Let f(x) = begin{cases} |4x^2 - 8x + 5|, & te | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) First,analyze the condition $8x^2 - 6x + 1 < 0$. Solving $(2x - 1)(4x - 1) < 0$,we get $x \in (1/4, 1/2)$.For $x \in (1/4, 1/2)$,$f(x) = [4x^2 - 8

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) First,analyze the condition $8x^2 - 6x + 1 For $x \in (1/4, 1/2)$,$f(x) = [4x^2 - 8x + 5]$. Let $g(x) = 4x^2 - 8x + 5 = 4(x-1)^2 + 1$. In $(1/4, 1/2)$,$g(x)$ decreases from $g(1/4) = 4(1/16) - 2 + 5 = 13/4 = 3.25$ to $g(1/2) = 4(1/4) - 4 + 5 = 2$.Thus,$f(x) = [g(x)]$ takes values $3$ for $x \in (1/4, x_1)$ where $g(x_1) = 3$,and $2$ for $x \in [x_1, 1/2)$.At $x = 1/4$,$g(1/4) = 3.25$. The function is continuous but has a sharp corner (non-differentiable).At $x = x_1$,$g(x_1) = 3$. The function jumps from $3$ to $2$,so it is discontinuous and non-differentiable.At $x = 1/2$,$g(1/2) = 2$. The function jumps from $2$ to $g(1/2) = 2$ (continuous),but the derivative changes abruptly,making it non-differentiable.Thus,the points of non-differentiability are $x = 1/4, x_1, 1/2$. The total number of points is $3$.

Similar Questions

Let $f:R \to R$ be a function defined by $f(x) = \text{Min}\{x + 1, |x| + 1\}$. Then which of the following is true?

If $f(x)=\frac{2x}{4+3|x|}, x \in R$,then $f^{\prime}(0)=$

If $f(x) = \begin{cases} \frac{x^2 \ln \cos x}{\ln(1 + x^2)}, & x \neq 0 \\ 0, & x = 0 \end{cases}$,then $f(x)$ is

Which of the following functions is differentiable at $x = 0$?

The function given by $y = ||x| - 1|$ is differentiable for all real numbers except the points

Vedclass Test Series

Exam Paper Generator

Online Exam Module