If f(x)=1+kx−1−kxxfor −1≤x<02x2+3x−2 for 0≤x≤1is continuous at x=0 then k

limx→0−(1+kx)12−(1−kx)12x⇒ltx→0−k21+kx+k21−kx=k2+k2=k⇒ltx→0+2x2+3x¯−2=−2Since f(x)is continuous at x=0⇒k=−2

If f(x)=1+kx−1−kxxfor −1≤x<02x2+3x−2 for 0≤x≤1is continuous at x=0 then k

limx→0−(1+kx)12−(1−kx)12x⇒ltx→0−k21+kx+k21−kx=k2+k2=k⇒ltx→0+2x2+3x¯−2=−2Since f(x)is continuous at x=0⇒k=−2

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

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

$\Rightarrow \underset{x \to 0^{-}}{l t} \frac{k}{2 \sqrt{1 + k x}} + \frac{k}{2 \sqrt{1 - k x}} = \frac{k}{2} + \frac{k}{2} = k$

$\Rightarrow \underset{x \to 0^{+}}{l t} 2 x^{2} + 3 \bar{x} - 2 = - 2$

Since $f (x)$ is continuous at $x = 0$

$\Rightarrow k = - 2$

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If f(x)=1+kx−1−kxxfor −1≤x<02x2+3x−2 for 0≤x≤1is continuous at x=0 then k

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