If f(x) = begin{cases} frac{sqrt{1+px} - sqrt | vedclass

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(A) For the function $f(x)$ to be continuous in $[-1, 1]$, it must be continuous at $x = 0$.This implies that $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^+

Vedclass · Accepted Answer

$-\frac{1}{2}$

Vedclass · Answer

(A) For the function $f(x)$ to be continuous in $[-1, 1]$, it must be continuous at $x = 0$.This implies that $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^+} f(x) = f(0)$.First, calculate the right-hand limit and the value of the function at $x = 0$:$f(0) = \frac{2(0) + 1}{0 - 2} = -\frac{1}{2}$.Now, calculate the left-hand limit:$\lim_{x 	o 0^-} \frac{\sqrt{1+px} - \sqrt{1-px}}{x} = \lim_{x 	o 0^-} \frac{(\sqrt{1+px} - \sqrt{1-px})(\sqrt{1+px} + \sqrt{1-px})}{x(\sqrt{1+px} + \sqrt{1-px})}$$= \lim_{x 	o 0^-} \frac{(1+px) - (1-px)}{x(\sqrt{1+px} + \sqrt{1-px})} = \lim_{x 	o 0^-} \frac{2px}{x(\sqrt{1+px} + \sqrt{1-px})} = \lim_{x 	o 0^-} \frac{2p}{\sqrt{1+px} + \sqrt{1-px}}$$= \frac{2p}{\sqrt{1} + \sqrt{1}} = \frac{2p}{2} = p$.Equating the limits: $p = -\frac{1}{2}$.

If $f(x) = \begin{cases} \frac{\sqrt{1+px} - \sqrt{1-px}}{x}, & -1 \leq x < 0 \\ \frac{2x+1}{x-2}, & 0 \leq x \leq 1 \end{cases}$ is continuous in $[-1, 1]$, then $p = $

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