Question

The greatest value of the function
$f\left(x\right)=\frac{\sin 2x}{\sin (x+\frac{\pi }{4})}$ on the interval $[0,\frac{\pi }{2}]$ is

• A. $\frac{1}{\sqrt{2}}$
• B. $\sqrt{2}$
• C. $1$
• D. $-\sqrt{2}$

Answer 1

The correct option is C $1$

$f\left(x\right)=\frac{\sin x}{\sin (x+\frac{\pi }{4})}$ on interval $[0,\frac{\pi }{2}]$

$f\left(x\right)=\frac{2\sin x\cos x}{(\sin x\times \frac{1}{\sqrt{2}}+\cos x\times \frac{1}{\sqrt{2}})}\Rightarrow \frac{2\sqrt{2}\sin x\cos x}{(\sin x+\cos x)}$

Now, $\frac{\sin x+\cos x}{\sin x\cos x}\Rightarrow (\frac{1}{\cos x}+\frac{1}{\sin x})$

Greatest value of $(\frac{1}{\sin x}+\frac{1}{\cos x})$ will be at $45°$

Greatest value$=(\frac{1}{1/\sqrt{2}}+\frac{1}{1/\sqrt{2}})=2\sqrt{2}$

$\therefore$ Greatest value of $f\left(x\right)=\frac{2\sqrt{2}}{2\sqrt{2}}=1$ (Answer)