Trigonometric Ratio and Identities — Page 6 | JEE Mathematics

JEE Mathematics · 51 questions · Page 6 of 6 · Click an option or "Show Solution" to reveal answer

Q51

If m and M are the minimum and the maximum values of 4 +

{1 \over 2}

sin2 2x

-

2cos4 x, x

\in

R, then M

-

m is equal to :

{{15} \over 4}

{{9} \over 4}

{{7} \over 4}

{{1} \over 4}

Correct Answer

Option B

Solution

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}