Question

$f\left(x\right)=\{\begin{array}{cc}\frac{\sqrt{1+px}-\sqrt{1-px}}{x}:& -1\le x<0\\ \frac{2x+1}{x-2}:& 0\le x\le 1\end{array}$
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

The correct option is B $-\frac{1}{2}$

$f(0+0)=\underset{h\to 0}{lim}f\left(x\right)\frac{2(0+h)+1}{(0+h)-2}=-\frac{1}{2}$
$f(0-0)=\underset{h\to 0}{lim}\frac{\sqrt{1-ph}-\sqrt{1+ph}}{0-h}$
$=\underset{x\to 0}{lim}f\left(x\right)\frac{-2ph}{-h[\sqrt{1-ph}+\sqrt{1+ph}]}=p$
f(x) is continuous at x = 0 in [-1, 1] if
$f(0+0)=f(0-0)=f\left(0\right)\Rightarrow p=-\frac{1}{2}$.