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JEE Main 20169 AprilOnline
Exam
JEE Main
Year
2016
Date
9 April
Mode
Online
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

JEE Main 2016 · 9 April · Online · Mathematics · Trigonometric Ratios & Identities

If m and M are the minimum and the maximum values of 4 + (1)/(2) ^...

If mm and MM are the minimum and the maximum values of 4+12sin⁡22x−2cos⁡4x,x∈R,4 + \dfrac{1}{2}\sin^{2} 2x - 2\cos^{4} x, x \in R, then M−mM - m is equal to:

Options

  1. A154\dfrac{15}{4}
  2. B94\dfrac{9}{4}
  3. C74\dfrac{7}{4}
  4. D14\dfrac{1}{4}

Correct answer

B. 94\dfrac{9}{4}

Step-by-step solution

Given expression is 4+12sin⁡22x−2cos⁡4x4 + \dfrac{1}{2}\sin^{2} 2x - 2\cos^{4} x

=4+2(1−cos⁡2x)cos⁡2x−2cos⁡4x4 + 2\left(1 - \cos^{2} x\right)\cos^{2} x - 2\cos^{4} x

Creating perfect squares, we get

=−4{(cos⁡2x−14)2−1716}= - 4\left\{{\left(\cos^{2} x - \dfrac{1}{4}\right)}^{2} - \dfrac{17}{16}\right\}

Since, 0≤cos⁡2x≤10 \le \cos^{2} x \le 1

⇒⇒ −14≤cos⁡2x−14≤34- \dfrac{1}{4} \le \cos^{2} x - \dfrac{1}{4} \le \dfrac{3}{4}

⇒⇒ 0≤(cos⁡2x−14)2≤9160 \le {\left(\cos^{2} x - \dfrac{1}{4}\right)}^{2} \le \dfrac{9}{16}

⇒⇒ −1716≤(cos⁡2x−14)2−1716≤916−1716- \dfrac{17}{16} \le {\left(\cos^{2} x - \dfrac{1}{4}\right)}^{2} - \dfrac{17}{16} \le \dfrac{9}{16} - \dfrac{17}{16}

⇒⇒ 174≥−4{(cos⁡2x−14)2−1716}≥2\dfrac{17}{4} \ge - 4\left\{{\left(\cos^{2} x - \dfrac{1}{4}\right)}^{2} - \dfrac{17}{16}\right\} \ge 2

⇒⇒ M=174M = \dfrac{17}{4} and m=2m = 2
⇒⇒ M−m=174−2=94M - m = \dfrac{17}{4} - 2 = \dfrac{9}{4}

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