JEE Main 2022 (Online) 28th July Morning ShiftQuestion 1
Let $f:[0,1]\to\mathbf{R}$ be a twice differentiable function in $(0,1)$ such that $f(0)=3$ and $f(1)=5$. If the line $y=2x+3$ intersects the graph of $f$ at only two distinct points in $(0,1)$, then the least number of points $x\in(0,1)$, at which $f''(x)=0$, is ________.
Differentiation and its Application
Answer$2$
Solution
Concept. $g(x)=f(x)-(2x+3)$ vanishes at both end points and at two points inside; apply Rolle twice. $g''=f''$. Formulas used. (1) Rolle: between two zeros of $g$ there is a zero of $g'$Step 1 - Zeros of $g$. $g(0)=3-3=0$, $g(1)=5-5=0$, and two more in $(0,1)$: four zeros in all. Step 2 - Rolle. By (1), $g'$ has at least $3$ zeros, and then $g''$ has at least $2$ zeros, all in $(0,1)$. Step 3 - Conclude. $f''=g''$, so $f''=0$ at at least $2$ points. Answer: 2