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Find the values of \(k\) so that the function \(f\) is continuous at the indicated point in $ f(x) = \left\{ \begin{array} {1 1} kx^2 ,& \quad\text{ if $ x $ \(\leq 2\)}\\ 3,& \quad \text{if $x$ > 2}\\ \end{array} \right. \qquad at \;x = 2 $

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  • If $f$ is a real function on a subset of the real numbers and $c$ be a point in the domain of $f$, then $f$ is continuous at $c$ if $\lim\limits_{\large x\to c} f(x) = f(c)$.
Step 1:
LHL=$\lim\limits_{\large x\to 2}(kx^2)$
Putting $x=2-h$
LHL=$\lim\limits_{\large h\to 0}(k(2-h)^2)$
$\quad\;\;=4k$
$f(2)=4k$
Step 2:
RHL=$\lim\limits_{\large x\to 2^+}f(x)=3$
$f$ is continuous if $LHL=RHL=f(2)$
Therefore $4k=3$
$\Rightarrow k=\large\frac{3}{4}$
Hence the value of k is $\large\frac{3}{4}$
answered May 29, 2013 by sreemathi.v
 

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