Question

An extreme value of $4si{n}^{2}x+3co{s}^{2}x-24sin\frac{x}{2}-24cos\frac{x}{2},$ where $0\le x\le \frac{\pi }{2}$, is

• A. $4+\sqrt{2}$
• B. $4(1-6\sqrt{2})$
• C. $-21$
• D. $4$

Answer 1

Answer: (A)

$4+\sqrt{2}$

The given equation is,

Let $y=4{\sin }^{2}x+3{\cos }^{2}x+\sin x/2+\cos x/2$

$=(3{\sin }^{2}x+3{\cos }^{2}x)+{\sin }^{2}x+\sin x/2+\cos x/2$

$=3+{\sin }^{2}x+\sin x/2+\cos x/\mathrm{2.....}\left(1\right)$

As $[{\sin }^{2}x+{\cos }^{2}x=1]$

Multiplying and dividing by $\sqrt{2}$

$\sin x/2+\cos x/2=\sqrt{2}[\frac{1}{\sqrt{2}}\sin \frac{x}{2}+\frac{1}{\sqrt{2}}\cos x/2]$

$=\sqrt{2}[\sin x/2\cos \pi /4+\cos x/2\sin \pi /4]$

$=\sqrt{2}\left(\sin \right(x/2+\pi /4\left)\right].......\left(2\right)$

$\therefore$ Putting $\left(2\right)$ in $\left(1\right)$

$3+{\sin }^{2}x+\sqrt{2}\left[\sin \right(x/2+\pi /4\left)\right].....\left(3\right)$

Now ${\sin }^{2}x\le 1$ as sin lies in $-1$ to $1$

and since $-1\le \sin \le 1$ then

$-\sqrt{2}\le \left[\sqrt{2}\sin \right(x/2+45\left)\right]\le \sqrt{2}$

putting these in $\left(3\right)$

$=3+1+\sqrt{2}$

$=4+\sqrt{2}$