Question

Let f be any continuously differentiable function on $[a,b]$ and twice differentiable on $(a,b)$ such that $f\left(a\right)=f^{\prime }\left(a\right)=0$ and $f\left(b\right)=0$. Then,

• A. $f^{\prime \prime }\left(a\right)=0$
• B. $f^{\prime }\left(x\right)=0$ for some $x\in (a,b)$
• C. $f^{\prime \prime }\left(x\right)\ne 0$ for some $x\in (a,b)$
• D. $f^{\prime \prime \prime }\left(x\right)=0$ for some $x\in (a,b)$

Answer 1

Answer: (B), (C)

The correct options are
B $f^{\prime }\left(x\right)=0$ for some $x\in (a,b)$
C $f^{\prime \prime }\left(x\right)\ne 0$ for some $x\in (a,b)$

We have, $f$ is continuous and differentiable function on $[a,b]$.

Also, $f\left(a\right)f\left(k\right)=0$

By Rolle's theorem, there exists $c\in (a,b)$ such that $f^{\prime }\left(c\right)=0$

Thus, there exists $x\in (a,b)$ such that $f^{\prime }\left(x\right)=0$

Let at $x=c\in (a,b),f^{\prime }\left(c\right)=0$

Now, $f$ is continuously differentiable on $[a,b].$

$\Rightarrow$ $f^{\prime }$ is continuous on $[a,b].$

Also, $f$ is twice differentiable on $(a,b).$

$\therefore f^{\prime }$ is differentiable on $(a,b).$

and $f^{\prime }\left(a\right)=0=f^{\prime }\left(c\right)$

By Rolle's theorem, there exists $k\in (a,c)$ such that $f"\left(k\right)=0$

Thus, there exists $x\in (a,c)$ such that $f^{\prime \prime }\left(x\right)=0$.

So, there exists $x\in (a,b)$ such that $f^{\prime \prime }\left(x\right)=0$.

Let us consider, $f\left(x\right)=(x-a{)}^2(x-b)$,

where $f\left(a\right)=f\left(b\right)=f^{\prime }\left(a\right)=0$ but

$f^{\prime \prime }\left(a\right)\ne 0$ and $f^{\prime \prime }\left(x\right)\ne 0$ for any $x\in (a,b)$