QJEEM16Maaa02: JEE Main 2016 Mathematics - Class 11 JEEM16Maaa02 Solution: If  m and M are the minimum and the maximum values of 4 +…

Given,

4 + ${1 \over 2}$ sin² 2x $-$ 2cos⁴ x

= 4 + ${1 \over 2}$ (2sinx cosx)² $-$ 2cos⁴x

= 4 + ${1 \over 2}$ $\times$ 4 sin²x cos²x $-$ 2cos⁴ x

= 4 + 2 (1 $-$ cos²x) cos²x $-$ 2cos⁴ x

= 4 + 2 cos²x $-$ 4cos⁴x

= $-$ 4 $\left\{ {\cos } \right.$ ⁴x $-$ $\left. {{{{{\cos }^2}x} \over 2} - 1} \right\}$

= $-$ 4 $\left\{ {\cos } \right.$ ⁴x $-$ 2 . ${1 \over 4}$ . cos²x + $\left. {{1 \over {16}} - {1 \over {16}} - 1} \right\}$

= $-$ 4 $\left\{ {{{\left( {{{\cos }^2}x - {1 \over 4}} \right)}^2} - {{17} \over {16}}} \right\}$

We know,

O $\le$ cos²x $\le$ 1

$\Rightarrow$     $- {1 \over 4}$ $\le$ cos²x $- {1 \over 4}$ $\le$ ${3 \over 4}$

$\Rightarrow$    O $\le$ ${\left( {{{\cos }^2}x - {1 \over 4}} \right)^2}$ $\le$ ${9 \over {16}}$

$\Rightarrow$     $-$ ${17 \over {16}}$ $\le$ ${\left( {{{\cos }^2}x - {1 \over 4}} \right)^2}$ $-$ ${{17} \over {16}}$ $\le$ ${9 \over {16}}$ $-$ ${{17} \over {16}}$

$\Rightarrow$     $-$ ${{17} \over {16}}$ $\le$ ${\left( {{{\cos }^2}x - {1 \over 4}} \right)^2}$ $-$ ${{17} \over {16}}$ $\le$ $-$ ${{1} \over {2}}$

$\Rightarrow$     ${{17} \over {4}}$ $\ge$ $-$ 4 $\left\{ {{{\left( {{{\cos }^2}x - {1 \over 4}} \right)}^2} - {{17} \over {16}}} \right\} \ge 2$

$\therefore$    Maximum value, M = ${{17} \over 4}$

      Minimum value, m = $2$

$\therefore$    M $-$ m = ${{17} \over 4}$ $-$ $2$ = ${{9} \over 4}$