JEE Main 2022 (Online) 25th July Morning ShiftQuestion 1
Let $f(x)=\begin{cases}|4x^2-8x+5|, & \text{if } 8x^2-6x+1\ge0\\ [4x^2-8x+5], & \text{if } 8x^2-6x+1<0\end{cases}$ where $[\alpha]$ denotes the greatest integer less than or equal to $\alpha$. Then the number of points in $\mathbf{R}$ where $f$ is not differentiable is ________.
Differentiation and its Application
Answer$3$
Solution
Concept. $g=4x^2-8x+5>0$, so the modulus does nothing. The floor is used only on $\left(\frac14,\frac12\right)$, where $8x^2-6x+1<0$. Formulas used. (1) $8x^2-6x+1=(4x-1)(2x-1)$, $\ g=4(x-1)^2+1$Step 1 - Pieces. By (1), $f=g$ for $x\le\frac14$ or $x\ge\frac12$, and $f=[g]$ on $\left(\frac14,\frac12\right)$, where $g$ decreases from $3.25$ to $2$. Step 2 - $x=\frac14$. $f\left(\frac14\right)=3.25$, but just to the right $f=3$. Jump: not differentiable. Step 3 - Inside. $g=3$ at $x=1-\frac{1}{\sqrt2}\approx0.29$; $[g]$ jumps from $3$ to $2$ there. Not differentiable. Step 4 - $x=\frac12$. $f\left(\frac12\right)=2$ and $f=2$ just to the left: continuous. Left slope $0$, right slope $g'\left(\frac12\right)=-4$. Not differentiable. Step 5 - Count. $3$ points. Answer: 3