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# Find the value of $k$ in the following so that the function $f$ is continuous at the indicated point $f(x)=\left \{\begin{array}{1 1}\large\frac{\sqrt {1+kx}-\sqrt{1-kx}}{x}, & if\;-1\leq x< 0\\\large\frac{2x+1}{x-2}, & if\;0\leq x\leq 1\end{array}\right.\; at\; x=0$

$\begin{array}{1 1}\frac{1}{2} \\\frac{-1}{2} \\ 1 \\ -1 \end{array}$

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## 1 Answer

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Toolbox:
• A function is said to be continuous if the LHL = RHL.
• (i.e) $\lim\limits_{\large x\to 0^+}f(x)=\lim\limits_{\large x\to 0^-}f(x)$
• $\lim\limits_{\large x\to a}\large\frac{x^{\Large n}-a^{\large n}}{x-a}=$$na^{\large n-1} Step 1: Given : f(x)=\left \{\begin{array}{1 1}\large\frac{\sqrt {1+kx}-\sqrt{1-kx}}{x}, & if\;-1\leq x< 0\\\large\frac{2x+1}{x-2}, & if\;0\leq x\leq 1\end{array}\right.\; at\; x=0 Consider the LHL : f(x)=\large\frac{\sqrt {1+kx}-\sqrt{1-kx}}{x} This can be written as \quad=\lim\limits_{\large x\to 0}\large\frac{(1+kx)^{\Large\frac{1}{2}}-(1-kx)^{\Large\frac{1}{2}}}{(1+kx)-(1-kx)}$$\times 2k$
We know that $\lim\limits_{\large x\to a}\large\frac{x^n-a^n}{x-a}$$=na^{\large n-1} \quad=2k.\large\frac{1}{2}$$\times (1)^{\large\frac{-1}{2}}$
$\quad=k$
Step 2:
The RHL is $f(x)=\large\frac{2x+1}{x-2}$ at $x=0$
$\Rightarrow \lim\limits_{\large x\to 0}\large\frac{2x+1}{x-2}$
$\Rightarrow \large\frac{-1}{2}$
Step 3:
Since the function is continuous then LHL=RHL
(i.e)$k=-\large\frac{1}{2}$
answered Jun 26, 2013

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