Mathematics - Functions Question with Solution | TestHub

$f(x)={\left(3-\sqrt{4-x^2}\right)}^2+{\left(1+\sqrt{4-x^2}\right)}^3$

Let $\mathrm{t}=\sqrt{4-{\mathrm{x}}^2}$, clearly $0\le \mathrm{t}\le 2$

$\Rightarrow \mathrm{F}(\mathrm{t})=(3-\mathrm{t}{)}^2+(1+\mathrm{t}{)}^3$

For maxima and minima,

$\Rightarrow {\mathrm{F}}^{\prime }(\mathrm{t})=0$

$\Rightarrow -2(3-\mathrm{t})+3(1+\mathrm{t}{)}^2=0$

$\Rightarrow 3{\mathrm{t}}^2+8\mathrm{t}-3=0$

$\Rightarrow (3\mathrm{t}-1)(\mathrm{t}+3)=0$

$\Rightarrow \mathrm{t}=-3,\frac{1}{3}$

Also ${F^{\prime \prime }(t)|}_{t=-3}=-10\Rightarrow$ maxima

And ${F^{\prime \prime }(t)|}_{t=1/3}=+10\Rightarrow$ minima

As $t\ne -3$, hence maximum value of $F(t)$ will occur at the end points for which $\mathrm{F}(0)=10,\mathrm{F}(2)=28$

Hence, maximum value of $F(t)=28$ for $t=2$

⇒ maximum value of $\mathrm{f}(\mathrm{x})=28$ for $\mathrm{x}=0$.