If f x - 1-kx - 1-kx x - -1- x-0 2 x 2 -3x-2- 0- x- 1 is contnuous at x-0 then k is equal to

The correct option is C 2 #160;LHL=lim _ x→ 0 ^ √( 1+kx ) √( 1 kx ) #160;/ x #160; #160;=lim _ x→ 0 ^ 2kx / x( √( 1+kx ) +√( 1 kx ) ...

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Byju's Answer

Standard XII

Mathematics

Rationalization Method to Remove Indeterminate Form

If f x =√ 1...

Question

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

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Solution

The correct option is C $- 2$
$L H L = lim x \to 0^{-} \frac{\sqrt{1 + k x} - \sqrt{1 - k x}}{x}$

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

$R H L = lim x \to 0^{+} ({2 x}^{2} + 3 x - 2) = - 2$

Since it is given that $f (x)$ is continuous,
$∴ L H L = R H L \Rightarrow k = - 2$

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Similar questions

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f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} \frac{\sqrt{1 + k x} - \sqrt{1 - k x}}{x} & f o r & - 1 \leq x < 0 2 x^{2} + 3 x - 2 & f o r & 0 \leq x < 1 \end{matrix}

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If f(x)=⎧⎨⎩√1+kx−√1−kxx,−1≤x<02x2+3x−2,0≤x≤1 is contnuous at x=0 then k is equal to

The correct option is C −2LHL=limx→0−√1+kx−√1−kxx=limx→0−2kxx(√1+kx+√1−kx)=kRHL=limx→0+(2x2+3x−2)=−2Since it is given that f(x) is continuous,∴LHL=RHL⇒k=−2

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