Question

If $a_1,a_2$ and $a_3$ are the three value of $‘a^{\prime }$ which satisfy the equation $\underset{0}{\overset{\pi /2}{\int }}(\sin x+a\cos x{)}^{3}dx-\frac{4a}{\pi -2}\underset{0}{\overset{\pi /2}{\int }}x\cos xdx=2$, then the value of $a_1^2+a_2^2+a_3^2$ is

• A. $\frac{21}{4}$
• B. $\frac{21}{2}$
• C. $21$
• D. $7$

Answer 1

The correct option is A $\frac{21}{4}$

Given : $\underset{0}{\overset{\pi /2}{\int }}(\sin x+a\cos x{)}^{3}dx-\frac{4a}{\pi -2}\underset{0}{\overset{\pi /2}{\int }}x\cos xdx=2$
Let $I_1=\underset{0}{\overset{\pi /2}{\int }}(\sin x+a\cos x{)}^{3}dx$
$\Rightarrow I_1=\underset{0}{\overset{\pi /2}{\int }}({\sin }^{3}x+a^3{\cos }^{3}x+3a\sin x\cos x(\sin x+a\cos x\left)\right)dx\Rightarrow I_1=\underset{0}{\overset{\pi /2}{\int }}{\sin }^{3}xdx+a^3\underset{0}{\overset{\pi /2}{\int }}{\cos }^{3}xdx+3a\underset{0}{\overset{\pi /2}{\int }}{\sin }^{2}x\cos xdx+3a^2\underset{0}{\overset{\pi /2}{\int }}\sin x{\cos }^{2}xdx$
Now,
$\underset{0}{\overset{\pi /2}{\int }}{\cos }^{3}xdx=\underset{0}{\overset{\pi /2}{\int }}{\sin }^{3}xdx=\frac{1}{4}\underset{0}{\overset{\pi /2}{\int }}(3\sin x-\sin 3x)dx=\frac{1}{4}[3-\frac{1}{3}]=\frac{2}{3}$

$\Rightarrow I_1=\frac{2(1+a^3)}{3}+3a\underset{0}{\overset{\pi /2}{\int }}(1-{\cos }^{2}x)\cos xdx+3a^2\underset{0}{\overset{\pi /2}{\int }}\sin x(1-{\sin }^{2}x)dx\Rightarrow I_1=\frac{2(1+a^3)}{3}+3a[1-\frac{2}{3}]+3a^2[1-\frac{2}{3}]\Rightarrow I_1=\frac{2a^3+3a^2+3a+2}{3}$

Let $I_2=\underset{0}{\overset{\pi /2}{\int }}x\cos xdx$
$\Rightarrow I_2=[x\sin x+\cos x{]}_0^{\pi /2}\Rightarrow I_2=\frac{\pi -2}{2}$
Now, given equation is
$I_1-\frac{4a}{\pi -2}{I}_2=2\Rightarrow \frac{2a^3+3a^2-3a+2}{3}=2\Rightarrow 2a^3+3a^2-3a-4=0$

So,
$a_1^2+a_2^2+a_3^2=(a_1+a_2+a_3{)}^2-2(a_1{a}_2+a_2{a}_3+a_1{a}_3)=\frac{9}{4}+3=\frac{21}{4}$