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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Find all the local maximum values and local minimum values of the function \( f(x)=\sin 2x-x, -\large\frac{\pi}{2} < x < \large\frac{\pi}{2} \)

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  • $f$ is said to have a maximum value in $I$ , if there exist a point c in I such that $f(c) \geq f (x)$ for all $x \in I$.The number $f( c)$ is called the maximum value of f in I and the point c is called a point of maximum value of f in I
  • $f$ is said to have a minimum value in $I$ , if there exist a point $c$ in I such that $f(c) \leq f (x)$ for all $x \in I$.The number $f(c)$ is called the minimum value of f in I and the point $c$ in this case is called a point of minimum value of $f$ in I
  • $f$ is said to have a extreme value in $I$ , if there exist a point $c$ in I such that f(c) is either a maximum value or minimum value of $f$ in $I$. The number $f (c)$ in this case is called the extreme value of $f$ in $I$ and the point $c$ is called the extreme point.
Step 1:
Let $f(x)=\sin 2x-x,\large\frac{-\pi}{2}$$\leq x\leq \large\frac{\pi}{2}$
Differentiating w.r.t $x$
$f'(x)=2\cos 2x-1=0$
$\cos 2x=\large\frac{1}{2}$
$\cos 2x=\cos\large\frac{\pi}{3}$$,\cos(-\large\frac{\pi}{3})$
$\Rightarrow 2x=\large\frac{\pi}{3},-\frac{\pi}{3}$
$x=\large\frac{\pi}{6},\frac{-\pi}{6}$
Step 2:
Consider $x=-\large\frac{\pi}{6}$
Case (i) when $x < -\large\frac{\pi}{6}\Rightarrow$$ 2x <-\large\frac{\pi}{3}$
$-2x < \large\frac{\pi}{3}$
$\cos (-2x) < \cos\large\frac{\pi}{3}$
($\cos x$ is decreasing when 0 < x < $\pi)$
$\cos 2x < \large\frac{1}{2}$
$\Rightarrow 2\cos 2x < 1$
$2\cos 2x-1 < 0$
$f'(x) < 0$
Step 3:
Case (ii) when $x > \large\frac{\pi}{6}$
$-2x > \large\frac{\pi}{3}$
$\cos (-2x) > \cos\large\frac{\pi}{3}$
($\cos x$ is decreasing when 0 < x < $\pi)$
$\cos 2x > \large\frac{1}{2}$
$\Rightarrow 2\cos 2x > 1$
$2\cos 2x-1 > 0$
$f'(x) > 0$
Since $f'(x)$ changes sign from -ve to +ve.
When it passes through $x=-\large\frac{\pi}{6}$
$x=-\large\frac{\pi}{6}$ is a point of local minima.
Local minimum value $f(-\large\frac{\pi}{6})$
$\Rightarrow \sin 2(-\large\frac{\pi}{6})-(\large\frac{-\pi}{6})$
$\Rightarrow -\sin\large\frac{\pi}{3}+\frac{\pi}{6}$
$\Rightarrow \large\frac{\sqrt 3}{2}+\frac{\pi}{6}$
$\Rightarrow \large\frac{\pi-3\sqrt 3}{6}$
Step 4:
Consider $x=\large\frac{\pi}{6}$
Case (i):When $x < \large\frac{\pi}{6}$
$\Rightarrow 2x < \large\frac{\pi}{3}$
$\cos 2x > \large\frac{1}{2}$
($\cos x$ decreasing when 0 < x < $\pi$)
$2\cos 2x <1\Rightarrow 2\cos 2x-1 <0$
$f'(x) >0$
Step 5:
Case (ii) When $x > \large\frac{\pi}{6}$
$\Rightarrow 2x > \large\frac{\pi}{3}$
$\cos 2x < \large\frac{1}{2}$
($\cos x$ decreasing when 0 < x < $\pi$)
$2\cos 2x <1\Rightarrow 2\cos 2x-1 <0$
$f'(x) >0$
Since f'(x) changes sign from +ve to -ve when it passes through $x=\large\frac{\pi}{6}$
$\therefore x=\large\frac{\pi}{6}$ is a point of local maxima.
Local maximum value $f(\large\frac{\pi}{6})$$=\sin 2(\large\frac{\pi}{6})-\frac{\pi}{6}$
$\Rightarrow \sin\large\frac{\pi}{3}-\frac{\pi}{6}$
$\Rightarrow \large\frac{\sqrt 3}{2}-\frac{\pi}{6}$
answered Nov 8, 2013 by sreemathi.v
 

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