Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Applications of Derivatives

Question:

The intervals of monotonicity of the function $y=x^2-\log_e|x|.(x≠0)$

Options:

$(-∞,\frac{-1}{\sqrt{2}})∪(0,\frac{1}{\sqrt{2}})$

$(∞,\frac{1}{\sqrt{2}})∪(0,\frac{-1}{\sqrt{2}})$

$(∞,\frac{1}{\sqrt{2}})∪(0,\frac{1}{\sqrt{2}})$

$(-∞,-\frac{1}{\sqrt{2}})∪(0,\frac{-1}{\sqrt{2}})$

Correct Answer:

$(-∞,\frac{-1}{\sqrt{2}})∪(0,\frac{1}{\sqrt{2}})$

Explanation:

Let $y=x^2-\log_e|x|$

$f(x)=\left\{\begin{matrix}x^2-\log_e(-x),&x<0\\x^2-\log(x),&x>0\end{matrix}\right.$

$⇒f'(x)=\left\{\begin{matrix}2x-\frac{1}{(-x)}(-1),&x<0\\2x-\frac{1}{x},&x>0\end{matrix}\right.$

$∴f'(x)=2x-\frac{1}{x}$

for all $x(x≠0)$

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

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

Using number line rule, as shown in figure

it gives $f'(x)>0$ when $x∈(-\frac{1}{\sqrt{2}},0)(\frac{1}{\sqrt{2}},∞)$

and $f'(x)<0$ when $x∈(-∞,-\frac{1}{\sqrt{2}})(0,\frac{1}{\sqrt{2}})$

∴ f(x) is increasing when $x∈(-\frac{1}{\sqrt{2}},0)(\frac{1}{\sqrt{2}},∞)$

and decreasing when $x∈(-∞,-\frac{1}{\sqrt{2}})(0,\frac{1}{\sqrt{2}})$