JEE Main 2016 — Mathematics Trigonometry

Given expression is $4+\frac{1}{2}{\mathrm{sin}}^{2}⁡2x-2{\mathrm{cos}}^{4}⁡x$4+21​sin2⁡2x−2cos4⁡x

$=$=$4+2(1-{\mathrm{cos}}^{2}⁡x){\mathrm{cos}}^{2}⁡x-2{\mathrm{cos}}^{4}⁡x$4+2(1−cos2⁡x)cos2⁡x−2cos4⁡x

Creating perfect squares, we get

$$=-4 {{({\mathrm{cos}}^{2}⁡x-\frac{1}{4})}^{2}-\frac{17}{16}}$$=−4(cos2⁡x−41​)2−1617​

Since, $0 \leq {\mathrm{cos}}^{2}⁡x \leq 1$0≤cos2⁡x≤1

$\Rightarrow$⇒ $-\frac{1}{4} \leq {\mathrm{cos}}^{2}⁡x-\frac{1}{4} \leq \frac{3}{4}$−41​≤cos2⁡x−41​≤43​

$\Rightarrow$⇒ $0 \leq {({\mathrm{cos}}^{2}⁡x-\frac{1}{4})}^{2} \leq \frac{9}{16}$0≤(cos2⁡x−41​)2≤169​

$\Rightarrow$⇒ $-\frac{17}{16} \leq {({\mathrm{cos}}^{2}⁡x-\frac{1}{4})}^{2}-\frac{17}{16} \leq \frac{9}{16}-\frac{17}{16}$−1617​≤(cos2⁡x−41​)2−1617​≤169​−1617​

$\Rightarrow$⇒ $\frac{17}{4} \geq -4 {{({\mathrm{cos}}^{2}⁡x-\frac{1}{4})}^{2}-\frac{17}{16}} \geq 2$417​≥−4(cos2⁡x−41​)2−1617​≥2

$\Rightarrow$⇒ $M=\frac{17}{4}$M=417​ and $m=2$m=2

$\Rightarrow$⇒ $M-m=\frac{17}{4}-2=\frac{9}{4}$M−m=417​−2=49​