If m and M are the minimum and the maximum values of 4 +1/2 sin22x - 2 cos4x, x ∈ R, then M - m is equal to

4+12×4(1-cos2x)−2cos4x=4+2cos2x-2cos4x=2(-2cos4x+cos2x+2)=-4(cos4x-cos2x2-1)=4(cos2x-14)2-1716=174-4((cos2x-14)2)yMax=M=174ymin=m=174-(4×916)=84=2Now M-m=174-2=94

If m and M are the minimum and the maximum values of 4 +1/2 sin22x - 2 cos4x, x ∈ R, then M - m is equal to

4+12×4(1-cos2x)−2cos4x=4+2cos2x-2cos4x=2(-2cos4x+cos2x+2)=-4(cos4x-cos2x2-1)=4(cos2x-14)2-1716=174-4((cos2x-14)2)yMax=M=174ymin=m=174-(4×916)=84=2Now M-m=174-2=94

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If m and M are the minimum and the maximum values of 4 +1/2 sin²2x - 2 cos⁴x, x $\in$ R, then M - m is equal to

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$\frac{7}{4}$

$\frac{9}{4}$

$\frac{15}{4}$

$\frac{1}{4}$

answer is C.

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Detailed Solution

$\begin{array}{l} 4 + \frac{1}{2} \times 4 (1 - \cos^{2} x) - 2 \cos^{4} x \\ = 4 + 2 \cos^{2} x - 2 \cos^{4} x \\ = 2 (- 2 \cos^{4} x + \cos^{2} x + 2) \\ = - 4 (\cos^{4} x - \frac{\cos^{2} x}{2} - 1) \\ = 4 [{(\cos^{2} x - \frac{1}{4})}^{2} - \frac{17}{16}] \\ = \frac{17}{4} - 4 ({(\cos^{2} x - \frac{1}{4})}^{2}) \\ y_{M a x} = M = \frac{17}{4} \\ y_{m i n} = m = \frac{17}{4} - (4 \times \frac{9}{16}) = \frac{8}{4} = 2 \\ N o w M - m = \frac{17}{4} - 2 = \frac{9}{4} \end{array}$