`f(x)` is continuous function for all real values of `x` and satisfies `int_0^xf(t)dt=int_x^1t^2f(t)dt+(x^(16))/8+(x^6)/3+adot` Then the value of `a` is equal to: `-1/(24)` (b) `(17)/(168)` (c) `1/7` (d) `-(167)/(840)`

Options

A. `-1/24`
B. `17/168`
C. `1/7`
D. `-167/840`

Answer

D

Explanation

`int_(0)^(x)f(t)dt=int_(x)^(1)t^(2)f(t)dt+(x^(16))/8+(x^(6))/3+a`……………..1
For `x=1, int_(0)^(1)f(t)dt=01/8+1/3+a=11/24+a`……………2
Differentiating both sides of w.r.t`x`
We get `f(x)=0-x^(2)f(x)+2x^(15)+2x^(5)`
`impliesf(x)=(2(x^(15)+x^(4)))/(1+x^(2))`
`implies2int_(0)^(1)(x^(15)+x^(5))/(1+x^(2))dx+11/24+a`(from a)
`implies2int_(0)^(1)(x^(13)-x^(11)+x^(9)-x^(7)+x^(5))dx=11/24+a`
`implies 2(1/14-1/12+1/10-1/8+1/6)=11/24+a`
`implies a=-167/840`