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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Show that the function given by $f (x) = \sin x$ is neither increasing nor decreasing in $(0, \pi)$

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Toolbox:
  • A function $f(x)$ is said to be a strictly increasing function on $(a,b)$ if $x_1 < x_2\Rightarrow f(x_1) < f(x_2)$ for all $x_1,x_2\in (a,b)$
  • If $x_1 < x_2\Rightarrow f(x_1) > f(x_2)$ for all $x_1,x_2\in (a,b)$ then $f(x)$ is said to be strictly decreasing on $(a,b)$
  • A function $f(x)$ is said to be increasing on $[a,b]$ if it is increasing (decreasing) on $(a,b)$ and it is increasing (decreasing) at $x=a$ and $x=b$.
  • The necessary sufficient condition for a differentiable function defined on $(a,b)$ to be strictly increasing on $(a,b)$ is that $f'(x) > 0$ for all $x\in (a,b)$
  • The necessary sufficient condition for a differentiable function defined on $(a,b)$ to be strictly decreasing on $(a,b)$ is that $f'(x) < 0$ for all $x\in (a,b)$
Step 1:
Given : $f(x)=\sin x$
Differentiating w.r.t $x$ on both sides we get,
$f'(x)=\cos x$
$f'(x)$ is positive in the interval $(0,\large\frac{\pi}{2})$ and $f'(x)$ is negative in the interval $(\large\frac{\pi}{2}$$,\pi)$
Step 2:
$\Rightarrow f'(x)$ does not have the same sign in the interval $(0,\pi)$.
Hence $f(x)$ is neither increasing nor decreasing in $(0,\pi)$.
answered Jul 8, 2013 by sreemathi.v
 

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