Question

The range of the function $f\left(x\right)=\sqrt{4-x^2}+\sqrt{x^2-1}$ is

• A. $\left[\sqrt{3,}\sqrt{7}\right]$
• B. $\left[\sqrt{3,}\sqrt{5}\right]$
• C. $\left[\sqrt{2,}\sqrt{3}\right]$
• D. $\left[\sqrt{3,}\sqrt{6}\right]$

Answer 1

Answer: (D)

$\left[\sqrt{3,}\sqrt{6}\right]$

As we are looking at $f\left(x\right)=\sqrt{4-x^2}+\sqrt{x^2-1}$

we cannot have $|x|<1or|x|>2$

For it $|x|<1,then\sqrt{x^2-1}$ is not defined and if $|x|>2,then\sqrt{4-x^2}$ is not defined.

Hence domain of x is the interval $[-2,-1]\cup [1,2]$

Note that $f(-2)=f(-1)=f\left(1\right)=f\left(2\right)=\sqrt{3}$

Further $f^{\prime }\left(x\right)=\frac{1}{2\sqrt{4-x^2}}.(-2x)+\frac{1}{2\sqrt{x^2-1}}.2x$

$=x(\frac{1}{\sqrt{x^2-1}}-\frac{1}{\sqrt{4-x^2}})$

It is zero when

$\sqrt{x^2-1}=\sqrt{4-x^2}$

$2x^2=5$

$x^2=2.5$

$x=\pm \sqrt{2.5}$

where $f\left(\sqrt{2.5}\right)=\sqrt{4-(\sqrt{2.5}{)}^2}+\sqrt{(\sqrt{2.5}{)}^2-1}=2\sqrt{1.5}=\sqrt{6}$

Hence range is $[\sqrt{3},\sqrt{6}]$