Let f:[-1,2] rightarrow mathbb{R} be given by | vedclass

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

Question

(B) The function is $f(x) = 2x^2 + x + [x^2] - [x]$. The points of discontinuity for $[x^2]$ in $[-1, 2]$ are where $x^2 \in \{0, 1, 2, 3, 4\}$,i.e.,$

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) The function is $f(x) = 2x^2 + x + [x^2] - [x]$. The points of discontinuity for $[x^2]$ in $[-1, 2]$ are where $x^2 \in \{0, 1, 2, 3, 4\}$,i.e.,$x \in \{0, 1, \sqrt{2}, \sqrt{3}, 2, -1\}$. The points of discontinuity for $[x]$ are $x \in \{0, 1, 2\}$. Thus,the candidate points for discontinuity are $\{-1, 0, 1, \sqrt{2}, \sqrt{3}, 2\}$.$1$. At $x = -1$: $f(-1) = 2(1) - 1 + [1] - [-1] = 2 - 1 + 1 + 1 = 3$. $\lim_{x 	o -1^+} f(x) = 2(1) - 1 + [1^+] - [-1^+] = 2 - 1 + 1 - (-1) = 3$. Since $f(-1) = \lim_{x 	o -1^+} f(x)$,$f$ is continuous at $x = -1$.$2$. At $x = 0$: $f(0) = 0$. $\lim_{x 	o 0^-} f(x) = 2(0) + 0 + [0^-] - [-1] = 0 + (-1) + 1 = 0$. $\lim_{x 	o 0^+} f(x) = 2(0) + 0 + [0^+] - [0] = 0 + 0 - 0 = 0$. Thus,$f$ is continuous at $x = 0$.$3$. At $x = 1$: $f(1) = 2 + 1 + 1 - 1 = 3$. $\lim_{x 	o 1^-} f(x) = 2 + 1 + 0 - 0 = 3$. $\lim_{x 	o 1^+} f(x) = 2 + 1 + 1 - 1 = 3$. Thus,$f$ is continuous at $x = 1$.$4$. At $x = \sqrt{2}$: $f(\sqrt{2}) = 2(2) + \sqrt{2} + 2 - 1 = 5 + \sqrt{2}$. $\lim_{x 	o \sqrt{2}^-} f(x) = 4 + \sqrt{2} + 1 - 1 = 4 + \sqrt{2}$. Since $5 + \sqrt{2} 
eq 4 + \sqrt{2}$,$f$ is discontinuous at $x = \sqrt{2}$.$5$. At $x = \sqrt{3}$: $f(\sqrt{3}) = 2(3) + \sqrt{3} + 3 - 1 = 8 + \sqrt{3}$. $\lim_{x 	o \sqrt{3}^-} f(x) = 6 + \sqrt{3} + 2 - 1 = 7 + \sqrt{3}$. Since $8 + \sqrt{3} 
eq 7 + \sqrt{3}$,$f$ is discontinuous at $x = \sqrt{3}$.$6$. At $x = 2$: $f(2) = 2(4) + 2 + 4 - 2 = 12$. $\lim_{x 	o 2^-} f(x) = 8 + 2 + 3 - 1 = 12$. Thus,$f$ is continuous at $x = 2$.The points of discontinuity are $\{\sqrt{2}, \sqrt{3}\}$. The number of points is $2$. (Note: Given the options,if the question implies checking all integer and non-integer points,the count is $2$. If the options provided are fixed,please re-verify the function definition).

Let $f:[-1,2] \rightarrow \mathbb{R}$ be given by $f(x)=2x^2+x+[x^2]-[x]$,where $[t]$ denotes the greatest integer less than or equal to $t$. The number of points where $f$ is not continuous is:

Similar Questions

Let $f:R \rightarrow R$ be defined as $f(x) = \begin{cases} 0, & x \text{ is irrational} \\ \sin |x|, & x \text{ is rational} \end{cases}$. Then,which of the following is true?

If $f(x) = \begin{cases} 1 + x^2, & \text{when } 0 \le x \le 1 \\ 1 - x, & \text{when } x > 1 \end{cases}$,then

Discuss the continuity of the function $f$ given by $f(x) = |x|$ at $x = 0$.

Is the function defined by $f(x) = x^{2} - \sin x + 5$ continuous at $x = \pi$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module