Please solve RD Sharma class 12 chapter 16 Increasing and Decreasing Function Excercise Fill in the blanks Question 2: Maths Textbook Solution.

Answer: The is function decreasing in the intervals $\left ( 1,\infty \right )$

Hint: The function is decreasing if $f'\left ( x \right )<0$

Given: $f\left ( x \right )=\frac{2x^{2}-1}{x^{4}}$

Explanation: We have,

$f\left ( x \right )=\frac{2x^{2}-1}{x^{4}},x>0$                                                        ….(i)

Differentiate (i) with respect to x

$f\left ( x \right )=\frac{x^{4}\left ( 2.2x \right )-\left ( 2x^{2}-1 \right )4x^{3}}{\left ( x^{4}\right )^{2}}$

$=\frac{x^{4}(4x)-8x^{5}+4x^{3}}{x^{8}}$

$=\frac{4x^{5}-8x^{5}+4x^{3}}{x^{8}}$

$=\frac{-4x^{5}}{x^{8}}+\frac{4x^{3}}{x^{8}}$

$=\frac{-4}{x^{3}}+\frac{4}{x^{5}}$

$=\frac{4(1-x^{2})}{x^{5}}$

$=\frac{4(1-x)(1+x)}{x^{5}}$

So,  $\frac{4(1-x)(1+x))}{x^{5}}<0$                         [  $\because f (x)$ is decreasing]

$\Rightarrow \frac{(1-x)(1+x)}{x^{5}}<0$

$\Rightarrow (1-x)(1+x)<0$

$\Rightarrow (1-x)<0$                                  [ as $(1+x)<0\Rightarrow x<-1$ is not possible.since $x>0$ is given ]

$\Rightarrow x>1$

$\therefore x\epsilon (1, \infty )$

Thus, the function decreases in the interval $(1, \infty )$