The number of points of discontinuity of the | vedclass

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

Question

(A) Let $g(x) = [\frac{x^2}{2}]$ and $h(x) = [\sqrt{x}]$. The function $f(x) = g(x) - h(x)$ is discontinuous at points where $g(x)$ or $h(x)$ is disco

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(A) Let $g(x) = [\frac{x^2}{2}]$ and $h(x) = [\sqrt{x}]$. The function $f(x) = g(x) - h(x)$ is discontinuous at points where $g(x)$ or $h(x)$ is discontinuous, provided the jumps do not cancel out.$g(x) = [\frac{x^2}{2}]$ is discontinuous when $\frac{x^2}{2} = k$ for $k \in \{1, 2, 3, 4, 5, 6, 7, 8\}$, which gives $x^2 \in \{2, 4, 6, 8, 10, 12, 14, 16\}$. Thus, $x \in \{\sqrt{2}, 2, \sqrt{6}, \sqrt{8}, \sqrt{10}, \sqrt{12}, \sqrt{14}, 4\}$.$h(x) = [\sqrt{x}]$ is discontinuous when $\sqrt{x} = k$ for $k \in \{1, 2\}$, which gives $x \in \{1, 4\}$.The set of all potential points of discontinuity is $S = \{1, \sqrt{2}, 2, \sqrt{6}, \sqrt{8}, 3, \sqrt{10}, \sqrt{12}, \sqrt{14}, 4\}$.However, at $x=4$, $g(x)$ jumps from $7$ to $8$ and $h(x)$ jumps from $1$ to $2$. Since $f(4) = [8] - [2] = 6$, and $\lim_{x 	o 4^-} f(x) = 7 - 1 = 6$, $f(x)$ is continuous at $x=4$.Removing $x=4$ from the set, we have $9$ points. Checking $x=2$, $g(x)$ jumps from $1$ to $2$ and $h(x)$ is continuous. Thus $f(x)$ is discontinuous at $x=2$. All other points are distinct points of discontinuity. Total points = $8$.

The number of points of discontinuity of the function $f(x) = [\frac{x^2}{2}] - [\sqrt{x}]$ for $x \in [0, 4]$, where $[\cdot]$ denotes the greatest integer function, is:

Similar Questions

The number of points where the function $f(x) = \begin{cases} |2x^2 - 3x - 7| & \text{if } x \leq -1 \\ [4x^2 - 1] & \text{if } -1 < x < 1 \\ |x+1| + |x-2| & \text{if } x \geq 1 \end{cases}$ is discontinuous,where $[t]$ denotes the greatest integer $\leq t$,is:

If $f(x) = \begin{cases} \frac{\sin 5x \tan kx}{x^2} & , x \neq 0 \\ 1 & , x = 0 \end{cases}$ is continuous at $x = 0$,then the value of $k$ is . . . . . . . $(\because k \neq 0)$

If $f(x) = [x]$ for $x \in (-1, 2)$,then $f$ is discontinuous at (where $[x]$ represents the floor function).

If $f(x) = \begin{cases} \frac{8^x - 4^x - 2^x + 1}{x^2} & , \text{if } x > 0 \\ e^x \sin x + kx + \lambda \log 4 & , \text{if } x \le 0 \end{cases}$ is continuous at $x = 0$,then the value of $500 e^\lambda$ is

The function $f(x) = \frac{2x^2 + 7}{x^3 + 3x^2 - x - 3}$ is discontinuous for

Vedclass Test Series

Exam Paper Generator

Online Exam Module