Let `F:RtoR` be a thrice differntiable function. Suppose that `F(1)=0,F(3)=-4` and `F'(x)lt0` for all `x epsilon(1//2,3)`. Let `f(x)=xF(x)` for all `x inR`. Then the correct statement(s) is (are)

Options

A. `f(1) lt 0 `
B. `f(2) lt 0 `
C. `f(x) ne 0 " for all x in (1,3)`
D. `f(x)=0 for some x in`

Answer

A, B, C

Explanation

we have f(x) = x F(x) for all `x in R`
`rArr f'(x)=F(x) +x " for all " x in R`
`rArr f(1) = F(1) + f' (1)`
`rArr f(1) = 0 + F' (1) lt 0 [ because F'(x) lt 0 " for all " x in (1/2,3]`
So statement (a) is correct
It is given that `F(x) lt 0` for all ` x in (1/2,3)`.So F(x) is decreasing on the interval (1/2),3.
` because F(2) lt F(1)`
`rArr F(2) lt 0`
`rArr 2 F(2) lt 0 `
`rArr f(2) lt 0 `
So ,statement (b) is correct .
Again f(x)=F(x) for all ` x in R`
`rArr f(x)=F(x)=F(x)+xF'(x) " for all " x in R `
Now `f(x) lt 0 for all x in (1//2,3)`
`rArr F'(x) lt 0 " for all " x in (1,3) and F(x )` s decreasing on (1,3)
`F(x) lt 0 and F(x) lt 0 "for all" x in (1,3)`
`f(x) lt 0 " for all " x in (1,3)`
`f(x) lt 0 for all x in (1,3)`
`rArr f(x) ne 0 "for all" x in (1,3)`
So ,statement ( c) is correct and statement (d) is incorrect.