Let f (x) = \ cases |4 x ^ 2 - 8x + 5|, & if8 x ^ 2 - 6x + 1 ≥ 0 [4 x ^ 2 - 8x + 5], & if8 x ^ 2 - 6x + 1 < 0 cases ), where [α] denotes the greatest integer less than or equal to α . Then the number of points in R

Correct answer: 0 Given, f(x) = \ cases |4 x ^ 2 - 8x + 5|, & if8 x ^ 2 - 6x + 1 ≥ 0 [4 x ^ 2 - 8x + 5], & if8 x ^ 2 - 6x + 1 = \ cases 4 x ^ 2 - 8x + 5, & ifx ∈ [- ∞ , (1)/(4)] [(1)/(2), ∞) [4 x ^ 2 - 8x + 5] & ifx ∈ ((1)/(4), (1)/(2)) cases ) ⇒ f(x) = \ cases 4 x ^ 2 - 8x + 5 & ifx ∈ (- ∞ , (1)/(4)] [(1)/(2), ∞) 3 & x ∈ ((1)/(4), 2 - 2 2 ) 2 & x ∈ [ 2 - 2 2 , (1)/(2)) cases ) Now plotting the graph of f(x) we get, We can see f (x) is not differentiable at (1)/(4), x _ 1 , (1)/(2) (where x _ 1 = 2 - 2 2 )

JEE Main 202225 JulyMorning Shift

Exam: JEE Main
Year: 2022
Date: 25 July
Shift: Morning Shift
Subject: Mathematics
Chapter: Continuity and Differentiability

JEE Main 2022 · 25 July · Morning Shift · Mathematics · Continuity and Differentiability

Let f (x) = \ cases |4 x ^ 2 - 8x + 5|, & if8 x ^ 2 - 6x + 1 ≥ 0 [4...

Let

f

\left(x\right) = \left\{

\begin{cases}\left|4{x}^{2} - 8x + 5\right|, & if8{x}^{2} - 6x + 1 \ge 0 \\ \left[4{x}^{2} - 8x + 5\right], & if8{x}^{2} - 6x + 1 < 0\end{cases}

\right), where

\left[\alpha\right]

denotes the greatest integer less than or equal to

\alpha

. Then the number of points in

R

where

f

is not differentiable is _____ . $Let f (x) = \ cases |4 x ^ 2 - 8x + 5|, & if8 x ^ 2 - 6x + 1 ≥ 0 [4 x ^ 2 - 8x + 5], & if8 x ^ 2 - 6x + 1 < 0 cases ),$

Correct answer

Step-by-step solution

Given,
$f\left(x\right) = \left\{$ \begin{cases}\left|4{x}^{2} - 8x + 5\right|, & if8{x}^{2} - 6x + 1 \ge 0 \\ \left[4{x}^{2} - 8x + 5\right], & if8{x}^{2} - 6x + 1 < 0\end{cases} $\right)$ = \left\{ $\begin{cases}4{x}^{2} - 8x + 5, & ifx \in \left[- \infty , \dfrac{1}{4}\right] \cup \left[\dfrac{1}{2}, \infty\right) \\ \left[4{x}^{2} - 8x + 5\right] & ifx \in \left(\dfrac{1}{4}, \dfrac{1}{2}\right)\end{cases}$ \right)
$⇒ f\left(x\right) = \left\{$ \begin{cases}4{x}^{2} - 8x + 5 & ifx \in \left(- \infty , \dfrac{1}{4}\right] \cup \left[\dfrac{1}{2}, \infty\right) \\ 3 & x \in \left(\dfrac{1}{4}, \dfrac{2 - \sqrt{2}}{2}\right) \\ 2 & x \in \left[\dfrac{2 - \sqrt{2}}{2}, \dfrac{1}{2}\right)\end{cases} $\right)Now plotting the graph of$ f\left(x\right) $we get,<img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/64bf9c97-51c3-4b6e-98e0-32d420433319-image.png" style="width: 390px; height: 249px;" />We can see$ f $\left(x\right)$ is not differentiable at $\dfrac{1}{4}, {x}_{1}, \dfrac{1}{2}$ (where ${x}_{1} = \dfrac{2 - \sqrt{2}}{2}$ )

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