f(x) = {[(√(1+px)-√(1-px))/x,-1≤x​<0][(2x+1)/(x-2) ,0≤x≤1 is continuous in the interval [–1, 1], then p is equal to : A. –1 B. -(1/2)

Question

\(f(x) = \begin{cases} \frac{\sqrt{1+px}\,-\sqrt{1-px}}{x} &, \quad-1≤ x <{0}\\ \frac{2+1}{x-2} &, \quad \text0≤ x≤{ 1} \end{cases} \) is continuous in the interval [–1, 1], then p is equal to :

A. –1 

B. \(-\frac{1}{2}\)

C. \(\frac{1}{2}\)

D. 1

Answer 1

Option : (B)

(i) A function f(x) is said to be continuous at a point x = a of its domain, if 

 \(\lim\limits_{x \to a}f(x)\) = f(a)

 \(\lim\limits_{x \to a^+}f(a+h)\) = \(\lim\limits_{x \to a^-}f(a-h)\) = f(a)

Given :-

\(f(x) = \begin{cases} \frac{\sqrt{1+px}\,-\sqrt{1-px}}{x} &, \quad-1≤ x <{0}\\ \frac{2+1}{x-2} &, \quad \text0≤ x≤{ 1} \end{cases} \)