Let f be any continuous function on [0, 2] and twice differentiable on (0, 2). If f(0) = 0, f(1) = 1 and f(2) = 2, then

$\mathrm{f}\left(0\right)=0, \mathrm{f}\left(1\right)=1\text{ and }\mathrm{f}\left(2\right)=2$

Let $\mathrm{h}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{x}\right)-\mathrm{x}$

Clearly h (x) is continuous and twice differentiable on (0, 2)

Also,  $\mathrm{h}\left(0\right)=\mathrm{h}\left(1\right)=\mathrm{h}\left(2\right)=0$

$\therefore \mathrm{h}\left(\mathrm{x}\right)$ satisfies all the condition of Rolle’s theorem.

$\therefore$there exist ${\mathrm{C}}_1\in (0,1)\text{ such that }{\mathrm{h}}^{\mathrm{\prime }}\left({\mathrm{c}}_1\right)=0$

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{f}}^{\mathrm{\prime }}\left({\mathrm{c}}_1\right)-1=0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{f}}^{\mathrm{\prime }}\left({\mathrm{c}}_1\right)=1\end{array}$

also there exist ${\mathrm{c}}_2\in (1,2)\text{ such that }{\mathrm{h}}^{\mathrm{\prime }}\left({\mathrm{c}}_2\right)=0$

$\Rightarrow {\mathrm{f}}^{\mathrm{\prime }}\left({\mathrm{c}}_2\right)=1$

Now, using Rolle’s theorem on $\left[{\mathrm{C}}_1,{\mathrm{C}}_2\right]\text{ for }{\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)$

We have ${\mathrm{f}}^{\prime \prime }\left(\mathrm{c}\right)=0,\mathrm{c}\in \left({\mathrm{c}}_1,{\mathrm{c}}_2\right)$

Hence, ${\mathrm{f}}^{\prime \prime }\left(\mathrm{x}\right)=0\text{ for some }\mathrm{x}\in (0,2)$