Question

If $f\left(0\right)=f\left(1\right)=f\left(2\right)=0$ & function $f\left(x\right)$ is twice differentiable in $(0,2)$ and continuous in $[0,2]$. Then which of the following is/are definitely true

• A. $f^{\prime \prime }\left(c\right)=0;\forall cϵ(0,2)$
• B. $f^{\prime }\left(c\right)=0;$ for atleast two $cϵ$ $(0,2)$
• C. $f^{\prime }\left(c\right)=0$; for exactly one $cϵ$ $(0,2)$
• D. $f^{\prime \prime }\left(c\right)=0;$ for atleast one $c$ $ϵ$ $(0,2)$

Answer 1

Answer: (B), (D)

The correct options are
B $f^{\prime }\left(c\right)=0;$ for atleast two $cϵ$ $(0,2)$
D $f^{\prime \prime }\left(c\right)=0;$ for atleast one $c$ $ϵ$ $(0,2)$

It is given that
$f\left(0\right)=f\left(1\right)=f\left(2\right)$
Now considering $f\left(0\right)=f\left(1\right)$.
Hence $f\left(x\right)$ is continuous on the interval $[0,1]$ and differentiable on the interval $(0,1).$
Also $f\left(0\right)=f\left(1\right)$.
Hence by applying Rolle's Theorem
$f^{\prime }\left(c_1\right)=0$ where $0<c_1<1$
Similarly
$f\left(1\right)=f\left(2\right)$.
Applying Rolle's theorem we get
$f^{\prime }\left(c_2\right)=0$ where $1<c_2<2$.
Hence $f^{\prime }\left(x\right)$ has atleast two zeros in the interval $[0,2].$