Question

The range of the function $f(x)=2\sqrt{x-2}+\sqrt{4-x}$ is

• A. $(\sqrt{2,}\sqrt{10})$
• B. $[\sqrt{2},\sqrt{10})$
• C. $(\sqrt{2},\sqrt{10}]$
• D. $[\sqrt{2},\sqrt{10}]$

Answer 1

Answer: (D)

Clearly, domain of the function is $[2,4]$. Now,
$f^{\prime }(x)=\frac{1}{\sqrt{x-2}}-\frac{1}{2\sqrt{4-x}}$
$f^{\prime }(x)=0$ or $\sqrt{x-2}=2\sqrt{4-x}$
or $x-2=16-4x$ or
$x=\frac{18}{5}$
Now, $f(2)=\sqrt{2},f\left(\frac{18}{5}\right)$
$=2\sqrt{\frac{18}{5}-2}+\sqrt{4-\frac{18}{5}}=\sqrt{10}$
$f(4)=2\sqrt{2}$
Hence, range of the function is $[\sqrt{2},\sqrt{10}]$.
Also, here $x=(18/5)$ is the point of global maxima.