Get Answers to all your Questions

header-bg qa

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)


  • Option 2)


  • Option 3)


  • Option 4)



Answers (1)


As we learnt in 

Method for maxima or minima -

First and second derivative method :


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




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)


This option is incorrect.

Option 2)


This option is incorrect.

Option 3)


This option is incorrect.

Option 4)


This option is correct.

Posted by


View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE