Question

The minimum value of the expression $\frac{{9x}^2{\sin }^{2}x+4}{x\sin x}$ for $x\in (0,\pi )$ is

• A. $\frac{16}{3}$
• B. $12$
• C. $6$
• D. None of these

Answer 1

The correct option is B $12$

$\frac{{9x}^2{\sin }^{2}x+4}{x\sin x}$ for $x\in (0,\pi )$

Let $y=x\sin x.$

Then the expression becomes, $\frac{9y^2+4}{y}=9y+\frac{4}{y}$.

Since, $x>0,$ and $\sin x>0$ because $0<x<\pi ,$ we have $y>0.$

So, we can apply $AM-GM:$

$\Rightarrow$ $9y+\frac{4}{y}\ge 2\sqrt{9y\times \frac{4}{y}}=12$

The equality holds when,

$9y=\frac{4}{y}⟺y^2=\frac{4}{9}⟺y=\frac{2}{3}$

$\therefore$ The minimum value is $12$.

This is reached when we have $x\sin x=\frac{2}{3}$ in the original equation ( since, $x\sin x$ is continuous and increasing on the interval $0\le x\le \frac{\pi }{2},$ and its range on that ineterval is from $0\le x\sin x\le \frac{\pi }{2},$ this value of $\frac{2}{3}$ is attainable by the Intermediate value theorem ).