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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Prove that the function $f$ given by $f (x) = \log \sin x$ is strictly increasing on$ \left(0, \: \large {\frac{\pi}{2}}\right)$ and strictly decreasing on $\left( \large {\frac{\pi}{2}}, \pi\right)$

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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:
Let $f(x)=\log \sin x$
Differentiating w.r.t $x$ we get,
$f'(x)=\large\frac{1}{\sin x}$$\cos x$
$\qquad=\cot x$
Step 2:
Consider the interval $(0,\large\frac{\pi}{2})$
$f'(x)=\cot x>0$
$\cot x$ is +ve in first quadrant.
Therefore $f$ is strictly increasing in $(0,\large\frac{\pi}{2})$
Step 3:
Consider the interval $(\large\frac{\pi}{2},$$\pi)$
$f'(x)=\cot x <0$
$\cot x$ is -ve in second quadrant.
Therefore $f$ is strictly decreasing in $(\large\frac{\pi}{2},$$\pi)$
Hence proved.
answered Jul 10, 2013 by sreemathi.v
 

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