Let F : R `to` R be a thrice differentiable function . Suppose that `F(1)=0,F(3)=-4 and F(x) lt 0 ` for all `x in (1,3)`. `f(x) = x F(x)` for all `x inR`.
If `int_(1)^(3)x^(2)F'(x)dx=-12 andint_(1)^(3)x^(3)F'' (x)dx =40`, then the correct expression (s) is //are

Options

A. `9f'(3)+f'(1)-32=0`
B. `underset(1)overset(3)int f(x)dx=12`
C. `9f'(3)-f'(1)+32=0`
D. `underset(1)overset(3)int f(x)dx=-12`

Answer

C, D

Explanation

We have, f(x)=xF(x).
`:. underset(1)overset(1)int f(x)dx=underset(1)overset(3)int x underset(II)(F)(x)underset(I)(d)x`
`=[(x^(2))/(2)F(x)]_(1)^(3)-(1)/(2)underset(1)overset(3)int x^(2)F'(x)dx`
`(9)/(2)F(3)-(1)/(2)F(1)-(1)/(2)xx(-12)`
`=(9)/(2)(-4)-(1)/(2)(0)-(1)/(2)xx(-12)=-18+6=-12`
It is given that
`underset(1)overset(3)int underset(I)(x^(3))F''underset(II)(x)dx=40`
`rArr [x^(3)F'(x)]_(1)^(3)-3 underset(1)overset(3)int x^(2)F'(x)dx=40`
`rArr 27F'(3)-F'(1)-3xx(-12)=40" "`........(i)
Now, f(x)=xF(x)
`rArr f'(x)=F(x)+xF'(x)`
`rArr f'(1)=F(1)+F'(1)" and "f'(3)=F(3)+3F'(3)`
`f'(1)=0+F'(1)" and "f'(3)=-4+3F'(3)`
`rArr F'(1)=f'(1) " and "F'(3)=(1)/(3)f'(3)+(4)/(3)`
Substituting these values in (i), we obtain
`9f'(3)+36-f'(1)+36=40`
`rArr 9f'(3)-f'(1)+32=0`
Hence, options (c ) and (d) are correct.