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

f(x) is increasing for $x \in[1,2 \sqrt{5}]$ and has a local minima at x = 1

It is given that f'(x) has a local minimum at x = 0.

∴ x = 0 is a zero of f''(x)

$\Rightarrow f''(x)=a x$                       [∵ f(x) is a cubic polynomial]

$\Rightarrow f'(x)=\frac{a x^2}{2}+b$ and $f(x)=\frac{a x^3}{6}+b x+c$

It is also given that $f(x)$ has a local maximum at x = -1

$f'(-1)=0 \Rightarrow \frac{a}{2}+b=0 \Rightarrow a=-2 b$

We have,

$f(1)=-1$ and $f(2)=18$

$\Rightarrow -1 -\frac{a}{6}+b+c$ and $18=\frac{4 a}{3}+2 b+c$

Solving these equations, we get $a=\frac{57}{2}, b=\frac{-57}{4}, c=\frac{17}{2}$

∴  $f(x)=\frac{57}{12} x^3-\frac{57}{4} x+\frac{17}{2}=\frac{1}{4}\left(19 x^3-57 x+34\right)$

It is given that $x=\alpha$ is the point of local minima.

∴  $\alpha=1$ and $f(\alpha)=-1$

Distance between $(-1,2)$ and $(1,-1)$ is $\sqrt{13}$.

So, option (a) is not correct.

Since f(x) is a continuous function having local maxima 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 $[1,2 \sqrt{5}]$ and f(x) has a local minima at x = 1. So, option (b) is correct.