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

           

We have

$\begin{array}{l}{I}_n=\underset{n\pi }{\overset{\left(n+1\right)\pi }{\int }}{e}^t\left({\sin }^{6}at+{\cos }^{4}at\right)dt\\ \text{substitute }t=n\pi +\theta \\ \Rightarrow I_n=e^{n\pi }\underset{0}{\overset{\pi }{\int }}{e}^{\theta }\left({\sin }^{6}a\theta +{\cos }^{4}a\theta \right)d\theta \\ \therefore \underset{0}{\overset{4\pi }{\int }}{e}^t\left({\sin }^{6}at+{\cos }^{4}at\right)dt=\underset{0}{\overset{\pi }{\int }}+\underset{\pi }{\overset{2\pi }{\int }}+\underset{2\pi }{\overset{3\pi }{\int }}+\underset{3\pi }{\overset{4\pi }{\int }}\\ =\left(1+e^{\pi }+e^{2\pi }+e^{3\pi }\right)\underset{0}{\overset{\pi }{\int }}{e}^t\left({\sin }^{6}at+{\cos }^{4}at\right)dt\\ =\frac{e^{4\pi }-1}{e^{\pi }-1}\underset{0}{\overset{\pi }{\int }}{e}^t\left({\sin }^{6}at+{\cos }^{4}at\right)dt\\ \therefore \frac{\underset{0}{\overset{4\pi }{\int }}{e}^t\left({\sin }^{6}at+{\cos }^{4}at\right)dt}{\underset{0}{\overset{\pi }{\int }}{e}^t\left({\sin }^{6}at+{\cos }^{4}at\right)dt}=\frac{e^{4\pi }-1}{e^{\pi }-1}\end{array}$

For any value of ‘a’ the above result holds.

Options (1),(3) are correct.