Question

Let $f:[0,2]\to R$ be continuous on $[0,2]$ and twice differentiable in $(0,2).$ If $f\left(0\right)=0,$ $f\left(1\right)=1$ and $f\left(2\right)=1,$ then

• A. $f^{\prime }\left(c\right)=\frac{1}{3}$ for at least one $c\in (0,2)$
• B. $2f^{\prime }\left(c\right)+2c=3$ for at least one $c\in (0,1)$
• C. $2f^{\prime }\left(c\right)+2c=3$ for at least one $c\in (0,2)$
• D. $f^{\prime \prime }\left(c\right)=-1$ for at least one $c\in (0,2)$

Answer 1

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

$f^{\prime \prime }\left(c\right)=-1$ for at least one $c\in (0,2)$

$\to$ By LMVT in the interval $[0,1],$
$\frac{f\left(1\right)-f\left(0\right)}{1}=f^{\prime }\left(c_1\right)=1$ for some $c_1\in (0,1)$

Similarly, $\frac{f\left(2\right)-f\left(1\right)}{1}=f^{\prime }\left(c_2\right)=0$ for some $c_2\in (1,2)$
Since $\frac{1}{3}\in (f^{\prime }\left(c_2\right),f^{\prime }\left(c_1\right))$ and $f^{\prime }$ is continuous,
$\Rightarrow f^{\prime }\left(c\right)=\frac{1}{3}$ for at least one $c\in (0,2)$ by IVT.

$\to$ Consider a function, $H\left(x\right)=2f\left(x\right)+x^2$
$H\left(0\right)=0,H\left(1\right)=3,H\left(2\right)=6$
By LMVT in $[0,1],$
$H^{\prime }\left(c\right)=2f^{\prime }\left(c\right)+2c=\frac{H\left(1\right)-H\left(0\right)}{1}=3$ for some $c\in (0,1)$

$\to$ By LMVT in $[0,2],$
$H^{\prime }\left(c\right)=2f^{\prime }\left(c\right)+2c=\frac{H\left(2\right)-H\left(0\right)}{2}=3$ for some $c\in (0,2)$

$\to$ Applying Rolle's theorem to $H^{\prime }\left(x\right)$ in $[c_1,c_2]$ where $0\le c_1<c_2\le 2$
$H^{\prime }\left(c_1\right)=2f^{\prime }\left(c_1\right)+2c_1=\frac{H\left(1\right)-H\left(0\right)}{1}=3$
$H^{\prime }\left(c_2\right)=2f^{\prime }\left(c_2\right)+2c_2=\frac{H\left(2\right)-H\left(1\right)}{2-1}=3$
$H^{\prime \prime }\left(c\right)=0$ for some $c\in (0,2)$
$\Rightarrow 2f^{\prime \prime }\left(c\right)+2=0$
$\Rightarrow f^{\prime \prime }\left(c\right)=-1$