Note: This is part 5th of a 5 part question, split as 5 separate questions here.

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

- (i) If $f'$ is positive on an open interval $I$. Then $f$ is strictly increasing on $I$
- (ii) If $f'$ is negative on an open interval $I$, then $f$ is strictly decreasing on $I$

$f(x)=x+\cos x \;in\;[0,\pi]$

Step 1:

$f'(x)=1-\sin x$

Step 2:

Since $-1 \leq \sin x \leq 1$

$1-\sin x \geq 0 =>f'(x) \leq 0$

$\therefore $ the function is increasing in $[0, \pi]$

Ask Question

Tag:MathPhyChemBioOther

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...