If `f(x)` is a cubic polynomial which as local maximum at `x=-1` . If `f(2)=18,` `f(1)=-1` and `f'(x)` has minimum at `x=0` then

Options

A. the distance between `(-1,2)` and `(alpha,f(alpha))` where `x=alpha ` is the point of local minima is `2sqrt5`
B. `f(x)` is increasing for `x in [1,2sqrt5]` and has a local minima at `x=1`
C. the value of `f(0)` is `5`
D. none of these

Answer

B

Explanation

It given that f'(x) has a local minimum at x=0
`therefore x=0 " is a zero of " f''(x)`
`rArr f''(x) =ax " " [therefore f(x) " is a cubic polynomial] "`
`rArr f'(x) =(ax^2)/2+b and f(x) =(ax^3)/6+bx+c`
It also given that f(x) has a local maximum at x=-1
`therefore f'(-1) =0 rArra/2+b=0rArra=-2b`
We have ,
f(1) and f(2)=18
`rArr -1=1/6+b+c and 18 =(4a)/3+2b+c`
Solving these eqations we get `a-(57)/2,b=(57)/4,c=(17)/2`
`thereforef(x) =(57)/(12)x^3-(57)/4x+(17)/2=1/4(19x^3-57x+34)` It is given that x=a is the point of local minima
`therefore a=1 and f(alpha)=-1`
Distance between (-1,2) and (-1,1) is `(sqrt13)`
So option (a) is not correct
Since f(x) is a continous function localmaxima at x=-1 and local minima at x=1 such that f(-1) =18 and f(1) =-1 so a rough sketch of f(x) is as shown below

Clearly ,f(x) is increasing in `[1,2sqrt(5)]` and f(x)has a local minima at x=1 .So option (b) is correct.