Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
0 votes

Find the intervals in which the function \(f\) given by $ f(x) =\large \frac{4sin\: x - 2x - xcos \: x}{2+cos\:x}$ is (i) increasing (ii) decreasing.

Can you answer this question?

1 Answer

0 votes
  • $\large\frac{d}{dx}$$(\sin\theta)=\cos\theta$
  • $\large\frac{d}{dx}$$(\cos\theta)=-\sin\theta$
  • $\large\frac{d}{dx}$$(x^n)=nx^{n-1}$
Step 1:
$f(x)=\large\frac{4\sin x-2x-x\cos x}{2+\cos x}$
$\qquad=\large\frac{4\sin x-x(2+\cos x)}{2+\cos x}$
$\qquad=\large\frac{4\sin x}{2+\cos x}-\frac{x(2+\cos x)}{2+\cos x}$
$\qquad=\large\frac{4\sin x}{2+\cos x}$$-x$
Step 2:
Differentiating with respect to x we get
$f'(x)=\large\frac{(2+\cos x)4\cos x-4\sin x(-\sin x)}{(2+\cos x)^2}$$-1$
$\qquad=\large\frac{8\cos x+4\cos^2x-4\sin x(-\sin x)}{(2+\cos x)^2}$$-1$
$\qquad=\large\frac{8\cos x+4\cos^2x+4\sin^2x}{(2+\cos x)^2}$
$\qquad=\large\frac{4\cos x-\cos^2x}{(2+\cos x)^2}$
$\qquad=\large\frac{\cos x(4-\cos x)}{(2+\cos x)^2}$
Step 3:
Since $-1 \leq\cos x\leq 1$
$\Rightarrow 4-\cos x>0$ and also $(2+\cos x)^2>0$
Therefore $f'(x) >0$ or $<0$ according as $\cos x>0$ or $\cos x<0$ respectively.
Therefore $f(x)$ is increasing when $0<\large\frac{\pi}{2}$<$\large\frac{3\pi}{2}$ and decreasing when $\large\frac{\pi}{2}<$$x<\large\frac{3\pi}{2}$
answered Aug 10, 2013 by sreemathi.v
edited Sep 2, 2013 by sharmaaparna1

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App