Question

If $g(x)=2\sqrt{2}\sin x+\cos x$ then maximum value of $P(x)=\sqrt{g(x)-2}+\sqrt{4-g(x)}$ is

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (C)

$g(x)=2\sqrt{2}\sin x+\cos x$
$g(x)\in [-3,3]$
$y=\sqrt{g(x)-2}+\sqrt{4-g(x)}$
$y^2=g(x)-2+4-g(x)+2\sqrt{g(x)-2}\sqrt{4-g(x)}$
$y^2=2+2\sqrt{4g-g^2-8+2g}=2+2\sqrt{-[g(x){]}^2+6(g(x))-8}=2+2\sqrt{1-[g(x)-3{]}^2}$
$\text{ where }g(x)=3\text{ then }{y}^2{\max }^2=4$
$y_{\max }=2$