Question

The option(s) with the value of $a$ and $L$ that satisfy the following equation is(are)
$\frac{\underset{0}{\overset{4\pi }{\int }}{e}^t({\sin }^{6}at+{\cos }^{4}at)dt}{\underset{0}{\overset{\pi }{\int }}{e}^t({\sin }^{6}at+{\cos }^{4}at)dt}=L?$

• A. $a=2,L=\frac{e^{4\pi }-1}{e^{\pi }-1}$
• B. $a=2,L=\frac{e^{4\pi }+1}{e^{\pi }+1}$
• C. $a=4,L=\frac{e^{4\pi }-1}{e^{\pi }-1}$
• D. $a=4,L=\frac{e^{4\pi }+1}{e^{\pi }+1}$

Answer 1

Answer: (A), (C)

$a=4,L=\frac{e^{4\pi }-1}{e^{\pi }-1}$

$f\left(t\right)=e^t({\sin }^{6}at+{\cos }^{4}at)f(k\pi +t)=e^{k\pi +t}({\sin }^6(ak\pi +at)+{\cos }^4(ak\pi +at))f(k\pi +t)=e^{k\pi }f\left(t\right)$
$\underset{0}{\overset{4\pi }{\int }}f\left(t\right)dt=\underset{0}{\overset{\pi }{\int }}f\left(t\right)dt+\underset{\pi }{\overset{2\pi }{\int }}f\left(t\right)dt+\underset{2\pi }{\overset{3\pi }{\int }}f\left(t\right)dt+\underset{3\pi }{\overset{4\pi }{\int }}f\left(t\right)dt=(e^0+e^{\pi }+e^{2\pi }+e^{3\pi })\underset{0}{\overset{\pi }{\int }}f\left(t\right)dt$
$\frac{\underset{0}{\overset{4\pi }{\int }}{e}^t({\sin }^{6}at+{\cos }^{4}at)dt}{\underset{0}{\overset{\pi }{\int }}{e}^t({\sin }^{6}at+{\cos }^{4}at)dt}=\frac{(e^0+e^{\pi }+e^{2\pi }+e^{3\pi })\underset{0}{\overset{\pi }{\int }}f\left(t\right)dt}{\underset{0}{\overset{\pi }{\int }}f\left(t\right)dt}=LL=(1+e^{\pi }+e^{2\pi }+e^{3\pi })=\frac{e^{4\pi }-1}{e^{\pi }-1}\text{for even value of 'a' i.e.}a=2,4$