Let f be any continuous function on [0, 2] and twice differentiable...

Given $f\left(0\right) = 0, f\left(1\right) = 1 and$f\left(2\right) = 2$</p><p>Let,$h\left(x\right) = f\left(x\right) - x$</p><p>Clearly$h\left(x\right)$, will be continuous and twice differentiable on \left(0, 2\right)$.

Now, $h\left(0\right) = h\left(1\right) = h\left(2\right) = 0$

By Rolle's mean value theorem in \left[0, 1\right]$, we get</p><p>${h}^{'}\left({c}_{1}\right) = 0$</p><p>$⇒ {f}^{'}\left({c}_{1}\right) - 1 = 0$</p><p>$⇒ {f}^{'}\left({c}_{1}\right) = 1$, where${c}_{1} \in \left(0, 1\right)$</p><p>Also, in the interval \left[1, 2\right]$

${h}^{'}\left({c}_{2}\right) = 0$

$⇒ {f}^{׀}\left({c}_{2}\right) - 1 = 0$

$⇒ {f}^{'}\left({c}_{2}\right) = 1$, where ${c}_{2} \in \left(1, 2\right)$

Now, use Rolle's theorem on \left[{c}_{1}, {c}_{2}\right]$for${h}^{'}\left(x\right),

We have, ${h}^{''} ( c ) = 0$

$⇒ {f}^{''} ( c ) = 0$, where $c \in \left({c}_{1}, {c}_{2}\right)$

Hence, ${f}^{''} ( c ) = 0$ for some $x \in \left(0, 2\right).</p>$