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Mathematics Question on Maxima and Minima

The difference between the greatest and least value of the function, f(x)=sin⁡2x−xf(x) = \sin 2x - x on [−π2,π2]\left[-\frac{\pi}{2},\frac{\pi}{2}\right] is

A

3+22\frac{\sqrt3+\sqrt2}{2}

B

3+22+π6\frac{\sqrt3+\sqrt2}{2}+\frac{\pi}{6}

C

32+π3\frac{\sqrt3}{2}+\frac{\pi}{3}

D

3+22−π3\frac{\sqrt3+\sqrt2}{2}-\frac{\pi}{3}

Answer

32+π3\frac{\sqrt3}{2}+\frac{\pi}{3}

Explanation

Solution

f′(x)=2 cos 2x−1f'\left(x\right)=2\,cos\,2x-1 ⇒f′(x)=0\Rightarrow f'\left(x\right)=0 ⇒2 cos 2x−1=0\Rightarrow 2\,cos\,2x-1=0 ⇒cos 2x=12\Rightarrow cos\,2x=\frac{1}{2} ⇒2x=π3,−π3\Rightarrow 2x=\frac{\pi}{3}, - \frac{\pi}{3} ⇒x=π6,−π6\Rightarrow x=\frac{\pi}{6}, -\frac{\pi}{6} Now, f′(−π2)=π2f'\left(-\frac{\pi}{2}\right)=\frac{\pi}{2}, f′(π2)=−π2f'\left(\frac{\pi}{2}\right)=-\frac{\pi}{2} f′(−π6)=−32+π6f'\left(-\frac{\pi}{6}\right)=-\frac{\sqrt{3}}{2}+\frac{\pi}{6} and f′(π6)=32−π6f'\left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}-\frac{\pi}{6} Clearly 32−π6\frac{\sqrt{3}}{2}-\frac{\pi}{6} is the greatest value of f(x)f\left(x\right) and its least value =−π2=-\frac{\pi}{2}. Hence the reqd. difference =32−π6−(−π2)=32+π3=\frac{\sqrt{3}}{2}-\frac{\pi}{6}-\left(-\frac{\pi}{2}\right)=\frac{\sqrt{3}}{2}+\frac{\pi}{3}