If f(x)=1+kx-1-kxx, for -1≤x≤02x2+3x-2, for 0≤x≤1is continuous at x=0, then k=

fx is continuous at x=0 ⇒limx→0+ f(x)=limx→0-f(x) ⇒limx→0+2x2+3x-2=limx→0-1+kx-1-kxx ⇒-2=limx→0- 1+kx-1-kxx1+kx+1-kx⇒-2=limx→0- 2kxx1+kx+1-kx ∴k=-2

If f(x)=1+kx-1-kxx, for -1≤x≤02x2+3x-2, for 0≤x≤1is continuous at x=0, then k=

fx is continuous at x=0 ⇒limx→0+ f(x)=limx→0-f(x) ⇒limx→0+2x2+3x-2=limx→0-1+kx-1-kxx ⇒-2=limx→0- 1+kx-1-kxx1+kx+1-kx⇒-2=limx→0- 2kxx1+kx+1-kx ∴k=-2

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

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Detailed Solution

$f (x) is continuous at x = 0 \Rightarrow \lim_{x \to 0^{+}} f (x) = \lim_{x \to 0^{-}} f (x) \Rightarrow \lim_{x \to 0^{+}} (2 x^{2} + 3 x - 2) = \lim_{x \to 0^{-}} \frac{\sqrt{1 + kx} - \sqrt{1 - kx}}{x} \Rightarrow - 2 = \lim_{x \to 0^{-}} (\frac{(1 + kx) - (1 - kx)}{x (\sqrt{1 + kx} + \sqrt{1 - kx})}) \Rightarrow - 2 = \lim_{x \to 0^{-}} (\frac{2 k x}{x (\sqrt{1 + kx} + \sqrt{1 - kx})}) ∴ k = - 2$