Question

If a is a positive integer, then the number of values of a satisfying $\displaystyle\int_0^{\pi/2}\begin{Bmatrix}a^2\left(\frac{\cos 3x}{4}+\frac{3}{4}\cos\,x\right)+a\,\sin\,x-20\,\cos\,x\end{Bmatrix}dx\leq-\frac{a^2}{3}$ is

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

Answer 1

Answer: (D)

L.H.S. $=\left|\frac{a^{2}}{4}\cdot\frac{sin\,3x}{3}+\frac{3a^{2}}{4} sin\,x-a\, cos\,x-20 sin\,x\right|_{0}^{\pi /2}$
$=-\frac{a^{2}}{12}+\frac{3 a^{2}}{4}-20+a=\frac{2a^{2}}{3}+a-20$
$\therefore $ by the given condition $\frac{2a^{2}}{3}+a-20 \le\, -\frac{a^{2}}{3}$
$\Rightarrow 2a^{2}+3a-60 \le-a^{2} \Rightarrow 3a^{2}+3a-60 \le0$
$\Rightarrow a^{2}+a-20 \le\,0$
$\Rightarrow \left(a+5\right)\left(a-4\right)\le0$
$\Rightarrow -5 \le\, a \le\, 4$
Since a is a $+ve$ integer.
$\therefore a=1, 2, 3,4$
$\therefore $ number of values of $a = 4$.