Let `f(x)` be a 4 egree polynomial satisfying `f(r)=(1)/(r)` for `r=1,2,3,4,5` then
Q. Which of the following is/are correct:-

Options

A. `f(x)` is always increasing
B. `f(x)=o` has no negative root
C. `f(0)=0`
D. `f^(')(x)=0` for atleast one `xepsilon(3,6)`

Answer

B, D

Explanation

`f(x)=(1)/(x)` for `x=1,2,3,4,5`
`impliesxf(x)-1=0` for `x=1,2,3,4,5`
`g(x)=xf(x)-1=k(x-1)(x-2)(x-3)(x-4)(x-5)`
`(becausef(x)` is a 4 degree polynomial)
`0.f(0)-1=k(-1)(-2)(-3)(-4)(-5)`
`k=(1)/(5!)`
`xf(x)-1=(1)/(5!)(x-1)(x-2)(x-3)(x-4)(x-5)`
`6f(6)-1=1implies(f)6=(1)/(3)=f(3)`
`implies` By Rolle's theorem, there exists atleast one `x in (3,6)` for which `f^(')(x)=0`
also, for any `x=-k(kgt0)`
`-kf(-k)-1=-ve` quantity less than `-1`
`implieskf(-k)+1=+ve` quantity greater than `1impliesf(-k)gt0`
`impliesf(x)=0` has no negative root.