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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Which of the following functions are strictly decreasing on \(\left(0, \: \large {\frac{\pi}{2}}\right)\)

$\begin{array}{1 1} (A) cos x , cos 2x \\ (B) none\; of \;these \\ (C) cos \;3x \\(D) tan\; x \end{array} $

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1 Answer

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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:
(A) Let $f(x)=\cos x$
Differentiating w.r.t $x$ we get,
$f'(x)=-\sin x$
Consider the interval $(0,\large\frac{\pi}{2})$
$f'(x)=-\sin x < 0$
Hence $f(x)=\cos x$ is strictly decreasing in interval $(0,\large\frac{\pi}{2})$
Step 2:
(B) Let $f(x)=\cos 2x$
Differentiating w.r.t $x$ we get,
$f'(x)=-2\sin 2x$
In the interval $0 < x < \large\frac{\pi}{2}$$\Rightarrow 0 < 2x < \pi$
$\Rightarrow \sin 2x > 0$
$\Rightarrow -2\sin 2x < 0$
Therefore $f'(x)=-2\sin 2x <0$ on $(0,\large\frac{\pi}{2})$
Hence $f(x)=\cos 2x$ is strictly decreasing in interval $(0,\large\frac{\pi}{2})$
Step 3:
(C) Let $f(x)=\cos 3x$
Differentiating w.r.t $x$ we get,
$f'(x)=-3\sin 3x$
$f'(x)=0$
$\Rightarrow \sin 3x=0$
$\Rightarrow 3x=\pi$
Therefore $[x\in (0,\large\frac{\pi}{2})]$
$x=\large\frac{\pi}{3}$
The point $x=\large\frac{\pi}{3}$ divides the interval $(0,\large\frac{\pi}{2})$ into two disjoint intervals.
(i.e) $(0,\large\frac{\pi}{3})$ and $(\large\frac{\pi}{3},\frac{\pi}{2})$
Consider the interval $(0,\large\frac{\pi}{3})$
$f'(x)=-3\sin 3x < 0$
$[0 < x < \large\frac{\pi}{3}$$\Rightarrow 0 < 3x < \pi]$
Hence $f(x)$ is strictly increasing in interval $(\large\frac{\pi}{3},\frac{\pi}{2})$
Hence $f(x)$ is neither increasing nor decreasing in interval $(0,\large\frac{\pi}{2})$
Step 4:
(D) Let $f(x)=\tan x$
Differentiating w.r.t $x$ we get,
$f'(x)=\sec^2 x$
Consider the interval $(0,\large\frac{\pi}{2})$
$\Rightarrow f'(x)=\sec^2 x > 0$
Hence $f(x)$ is strictly increasing in interval $(0,\large\frac{\pi}{2})$
Step 5:
Therefore functions $\cos x$ and $\cos 2x$ are strictly decreasing in $(0,\large\frac{\pi}{2})$
Hence the correct options are $A$ and $B$.
answered Jul 9, 2013 by sreemathi.v
 

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