Let f(x) = |2x^2 + 5|x| - 3|,x in R. If m and | vedclass

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

Question

(D) Given $f(x) = |2x^2 + 5|x| - 3|$.Since $f(x)$ is a composition of continuous functions (polynomials and absolute value),it is continuous everywher

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(D) Given $f(x) = |2x^2 + 5|x| - 3|$.Since $f(x)$ is a composition of continuous functions (polynomials and absolute value),it is continuous everywhere. Thus,the number of points of discontinuity is $m = 0$.To find the points of non-differentiability,we analyze $g(x) = 2x^2 + 5|x| - 3$. For $x \ge 0$,$g(x) = 2x^2 + 5x - 3 = (2x - 1)(x + 3)$. The roots are $x = 1/2$ and $x = -3$. Since we consider $x \ge 0$,the root is $x = 1/2$.For $x Also,the function $f(x)$ involves $|x|$,which is non-differentiable at $x = 0$.Thus,$f(x)$ is non-differentiable at the roots of $2x^2 + 5|x| - 3 = 0$ (where the graph touches the x-axis) and at $x = 0$ (due to $|x|$).The points are $x = 1/2$,$x = -1/2$,and $x = 0$.So,the number of points of non-differentiability is $n = 3$.Therefore,$m + n = 0 + 3 = 3$.

Let $f(x) = |2x^2 + 5|x| - 3|$,$x \in R$. If $m$ and $n$ denote the number of points where $f$ is not continuous and not differentiable respectively,then $m + n$ is equal to:

Similar Questions

If $f(x) = \begin{cases} \frac{x - 1}{2x^2 - 7x + 5} & \text{for } x \neq 1 \\ -\frac{1}{3} & \text{for } x = 1 \end{cases}$,then $f'(1) = $

If $f(x) = \begin{cases} e^x + ax, & x < 0 \\ b(x - 1)^2, & x \ge 0 \end{cases}$ is differentiable at $x = 0$,then $(a, b)$ is

The function $y = e^{-|x|}$ is

Let $f(x)=a_0+a_1|x|+a_2|x|^2+a_3|x|^3$,where $a_0, a_1, a_2, a_3$ are real constants. Then $f(x)$ is differentiable at $x=0$ if and only if:

The domain of the derivative of the function $f(x) = \frac{x}{1+|x|}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module