JEE Main 2024 (Online) 30th January Morning ShiftQuestion 1
Let $g:\mathbf{R}\to\mathbf{R}$ be a non constant twice differentiable function such that $g'\left(\frac{1}{2}\right)=g'\left(\frac{3}{2}\right)$. If a real valued function $f$ is defined as $f(x)=\frac{1}{2}\left[g(x)+g(2-x)\right]$, then
Differentiation and its Application
Solution
Concept. $f'$ vanishes at $\frac12$, $1$, $\frac32$; Rolle's theorem gives two zeros of $f''$. Formulas used. (1) $f'(x)=\frac12\left[g'(x)-g'(2-x)\right]$(2) Rolle: between two zeros of $f'$ there is a zero of $f''$Step 1 - Zeros of $f'$. By (1): $$f'\left(\tfrac12\right)=\tfrac12\left[g'\left(\tfrac12\right)-g'\left(\tfrac32\right)\right]=0,\quad f'(1)=0,\quad f'\left(\tfrac32\right)=0.$$ Step 2 - Rolle. By (2), $f''=0$ somewhere in $\left(\frac12,1\right)$ and somewhere in $\left(1,\frac32\right)$: at least two points in $(0,2)$. Step 3 - (D). $f'\left(\frac32\right)+f'\left(\frac12\right)=0\ne1$. Answer: (A)