Ask Questions, Get Answers

Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives

The maximum value of sin x,cos x is

$(A)\;\frac{1}{4}\quad(B)\;\frac{1}{2}\quad(C)\;\sqrt 2\quad(D)\;2\sqrt 2$

1 Answer

  • To obtain the absolute maxima or minima for the function $f(x)$
  • (i) Find $f'(x)$ and put $f'(x)=0$
  • (ii) Obtain the points from $f'(x)=0$
  • (iii) By phthogoras theorem, find $f''(x)$ and check the value of $f''(x)$ for each of the points obtained if $f''(x)>0$ it has a minimum at that point. If $f''(x)<0$ then it has a maximum at that point.
Step 1
$y=\sin x \cos x$
$ = \large\frac{1}{2} \sin2x$
On differentiating w.r.t $x$ we get,
$ y'=\large\frac{1}{2}.2 \cos 2x = \cos2x$
Again differentiating w.r.t $x$ we get,
$ y''=-2 \sin 2x <0$
Hence the function has a maximum value.
Step 2
To obtain the maximum value
$ \large\frac{dy}{dx}=0$
$ \Rightarrow \cos 2x=0$
$ \Rightarrow 2x=\large\frac{\pi}{2}$
$ x=\large\frac{\pi}{4}$
when $ x = \large\frac{\pi}{4}$
$ y= \sin \large\frac{\pi}{4}. \cos \large\frac{\pi}{4}$
$ = \large\frac{1}{\sqrt 2} \times \large\frac{1}{\sqrt 2}$
$ = \large\frac{1}{2}$
Hence the maximum value of the function is $ \large\frac{1}{2}.$
Hence the correct option is B.
answered Aug 12, 2013 by thanvigandhi_1