Question

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

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

$E=9xsinx+\frac{4}{xsinx}$ [note that x sin x > 0 in $(0,\pi )$]
$E={(3\sqrt{xsinx}-\frac{2}{\sqrt{xsinx}})}^2+12$
$\therefore E_{min}$=12 which occurs when 3 x sin x = 2 $\Rightarrow xsinx=\frac{2}{3}]$
note that x sin x is continuous at x = 0 and attains the value $\frac{\pi }{2}$ which is greater than $\frac{2}{3}atx=\frac{\pi ]}{2}$ hence it must take the
$\frac{2}{3}in(0,\frac{\pi }{2})$]