Question

The maximum value of $f\left(x\right)=ta{n}^{-1}\left(\frac{(\sqrt{12}-2)x^2}{x^4+2x^2+3}\right)$ is

• A. ${18}^{\circ }$
• B. ${36}^{\circ }$
• C. ${22.5}^{\circ }$
• D. ${15}^{\circ }$

Answer 1

The correct option is D ${15}^{\circ }$

$f\left(x\right)=ta{n}^{-1}\left(\frac{(\sqrt{12}-2)x^2}{x^4+2x^2+3}\right)$
$f^{{}^{\prime }}\left(x\right)=\frac{4(\sqrt{3}-1)x(x^4-3)}{x^8+4x^6+(26-8\sqrt{3})x^4+12x^2+9}$
Critical points are $0,-\sqrt[4]{43},\sqrt[4]{43}$
And for $x=-\sqrt[4]{43},\sqrt[4]{43}$ $f\left(x\right)$ attain maximum
And $f\left(x\right)={15}^0$