If f ( x ) = { 1 + k x - 1 - k x x , for - 1 x 0 2 x 2 + 3 x - 2 for 0 x 1 is continuous at x = 0 , then k is equal to:

Question

If $f (x) =$ ${\begin{matrix} \frac{\sqrt{1 + k x} - \sqrt{1 - k x}}{x}, & for - 1 \leq x < 0 \\ 2 x^{2} + 3 x - 2 & for 0 \leq x \leq 1 \end{matrix}$ is continuous at $x = 0$ , then $k$ is equal to:

Smart Achievers · Accepted Answer

(3)

Ans. - 2

Explanation:

$LHL = \lim_{x \to 0^{-}} \frac{\sqrt{1 + k x} - \sqrt{1 - k x}}{x}$

$= \lim_{x \to 0^{-}} \frac{2 k x}{x (\sqrt{1 + k x + \sqrt{1 - k x}})} = k$

$RHL = \lim_{x \to 0^{+}} (2 x^{2} + 3 x - 2) = - 2$

$f (0) = - 2$

$∵$ It is given that $f (x)$ is continuous at $x = 0$

$∴$ $LHL = RHL = f (0) \Rightarrow k = - 2$

Solution

Q.
If $f (x) =$ ${\begin{matrix} \frac{\sqrt{1 + k x} - \sqrt{1 - k x}}{x}, & for - 1 \leq x < 0 \\ 2 x^{2} + 3 x - 2 & for 0 \leq x \leq 1 \end{matrix}$ is continuous at $x = 0$ , then $k$ is equal to:

1 - 4

2 - 3

3 - 2

4 - 1

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