If $f(x) = \begin{cases} \frac{\log (1 + 2ax) - \log (1 - bx)}{x}, & x \neq 0 \\ k, & x = 0 \end{cases}$ is continuous at $x = 0$,then the value of $k$ is

  • A
    $b + a$
  • B
    $b - 2a$
  • C
    $2a - b$
  • D
    $2a + b$

Explore More

Similar Questions

Let $f(x) = \begin{cases} 2 - |x^2 + 5x + 6|, & x \neq -2 \\ a^2 + 1, & x = -2 \end{cases}$. Then the range of $a$ such that $f(x)$ has a maximum at $x = -2$ is

If $f(x) = \frac{4^{x-\pi} + 4^{\pi-x} - 2}{(x-\pi)^2}$ for $x \neq \pi$ is continuous at $x = \pi$,then $f(\pi) = k$. Find the value of $k$.

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

Show that every polynomial function is continuous.

Statement $1$: $A$ function $f: R \to R$ is continuous at $x_0$ if and only if $\lim_{x \to x_0} f(x)$ exists and $\lim_{x \to x_0} f(x) = f(x_0)$.
Statement $2$: $A$ function $f: R \to R$ is discontinuous at $x_0$ if and only if $\lim_{x \to x_0} f(x)$ exists and $\lim_{x \to x_0} f(x) \neq f(x_0)$.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo