\ f (1) = 1 > 0 \\ f (e^{2}) = e^{2} + 2 - 2 e^{2} = 2 - e^{2} < 0 \end{aligned}$

$\Rightarrow f (x) = 0$ for some $x \in (1, e^{2})$

$∴ I$ is correct

$f^{'} (x) = 1 + \frac{1}{x} - \ln x - 1 = \frac{1}{x} - \ln x$

$f^{'} (x) > 0$ for $(0, 1)$

$f^{'} (x) < 0$ for $(e, \infty)$

$∴ P$ and $Q$ are correct, II is correct, III is incorrect.

$f^{''} (x) = \frac{- 1}{x^{2}} - \frac{1}{x}$

$f^{''} (x) < 0$ for $(0, \infty)$

$∴ S$ , is correct, $R$ is incorrect.

IV is incorrect.

$\begin{aligned} lim_{x \to \infty} f (x) & = - \infty \\ lim_{x \to \infty} f^{'} (x) & = - \infty \\ lim_{x \to \infty} f^{''} (x) & = 0 \end{aligned}$

$∴$ ii, iii, iv are correct.

Application of Derivatives 2 Question 21

Match the Columns

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

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Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Application of Derivatives 2 Question 21

Match the Columns

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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