11lend cases is continuous at x -0- then the value of k isA- 1B- -1C- 2D- 0

The correct option is B 1 nbsp; nbsp;For x lt;0, f(x)=√(1+kx) √(1 kx)/x×√(1+kx)+√(1 kx)/√(1+kx)+√(1 kx) =2k/√(1+kx)+√(1 kx) ∴LHL at x=0, f(0^ )=2k/√(1) ...

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Byju's Answer

Standard XII

Mathematics

How to Find the Inverse of a Function

11lend cases ...

Question

$If f (x) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} \frac{\sqrt{1 + k x} - \sqrt{1 - k x}}{x} if - 1 \leq x < 0 \frac{2 x + 1}{x - 1} if 0 \leq x \leq 1 \end{matrix}$ is continuous at $x = 0$ , then the value of $k$ is

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Solution

The correct option is B $- 1$
For $x < 0$ ,
$f (x) = \frac{\sqrt{1 + k x} - \sqrt{1 - k x}}{x} \times \frac{\sqrt{1 + k x} + \sqrt{1 - k x}}{\sqrt{1 + k x} + \sqrt{1 - k x}}$
$= \frac{2 k}{\sqrt{1 + k x} + \sqrt{1 - k x}}$

$∴ LHL at x = 0$ ,
$f (0^{-}) = \frac{2 k}{\sqrt{1} + \sqrt{1}} = k$

$RHL at x = 0$ ,
$f (0^{+}) = \frac{1}{- 1} = - 1$
$∵ f$ is continuous at $x = 0$ ,
$\Rightarrow k = - 1$

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If f(x)=⎧⎪ ⎪⎨⎪ ⎪⎩√1+kx−√1−kxx if −1≤x<0 2x+1x−1 if 0≤x≤1 is continuous at x=0, then the value of k is

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