Let $f(x)=\left\{\begin{array}{l}\left|4 x^{2}-8 x+5\right|, \text { if } 8 x^{2}-6 x+1 \geq 0 \\ \left[4 x^{2}-8 x+5\right], \text { if } 8 x^{2}-6 x+1<0\end{array}\right.$, where $[\alpha]$ denotes the greatest integer less than or equal to $\alpha$. Then the number of points in $\mathbb{R}$ where $f$ is not differentiable is ______.

JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Continuity and Differentiability

Let

f (x) = {4 x^{2} - 8 x + 5, if 8 x^{2} - 6 x + 1 \geq 0 [4 x^{2} - 8 x + 5], if 8 x^{2} - 6 x + 1 < 0

,
where

[α]

denotes the greatest integer less than or equal to

α

. Then the number of points in

R

where

f

is not differentiable is ______.

Answer: 3

First, solve

8 x^{2} - 6 x + 1 \geq 0

. The roots are

x = \frac{1}{4}

and

x = \frac{1}{2}

. So the condition holds for

x \leq \frac{1}{4}

x \geq \frac{1}{2}

.

For

x \in [\frac{1}{4}, \frac{1}{2}]

, we use the greatest integer function

[4 x^{2} - 8 x + 5]

. The expression

4 x^{2} - 8 x + 5 = 4 (x - 1)^{2} + 1

has minimum value 1 at

x = 1

. In the interval

[\frac{1}{4}, \frac{1}{2}]

, evaluate: at

x = \frac{1}{4}

: value is 1; at

x = \frac{1}{2}

: value is 2. The function

4 x^{2} - 8 x + 5

is continuous and takes values in

[1, 2)

on this interval.

For

x \in / [\frac{1}{4}, \frac{1}{2}]

, we use

∣4 x^{2} - 8 x + 5∣

. The expression

4 x^{2} - 8 x + 5

has roots where it equals zero, but checking the discriminant:

64 - 80 < 0

, so it's always positive, meaning the absolute value has no effect.

Non-differentiability occurs at:

$x = \frac{1}{4}$ (transition point)
$x = \frac{1}{2}$ (transition point)
One point inside $[\frac{1}{4}, \frac{1}{2}]$ where the greatest integer function jumps (where $4 x^{2} - 8 x + 5$ crosses an integer).

Total: 3 points.

JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Integer answer