\(f(x)=\begin{cases}\displaystyle\frac{\sqrt{1+px}-\sqrt{1-px}}{x} & -1 \le x \lt 0\\ \displaystyle\frac{2x+1}{x-2} & 0 \le x \le 1\end{cases}\)
is continuous in the interval [−1, 1], then p is equal to :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(-\displaystyle\frac{1}{2}\)
Each piece is continuous on its own interval (the denominator x − 2 is not zero on [0, 1]), so the only point to check is x = 0, where the formula changes. Continuity at x = 0 needs the left-hand limit to equal f(0).
Value at 0, from the second piece:
\( \displaystyle f(0) = \frac{0 + 1}{0 - 2} = -\frac{1}{2} \)
Left-hand limit: multiply the numerator and the denominator by \(\sqrt{1+px}+\sqrt{1-px}\):
\( \displaystyle \frac{\sqrt{1+px}-\sqrt{1-px}}{x} = \frac{(1+px)-(1-px)}{x\left(\sqrt{1+px}+\sqrt{1-px}\right)} \)
\( \displaystyle = \frac{2p}{\sqrt{1+px}+\sqrt{1-px}} \)
\( \displaystyle \lim_{x \to 0^-} \frac{2p}{\sqrt{1+px}+\sqrt{1-px}} = \frac{2p}{1+1} = p \)
Equating the limit with f(0):
\( \displaystyle p = -\frac{1}{2} \)