Question

$f(x)=\frac{\sqrt{1+px}\,-\,\sqrt{1-px}}{x}, -1\leq x\leq 0=\frac{2x+1}{x-2}, 0 \leq x \leq 1$ is continuous in the interval [-1, 1], thenp equals.

• A. 2
• B. $-\frac{1}{2}$
• C. $\frac{1}{2}$
• D. 1

Answer 1

Answer: (B)

$-\frac{1}{2}$

Answer 2

$\lim_{x\to0-}f\left(x\right)= \lim _{x\to 0} \frac{\sqrt{ 1 -px} - \sqrt{1 -px}}{x}$ = $\lim _{x\to 0} \frac{\left(1 +px\right) -\left(1-px\right)}{x\left[\sqrt{1+px}+\sqrt{1-px}\right]} =\frac{2p}{1+1} =p$ $\lim _{x\to 0+} f\left(x\right) = \lim _{x\to 0} \frac{2x+1}{x-2} =\frac{0+1}{0-2} =\frac{1}{2}$ Since $f(x)$ is continuous in [-1, 1] $therefore \, f(x)$ is continuous at $x$ = 0 $\therefore$ $\lim_{x\to0-} f(x)$ exists and $f(0) \, \therefore \, p = - \frac{1}{2}$