Question

$f\left(x\right)$ is cubic polynomial which has local maximum at $x=-1,$ If $f\left(2\right)=18,f\left(1\right)=-1$ and $f^{\prime }\left(x\right)$ has local minima at $x=0,$ then

• A. the distance between point of maxima and minma is $2\sqrt{5}$
• B. $f\left(x\right)$ is increasing for $x\in [1,2\sqrt{5}]$
• C. $f\left(x\right)$ has local minima at $x=1$
• D. the value of $f\left(0\right)=5$

Answer 1

Answer: (B), (C)

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

Since $f\left(x\right)$ has local maxima at $x=-1$ and $f\left(x\right)$ has local minima at $x=0$

$\therefore f^{\prime \prime }\left(x\right)=\lambda x\Rightarrow f^{\prime }\left(x\right)=\lambda \frac{x^2}{2}+c$

As $f^{\prime }(-1)=0\Rightarrow \frac{\lambda }{2}+c=0\Rightarrow \lambda =-2c$ ...(1)

again, integrating both sides, we get

$f\left(x\right)=\lambda \frac{x^3}{6}+cx+d$ ...(2)

$f\left(2\right)=\lambda \left(\frac{8}{6}\right)+2c+d=18$ and $f\left(1\right)=\frac{\lambda }{6}+c+d=-1$ ...(3)

Using (1),(2) and (3), we get

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

$\Rightarrow f^{\prime }\left(x\right)=\frac{1}{4}(57x^3-57)=\frac{57}{4}(x-1)(x+1)$

using number line rule

$\therefore f\left(x\right)$ is increasing for $[1,2\sqrt{5}]$ and $f\left(x\right)$ has local maximum at $x=-1$ and $f\left(x\right)$ has local minimum at $x=1$

also, $f\left(0\right)=\frac{34}{4}$

Hence option (B) and (C) are correct