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Let f:R to R be a thrice differentiable ...

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//2,3)`.
Let f(x)=xF(x) for all `x in R`. If `int_1^3 x^(2)F'(x)dx=-12` and `int_(1)^(3) F''(x)dx=40`, then which of the following is (are) correct?

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`

Text Solution

Verified by Experts

The correct Answer is:
C, D
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.
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