First slide

Continuity

Question

$f (x) = \{\begin{array}{ll} \frac{\sqrt{(1 + p x)} - \sqrt{(1 - p x)}}{x}, - 1 \leq x < 0 \\ \frac{2 x + 1}{x - 2}, 0 \leq x \leq 1 \end{array} is continuous in the interval [- 1, 1]$ $then is equal to$

Moderate

A

$- 1$
B

$- \frac{1}{2}$
C

$\frac{1}{2}$
D

$1$

Solution

$\begin{array}{l} f (0 + 0) = \lim_{h \to 0} \frac{2 (0 + h) + 1}{(0 + h) - 2} = - \frac{1}{2} \\ f (0 - 0) = \lim_{h \to 0} \frac{\sqrt{(1 - p h)} - \sqrt{1 + p h}}{0 - h} \\ = \lim_{h \to 0} \frac{- 2 p h}{- h [\sqrt{(1 - p h)} + \sqrt{(1 + p h)}]} = p \\ f(x) is continuous at x = 0 i n [1, 1] i f \\ f (0 + 0) = f (0 - 0) = f (0) \Rightarrow p = - 1 / 2. \\ Hence (2) is the correct answer. \end{array}$

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