The option(s) with the values of a and L that satisfy the following equation is (are) `(int_(0)^(4pi)e^(1)(sin^(6)at+cos^(4)at)dt)/(int_(0)^(pi)e^(1)(sin^(6)at+cos^(4)at)dt) = L ?`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)=0`

The option(s) with the values of a and L that satisfy the following eq

1.	The option(s) with the values of a and L that satisfy the following equation is (are) `(int_(0)^(4pi)e^(1)(sin^(6)at+cos^(4)at)dt)/(int_(0)^(pi)e^(1)(sin^(6)at+cos^(4)at)dt) = L ?`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)=0`
Answer» Correct Answer - AC

Consider the integral `I=int_(0)^(2pi)(dx)/(5-2cosx)` Making the substitution `"tan"1/2x=t`, we have `I=int_(0)^(2pi)(dx)/(5-2cosx)=int_(0)^(0)(2dt)/((1+t^(2))[5-2(1-t^(2))//(1+t^(2))])=0` The result is obviously wrong, since the integrand is positive and consequently the integral of this function cannot be equal to zero. Find the mistake.
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Given that `lim_(nrarroo) sum_(r=1)^(n) (log_(e)(n^(2)+r^(2))-2log_(e)n)/n = log_(e)2+pi/2-2`, then evaluate : ` lim_(nrarroo) (1)/(n^(2m))[(n^(2)+1^(2))^(m)(n^(2)+2^(2))^(m) "......"(2n^(2))^(m)]^(1//n)`.
`lim_(nrarroo) sum_(r=0)^(n-1) (1)/(sqrt(n^(2)-r^(2)))`
`lim_(nrarroo) (((n+1)(n+2)"........"3n)/(n^(2n)))^(1//n)` is equal to :A. `(27)/(e^(2))`B. `(9)/(e^(2))`C. `3 log3-2`D. `(18)/(e^(4))`