JEE Main 2021 (Online) 31st August Evening ShiftQuestion 1
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 :
Differentiation and its Application
Solution
Concept. The mean value theorem on $[0,1]$ and $[1,2]$ gives two points with $f'=1$; Rolle on $f'$ gives $f''=0$. Formulas used. (1) MVT: $f'(c)=\dfrac{f(q)-f(p)}{q-p}$ for some $c\in(p,q)$(2) Rolle: $f'(c_1)=f'(c_2)$ $\Rightarrow$ $f''(c)=0$ for some $c\in(c_1,c_2)$Step 1 - MVT. By (1), $f'(c_1)=\frac{1-0}{1}=1$ with $c_1\in(0,1)$ and $f'(c_2)=\frac{2-1}{1}=1$ with $c_2\in(1,2)$. Step 2 - Rolle. By (2), $f''(c)=0$ for some $c\in(c_1,c_2)\subset(0,2)$. Step 3 - Others. $f(x)=x$ fits the data with $f'\ne0$ and $f''=0$; other functions fit with $f''\ne0$ somewhere. So (A), (B), (D) need not hold. Answer: (C)