Let f: R rightarrow R be defined as f(x) = be | vedclass

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

Question

(B) To check for continuity,we examine the points $x = 0, 1, 2$.At $x = 0$:$f(0^-) = \lim_{x 	o 0^-} [e^x] = 0$ (since $e^x < 1$ for $x < 0$).$f(0^+)

Vedclass · Accepted Answer

If $f$ is discontinuous at exactly one point,then $a + b + c = 1$.

Vedclass · Answer

(B) To check for continuity,we examine the points $x = 0, 1, 2$.At $x = 0$:$f(0^-) = \lim_{x 	o 0^-} [e^x] = 0$ (since $e^x $f(0^+) = a e^0 + [0 - 1] = a - 1$.For continuity at $x = 0$,$a - 1 = 0 \implies a = 1$.At $x = 1$:$f(1^-) = a e^1 + [1 - 1] = a e + 0 = a e$.$f(1^+) = b + [\sin(\pi)] = b + 0 = b$.Since $a = 1$,$f(1^-) = e \approx 2.718$ and $f(1^+) = b$. Since $e$ is not an integer,$f$ is discontinuous at $x = 1$ for any $b$.At $x = 2$:$f(2^-) = b + [\sin(2\pi)] = b + 0 = b$.$f(2^+) = [e^{-2}] - c = 0 - c = -c$.For continuity at $x = 2$,$b = -c \implies b + c = 0$.Thus,$f$ is discontinuous at $x = 1$ regardless of $a, b, c$. If we set $a = 1$ and $b + c = 0$,$f$ is discontinuous only at $x = 1$. In this case,$a + b + c = 1 + (b + c) = 1 + 0 = 1$.

Similar Questions

The function $f(x) = \frac{\log(1+ax) - \log(1-bx)}{x}$ is not defined at $x=0$. The value which should be assigned to $f$ at $x=0$ so that it is continuous at $x=0$ is

Let $a$ be a positive real number. If a real valued function $f(x) = \begin{cases} \frac{6^x-3^x-2^x+1}{1-\cos \left(\frac{x}{a}\right)} & \text{if } x \neq 0 \\ \log 3 \log 4 & \text{if } x=0 \end{cases}$ is continuous at $x=0$,then $a=$

On the interval $I = [-2, 2]$,the function $f(x) = \begin{cases} (x + 1) e^{-\left[ \frac{1}{|x|} + \frac{1}{x} \right]} & x \neq 0 \\ 0 & x = 0 \end{cases}$ is given. Which one of the following does not hold good?

The value of $k$,for which the function $f(x) = \begin{cases} (\frac{4}{5})^{\frac{\tan 4x}{\tan 5x}}, & 0 < x < \frac{\pi}{2} \\ k + \frac{2}{5}, & x = \frac{\pi}{2} \end{cases}$ is continuous at $x = \frac{\pi}{2}$,is:

If the function $f(x) = \frac{4\sqrt{2}(\sin 3x + \sin x)}{2 \sin 2x \sin \frac{3x}{2} + \cos \frac{5x}{2} - \cos \frac{3x}{2}}$ for $x \neq \frac{\pi}{2}$ is continuous at $x = \frac{\pi}{2}$, then the value of $f(\frac{\pi}{2})$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module