If $f(x) = \frac{1 - \sin x + \cos x}{1 + \sin x + \cos x}$ for $x \neq \pi$ is continuous at $x = \pi$,then $f(\pi) =$

  • A
    $-1$
  • B
    $2$
  • C
    $0$
  • D
    $1$

Explore More

Similar Questions

Let $a, b \in R, (a \ne 0)$. If the function $f$ defined as
$f(x) = \begin{cases} \frac{2x^2}{a}, & 0 \le x < 1 \\ a, & 1 \le x < \sqrt{2} \\ \frac{2b^2 - 4b}{x^3}, & \sqrt{2} \le x < \infty \end{cases}$
is continuous in the interval $[0, \infty)$,then an ordered pair $(a, b)$ is

If the function $f(x)$,defined below,is continuous in the interval $[0, \pi]$,then find the values of $a$ and $b$.
$f(x) = \begin{cases} x + a\sqrt{2}(\sin x), & 0 \le x < \frac{\pi}{4} \\ 2x(\cot x) + b, & \frac{\pi}{4} \le x \le \frac{\pi}{2} \\ a(\cos 2x) - b(\sin x), & \frac{\pi}{2} < x \le \pi \end{cases}$

If $f(x) = \begin{cases} \frac{\sin x}{x} + \cos x, & x \ne 0 \\ 2, & x = 0 \end{cases}$,then which of the following is true?

The function $f(x) = |x-2| + x$ is

If $f(x) = \begin{cases} x, & x \le 0 \\ 0, & x > 0 \end{cases}$ then the function $f(x)$ at $x = 0$ 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