Question

Let $\alpha \in \left(0 , \frac{\pi }{2}\right)$ and $f\left(x\right)=\sqrt{x^{2} + x}+\frac{\left(tan\right)^{2} \alpha }{\sqrt{x^{2} + x}},x>0$ . If the least value of $f\left(x\right)$ is $2\sqrt{3},$ then $\alpha $ is equal to

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{8}$
• C. $\frac{\pi }{6}$
• D. $\frac{\pi }{4}$

Answer 1

Answer: (A)

By AM-GM inequality
$\sqrt{x^{2} + x}+\frac{tan^{2} \alpha }{\sqrt{x^{2} + x}}\geq 2\cdot \sqrt{\sqrt{x^{2} + x} \cdot \frac{tan^{2} \alpha }{\sqrt{x^{2} + x}}}=2tan\alpha $
Since the least value of $f\left(x\right)$ is $2tan \alpha =2\sqrt{3}$ (given)
Hence, $\alpha =\frac{\pi }{3}$