If the function $f(x) = \begin{cases} [\tan(\frac{\pi}{4} + x)]^{\frac{1}{x}}, & x \neq 0 \\ K, & x = 0 \end{cases}$ is continuous at $x = 0$,then $K = ?$

  • A
    $e$
  • B
    $e^{-1}$
  • C
    $e^{2}$
  • D
    $e^{-2}$

Explore More

Similar Questions

Given $f(x) = b ([x]^2 + [x]) + 1$ for $x \geq -1$ and $f(x) = \sin(\pi(x+a))$ for $x < -1$,where $[x]$ denotes the greatest integer function,for what values of $a$ and $b$ is the function continuous at $x = -1$?

If $f(x) = \begin{cases} mx^2 + n, & x < 0 \\ nx + m, & 0 \leq x \leq 1 \\ nx^3 + m, & x > 1 \end{cases}$
For what integers $m$ and $n$ do $\lim_{x \to 0} f(x)$ and $\lim_{x \to 1} f(x)$ exist?

If $[x]$ denotes the greatest integer not exceeding $x$ and if the function $f$ defined by $f(x)= \begin{cases} \frac{a+2 \cos x}{x^2} & , x < 0 \\ b \tan \frac{\pi}{[x+4]} & , x \geq 0 \end{cases}$ is continuous at $x=0$, then the ordered pair $(a, b)$ is equal to

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 $f: R \rightarrow R$ is a function defined by $f(x)=[x-1] \cos \left(\frac{2 x-1}{2}\right) \pi,$ where $[.]$ denotes the greatest integer function,then $f$ is

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