Question

$f\left(x\right)$ is a polynomial of the third degree which has a local maximum at $x=-1.$ If $f\left(1\right)=-1,f\left(2\right)=18$ and $f^{\prime }\left(x\right)$ has a local minimum at x=0 then

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

Answer 1

Answer: (B), (C)

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

Let $f\left(x\right)=a{x}^3+b{x}^2+cx+d$
Now
$f^{\prime }(x{)}_{x=-1}=0$
$3a{x}^2+2bx+c=0$ at x=-1.
$3a-2b+c=0$ ...(i)
Also
$f^{\prime \prime }(x{)}_{x=0}=0$
$6ax+2b=0$ at $x=0$
$b=0$, $\therefore f\left(1\right)=-1$
$a+b+c+d=-1$ ...(ii)
Since b=0, we get
$a+c+d=-1$ and from i
$3a+c=0$
$d-2a=-1$
Now
$f\left(2\right)=18$
$8a+2c+d=18$
$c=-3a$ and $d=2a-1$
Thus
$8a-6a+2a-1=18$
$4a=19$
$\therefore a=\frac{19}{4}$
$c=\frac{-57}{4}$
$d=\frac{34}{4}$
Hence
$f\left(x\right)=\frac{1}{4}[19x^3-57x+34]$
$f^{\prime }\left(x\right)=\frac{1}{4}[57x^2-57]$
Hence
$f^{\prime }(x{)}_{x=1}=0$
and
$f^{\prime \prime }\left(x\right)=\frac{1}{4}\left[114.x\right]$
$f"\left(x\right)>0$ at $x=1$.
Hence it has a local minimum at $x=1$.
Now $f^{\prime }\left(x\right)>0$ implies
$\frac{1}{4}[57x^2-57]>0$
$x^2-1>0$
$|x|>1$
$xϵ(-\infty ,-1]\cup [1,\infty )$
Hence $f\left(x\right)$ is increasing in the above interval.