Note: This is part 4th of a 4 part question, split as 4 separate questions here.

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

- If $f(x)$ is defined on an open interval $I$ and $f'(x)$ takes both positive and negative values on $I$, then $f$ is not monotonic on $I$

$f(x)=\tan x +\cot x$ on $ (0, \large\frac{\pi}{2})$

Step 1:

$f'(x)=\sec^2 x -cosec ^2 x$

$\qquad=\large\frac{1}{\cos ^2 x}-\frac{1}{\sin ^2 x}$

$\qquad=\large\frac{\sin ^2 x -\cos ^2 x}{\sin ^2 x \cos ^2 x}$

$ \qquad=(\sin x + \cos x)(\sin x-\cos x)$

Step 2:

When $ 0 < x < \large\frac{\pi}{4} \; $$\sin x + \cos x >0$ and $ \sin x -\cos x <0$

$\therefore\; f'(x) <0$

When $ \large\frac{\pi}{4}$$ < x < \large\frac{\pi}{2} \; $$\sin x + \cos x >0$ and $ f'(x) >0$

$f'(x)$ changes in sigh in $(0, \large\frac{\pi}{2})$

$\therefore $ it is not monotonic on $(0, \large\frac{\pi}{2})$

Ask Question

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...