Question 282 of 298MediumMCQ
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 :
Solutions · 1
Official 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 = \[{{9} \over 4}\]
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