JEE MAIN - Mathematics (2022 - 25th July Morning Shift - No. 20)
Let $$f(x)=\left\{\begin{array}{l}\left|4 x^{2}-8 x+5\right|, \text { if } 8 x^{2}-6 x+1 \geqslant 0 \\ {\left[4 x^{2}-8 x+5\right], \text { if } 8 x^{2}-6 x+1<0,}\end{array}\right.$$ 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 ___________.
Answer
3
Explanation
$f(x)= \begin{cases}\left|4 x^{2}-8 x+5\right|, & \text { if } 8 x^{2}-6 x+1 \geq 0 \\ {\left[4 x^{2}-8 x+5\right],} & \text { if } 8 x^{2}-6 x+1<0\end{cases}$
$$ = \begin{cases}4 x^{2}-8 x+5, & \text { if } x \in\left[-\infty, \frac{1}{4}\right] \cup\left[\frac{1}{2}, \infty\right) \\ {\left[4 x^{2}-8 x+5\right]} & \text { if } x \in\left(\frac{1}{4}, \frac{1}{2}\right)\end{cases} $$
$$f(x)=\left\{\begin{array}{cc} 4 x^2-8 x+5 & \text { if } x \in\left(-\infty, \frac{1}{4}\right] \cup\left[\frac{1}{2}, \infty\right) \\ 3 & x \in\left(\frac{1}{4}, \frac{2-\sqrt{2}}{2}\right) \\ 2 & x \in\left[\frac{2-\sqrt{2}}{2}, \frac{1}{2}\right) \end{array}\right.$$
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$\therefore \quad$ Non-diff at $x=\frac{1}{4}, \frac{2-\sqrt{2}}{2}, \frac{1}{2}$
$$ = \begin{cases}4 x^{2}-8 x+5, & \text { if } x \in\left[-\infty, \frac{1}{4}\right] \cup\left[\frac{1}{2}, \infty\right) \\ {\left[4 x^{2}-8 x+5\right]} & \text { if } x \in\left(\frac{1}{4}, \frac{1}{2}\right)\end{cases} $$
$$f(x)=\left\{\begin{array}{cc} 4 x^2-8 x+5 & \text { if } x \in\left(-\infty, \frac{1}{4}\right] \cup\left[\frac{1}{2}, \infty\right) \\ 3 & x \in\left(\frac{1}{4}, \frac{2-\sqrt{2}}{2}\right) \\ 2 & x \in\left[\frac{2-\sqrt{2}}{2}, \frac{1}{2}\right) \end{array}\right.$$
$\therefore \quad$ Non-diff at $x=\frac{1}{4}, \frac{2-\sqrt{2}}{2}, \frac{1}{2}$
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