The option(s) with the values of `aa n dL` that satisfy the following equation is (are) `(int0 4pie^t(s in^6a t+cos^4a t)dt)/(int0pie^t(s in^6a t+cos^4a t)dt)=L` `a=2,L=(e^(4pi)-1)/(e^(pi)-1)` (b) `a=2,L=(e^(4pi)+1)/(e^(pi)+1)` `a=4,L=(e^(4pi)-1)/(e^(pi)-1)` (d) `a=4,L=(e^(4pi)+1)/(e^(pi)+1)`

Options

A. `a=2,L=(e^(4pi)-1)/(e^(pi)-1)`
B. `a=2,L=(e^(4pi)+1)/(e^(pi)+1)`
C. `a=4,L=(e^(4pi)-1)/(e^(pi)-1)`
D. `a=4,L=(e^(4pi)+1)/(e^(pi)+1)`

Answer

A, C

Explanation

Let f(t)`=e^(t)(sin^(6)at+cos^(4)at)`.Then
`f(kpi+t)=r^(kpi+t){sin^(6)(akpi+at)+cos^(4)(akpi+at)}`
`=e^(kpi)e^(t)(sin^(6)at+cos^(4)at)`for even value of a
`=e^(kpi)f(t)`
`:. underset(0)overset(4pi)int (sin^(6)at+cos^(4)at)dt`
`=underset(0)overset(4pi)int f(t)dt=underset(r=1)overset(4)sum overset(r pi)underset((r-1)pi)int f(t)dt`
`=underset(r=1)overset(4)sum underset(0)overset(pi)int ((r-1)pi+u)du`, where `t=(r-1)pi+u`
`=underset(r=1)overset(4)sum overset(pi)underset(0)int e^((r-1)pi)f(u)du`
`=underset(r=1)overset(4)sum e((r-1)pi)overset(pi)underset(0)int f(u)du`
`=(1+e^(pi)+e^(2pi)+e^(3pi))overset(pi)underset(0)intf(u)du`
`=((e^(4pi-1))/(e^(pi)-1))underset(0)overset(pi)intf(t)dt`
`rArr (underset(0)overset(4pi)int e^(t)(sin^(6)at+cos^(4)at)dt)/(underset(0)overset(pi)int e^(t)(sin^(6)at+cos^(4)at)dt)=(e^(4pi-1))/(e^(pi)-1)` for a=2,4