Question

$f\left(x\right)$ is cubic polynomial with $f\left(2\right)=18$ and $f\left(1\right)=-1.$ Also $f\left(x\right)$ has local maxima at $x=-1$ and $f\text{'}\left(x\right)$ has local minima at $x=0$, then

• A. the distance between $(-1,2)$ and $(a,f(a\left)\right)$, where $x=a$ is the point of local minima 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 $f\left(0\right)=15$

Answer 1

Answer: (B), (C)

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

Let $f\left(x\right)=a{x}^3+b{x}^2+cx+d$
$f\left(2\right)=18\Rightarrow 8a+4b+2c+d=18\cdots \left(1\right)$

$f\left(1\right)=-1\Rightarrow a+b+c+d=-1\cdots \left(2\right)$

$f\left(x\right)$ has local maxima at $x=-1$,
$f^{\prime }(-1)=0\Rightarrow 3a-2b+c=0\cdots \left(3\right)$

$f^{\prime }\left(x\right)$ has local minima at $x=0$,
$f^{\prime \prime }\left(0\right)=0\Rightarrow b=0\cdots \left(4\right)$

Solving $\left(1\right),\left(2\right),\left(3\right)$ and $\left(4\right)$, we get
$f\left(x\right)=\frac{1}{4}(19x^3-57x+34)$
or $f\left(0\right)=\frac{17}{2}$

$f^{\prime }\left(x\right)=\frac{57}{4}(x^2-1)>0\forall x>1$
$f^{\prime }\left(x\right)=0\Rightarrow x=1,-1$
$f^{\prime \prime }(-1)<0,f^{\prime \prime }\left(1\right)>0$
Thus, $x=-1$ is a point of local maximum and $x=1$ is a point of local minimum.

So $a=1$,
$(a,f(a\left)\right)=(1,f(1\left)\right)=(1,-1)$
The distance between $(-1,2)$ and $(1,f(1\left)\right),$ i.e., $(1,-1)$ is
$=\sqrt{2^2+3^2}=\sqrt{13}\ne 2\sqrt{5}$