Question

The option(s) with the values of $\alpha$ and $L$ that satisfies the following equation is (are)
$\frac{{\int }_0^{4\pi }{e}^t({\sin }^6\alpha t+{\cos }^4\alpha t)dt}{{\int }_0^{\pi }{e}^t({\sin }^6\alpha t+{\cos }^4\alpha t)dt}=L,$ is/are

• 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)

The correct options are
A $a=2,L=\frac{e^{4\pi }-1}{e^{\pi }-1}$
C $a=4,L=\frac{e^{4\pi }-1}{e^{\pi }-1}$

Let $f\left(t\right)=e^t({\sin }^6\alpha t+{\cos }^6\alpha t)$
$\therefore f(k\pi +t)=e^{k\pi +t}\left({\sin }^6\alpha \right(k\pi +t)+{\cos }^6\alpha (k\pi +t\left)\right)=e^{k\pi }f\left(t\right)$
$\therefore \frac{{\int }_0^{4\pi }{e}^t({\sin }^6\alpha t+{\cos }^4\alpha t)dt}{{\int }_0^{\pi }{e}^t({\sin }^6\alpha t+{\cos }^4\alpha t)dt}=\frac{(1+e^{\pi }+e^{2\pi }+e^{3\pi }){\int }_0^{\pi }{e}^t({\sin }^6\alpha t+{\cos }^4\alpha t)dt}{{\int }_0^{\pi }{e}^t({\sin }^6\alpha t+{\cos }^4\alpha t)dt}$
$=1+e^{\pi }+e^{2\pi }+e^{3\pi }=\frac{e^{4\pi }-1}{e^{\pi }-1}$
Hence option A and C are correct.