The function f(x) = p[x + 1] + q[x - 1], wher | vedclass

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

Question

(B) Given $f(x) = p[x + 1] + q[x - 1]$.At $x = 1$,$f(1) = p[1 + 1] + q[1 - 1] = p[2] + q[0] = 2p + 0 = 2p$.For $f(x)$ to be continuous at $x = 1$,the

Vedclass · Accepted Answer

$p + q = 0$

Vedclass · Answer

(B) Given $f(x) = p[x + 1] + q[x - 1]$.At $x = 1$,$f(1) = p[1 + 1] + q[1 - 1] = p[2] + q[0] = 2p + 0 = 2p$.For $f(x)$ to be continuous at $x = 1$,the left-hand limit $(LHL)$ must equal the right-hand limit $(RHL)$ and the value of the function $f(1)$.$LHL$: $\lim_{h 	o 0} f(1 - h) = \lim_{h 	o 0} (p[1 - h + 1] + q[1 - h - 1]) = \lim_{h 	o 0} (p[2 - h] + q[-h]) = p(1) + q(-1) = p - q$.$RHL$: $\lim_{h 	o 0} f(1 + h) = \lim_{h 	o 0} (p[1 + h + 1] + q[1 + h - 1]) = \lim_{h 	o 0} (p[2 + h] + q[h]) = p(2) + q(0) = 2p$.Equating $LHL$ and $RHL$: $p - q = 2p$.This simplifies to $-q = p$,or $p + q = 0$.

The function $f(x) = p[x + 1] + q[x - 1],$ where $[x]$ is the greatest integer function,is continuous at $x = 1$ if:

Similar Questions

If $f(x) = \frac{1+\cos \pi x}{\pi(1-x)^2}$ for $x \neq 1$ is continuous at $x=1$,then $f(1)$ is equal to

The points at which the function $f(x) = \frac{x + 1}{x^2 + x - 12}$ is discontinuous,are

If $f(x) = \frac{1 - \sin x}{\log(1 + \pi^2 - 4\pi x + 4x^2)}$ is continuous at $x = \frac{\pi}{2}$,then $f\left(\frac{\pi}{2}\right) = $

If $f(x) = \begin{cases} x^2, & \text{when } x \le 1 \\ x + 5, & \text{when } x > 1 \end{cases}$,then

If $f(x) = \begin{cases} x^2, & x \ne 1 \\ 2, & x = 1 \end{cases}$,then which of the following is true?

Vedclass Test Series

Exam Paper Generator

Online Exam Module