Question

Consider a function $f\left(x\right)$ on real line defined such that $f^{\prime }\left(x\right)$ & $f^{\prime \prime }\left(x\right)$ exists for all $x$ and that $f\left(0\right)=0,f\left(1\right)=2,f\left(2\right)=1$, and $f\left(3\right)=-3$, then which of the following is/are correct.

• A. there exists atleast two values of $c$ in $(0,3)$ such that $f^{\prime }\left(c\right)=0$
• B. there exists atleast two values of $c$ in $(0,3)$ such that $f^{\prime }\left(c\right)=2$
• C. there exists atleast one values of $c$ in $(0,3)$ such that $f^{\prime \prime }\left(c\right)=0$
• D. there exists atleast $3$ roots of the equation $2f\left(x\right)=3$ in $(0,3)$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A there exists atleast two values of $c$ in $(0,3)$ such that $f^{\prime }\left(c\right)=0$
B there exists atleast two values of $c$ in $(0,3)$ such that $f^{\prime }\left(c\right)=2$
C there exists atleast one values of $c$ in $(0,3)$ such that $f^{\prime \prime }\left(c\right)=0$
D there exists atleast $3$ roots of the equation $2f\left(x\right)=3$ in $(0,3)$

Let $f\left(x\right)=a{x}^2+bx+c$
Now $f\left(0\right)=0$ implies $c=0$.
Now$f\left(1\right)=2$Or $a+b=2$ ...(i)
And
$f\left(2\right)=1$Or $4a+2b=1$
Or
$2a+b=\frac{1}{2}$
Hence $a=\frac{-3}{2}$ and $b=\frac{7}{2}$.
Thus $f\left(x\right)=\frac{1}{2}[7x-3x^2]$
Thus
$f\left(3\right)=-3$.
Now
$f^{\prime }\left(x\right)=\frac{1}{2}[7-6x]$$=0$
$\Rightarrow x=\frac{7}{6}$.
Which lies between $(0,3)$.
$f^{\prime }\left(x\right)=2$
$\Rightarrow \frac{1}{2}[7-6x]=2$
$4=7-6x$
$-3=-6x$
$x=\frac{1}{2}$.
Which also lies between $(0,3)$.