Find the maximum/minimum value of y in the functions given below.
$y = (\sin 2 x - x), where - \frac{π}{2} \leq x \leq \frac{π}{2}$

Moderate

A

$y_{\min} = - (\frac{\sqrt{3}}{4} - \frac{π}{6}) at x = - π / 6 and y_{\max} = (\frac{\sqrt{5}}{2} - \frac{π}{6}) at x = π / 6$
B

$y_{\min} = - (\frac{\sqrt{3}}{2} - \frac{π}{4}) at x = - π / 4 and y_{\max} = (\frac{\sqrt{3}}{2} - \frac{π}{2}) at x = π / 2$
C

$y_{\min} = - (\frac{\sqrt{3}}{2} - \frac{π}{6}) at x = - π / 6 and y_{\max} = (\frac{\sqrt{3}}{2} - \frac{π}{6}) at x = π / 6$
D

None of the above

Solution

$y = (\sin 2 x - x)$
$\frac{d y}{d x} = 2 \cos 2 x - 1$
and
$\frac{d^{2} y}{d x^{2}} = - 4 \sin 2 x$
Putting
$\frac{d y}{d x} = 0, we get$
$2 \cos 2 x - 1 = 0$
$\cos 2 x = \frac{1}{2}$
or $2 x = \pm 60 °$
$At 2 x = + 60 °, \frac{d^{2} y}{d x^{2}} is - ve, so value of y is$ $maximum. At 2 x = - 60 °, \frac{d^{2} y}{d x^{2}} is positive. So value$ $of y is minimum.$
$y_{\max} = \sin (+ 60 °) - \frac{π}{6}$
$= (\frac{\sqrt{3}}{2} - \frac{π}{6}) at 2 x = \frac{π}{3}$
or $x = π / 6$
and $y_{\min} = \sin (- 60 °) + \frac{π}{6}$
at $x = - \frac{π}{6}$
$= (\frac{π}{6} - \frac{\sqrt{3}}{2})$

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