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

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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

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(c) : Let $h(x)=f(x)-x$ be any function which is continuous and twice differentiable\ $h(0)=f(0)-0=0-0=0$\ $h(2)=f(2)-2=2-2=0$\ $\Rightarrow \quad h(0)=h(2)$\ $	herefore \quad$ There exists atleast one $c \in(0,2)$ such that $h^{\prime}(c)=0$\ $\Rightarrow f^{\prime}(c)-1=0 \Rightarrow f^{\prime}(c)=1$\ $\Rightarrow \quad f^{\prime}(x)=1$ for some $x \in(0,2)$\ Again, $f^{\prime \prime}(c)-0=0 \Rightarrow f^{\prime \prime}(c)=0$\ $	herefore \quad f^{\prime \prime}(x)=0$ for some $x \in(0,2)$\