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 Mm 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)

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.

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