If `int_0^xf(t)dt=x^2+int_x^1t^2f(t)dt,` then `f^(prime)(1/2)` is

Options

A. `(24)/(25)`
B. `(18)/(25)`
C. `(6)/(25)`
D. `(4)/(5)`

Answer

A

Explanation

Given , `int_(0)^(x) f (t) dt = x^(2)+int_(x)^(1)t^(2)f (t) dt`
On differentiating both sides , W. r. t . 'x' we get
`f(x) =2x+0-x^(2)f(x)`
`[ :' (d)/(dx)[int_(phi(x))^(Psi(x))f(t)dt ] = f (Psi(x))(d)/(dx)Psi (x) - f (phi(x))(d)/(dx)phi(x)]`
`rArr (1+x^(2))f(x) = 2x rArr f(x) = (2x)/(1+x^(2))`
On differentiating W . r. t . 'x' we get
`f'(x)((1+x^(2))(2)-(2x)(0+2x))/((1+x^(2))^(2))`
`= (2+2x^(2)-4x^(2))/((1+x^(2))^(2))=(2-2x^(2))/((1+x^(2))^(2))`
`:. f ' ((1)/(2))=(2-2((1)/(2))^(2))/((1+((1)/(2))^(2))^(2))=(2-2((1)/(4)))/((1+(1)/(2))^(2))=(2-(1)/(2))/((5)/(4))^(2)=((3)/(2))/((25)/(16))=(24)/(25)`