Let `f: R -> R` be a differentiable function such that `f(0) = 0,f(pi/2)=3a n df^(prime)(0)=1.` If `g(x)=int_x^(pi/2)[f^(prime)(t)cos e ct-cottcos e ctf(t)]dt` for `x(0,pi/2],` then `(lim)_(x -> 0)g(x)=`

Answer

2

Explanation

`g(x)=int_(x)^((pi)/2)(f'(t)cosect-f(t)cosectcott)dt`
`=int_(x)^((pi)/2)d/(dt)(f(t)cosect)dt`
`f((pi)/2)cosec((pi)/2)-(f(x))/(sinx)=3-(f(x))/(sinx)`
`:.lim_(xto0) g(x)=3-log_(xto 0)(f(x))/(sinx)`
`=lim_(xto 0)g(x)`
`=3-log(f'(x))(cosx)` (using L'Hopital Rule)
`=3-1 ( :' f'(0)=1)`
`=2`