Question

If a is a positive integer, then the number of values of a satisfying ${\int }_0^{\pi /2}\left\{\begin{array}{c}{a}^2\left(\frac{\cos 3x}{4}+\frac{3}{4}\cos x\right)+a\sin x-20\cos x\end{array}\right\}dx\le -\frac{a^2}{3}$ is

• A. one
• B. two
• C. three
• D. four.

Answer 1

Answer: (D)

L.H.S. $={\left|\frac{a^2}{4}\cdot \frac{sin3x}{3}+\frac{3a^2}{4}sinx-acosx-20sinx\right|}_0^{\pi /2}$
$=-\frac{a^2}{12}+\frac{3a^2}{4}-20+a=\frac{2a^2}{3}+a-20$
$\therefore$ by the given condition $\frac{2a^2}{3}+a-20\le -\frac{a^2}{3}$
$\Rightarrow 2a^2+3a-60\le -a^2\Rightarrow 3a^2+3a-60\le 0$
$\Rightarrow a^2+a-20\le 0$
$\Rightarrow \left(a+5\right)\left(a-4\right)\le 0$
$\Rightarrow -5\le a\le 4$
Since a is a $+ve$ integer.
$\therefore a=1,2,3,4$
$\therefore$ number of values of $a=4$.