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If m and M are the minimum and the maximum values of

$4+\frac{1}{2}\sin ^{2}2x-2\cos ^{4}x\, ,x\, \epsilon \, R,$

then M−m is equal to :

Option 1)
$\frac{15}{4}$

Option 2)
$\frac{9}{4}$

Option 3)
$\frac{7}{4}$

Option 4)
$\frac{1}{4}$

Answers (1)

best_answer

As we learnt in

Method for maxima or minima -

First and second derivative method :

$Step\:1.\:\:find\:values\:of\:x\:for\:\frac{dy}{dx}=0$

$Step\:2.\:\:x=x_{\circ }\:\:is\:a\:point\:of\:local\:maximum\:if\:f'(x)>0$ $and\:local\:minimum\:if\;f'(x)<0.$

$Step\:\:3.\:\:\:x=x_{\circ }\:\:is\:a\:point\:of\:local\:miximum\:if$ $f''(x)<0\:\:and\:local\:minimum\:if\:f''(x)>0$

- wherein

$Where\:\:y=f(x)$

$\frac{dy}{dx}=f'(x)$

Let, $y=4+\frac{1}{2}sin^{2}2x-2cos^{4}x$

$\frac{dy}{dx}=0+\frac{1}{2}\cdot 2sin2x\cdot cos2x\times 2+8cos^{3}x\cdot sinx$

$=sin4x+8cos^{3}x\ sinx$

$\Rightarrow sin2x\left (4 cos^{2}x-1 \right )$

$0=sin2x\left (4cos^{2}x-1 \right )$ So that $M=\frac{17}{4}$

$\therefore 2x=n\pi$ m = 4

$\therefore x=\frac{\pi }{2}$

and $cosx=\pm \frac{1}{2}$

$\therefore M-m=\frac{1}{4}$

Correct option is 4.

Option 1)

$\frac{15}{4}$

This option is incorrect.

Option 2)

$\frac{9}{4}$

This option is incorrect.

Option 3)

$\frac{7}{4}$

This option is incorrect.

Option 4)

$\frac{1}{4}$

This option is correct.

Posted by

divya.saini

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