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Determine the value of k so that the function $ f(x) = \left\{ \begin{array}{l l}kx^2, & \quad if { x \leq 2 } \\ 3, & \quad if { x > 2 } \end{array} \right. $ is continuous

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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:
$ f(x) = \left\{ \begin{array}{l l}kx^2, & \quad if { x \leq 2 } \\ 3, & \quad if { x > 2 } \end{array} \right. $
We have,
LHL at $x=2$
$\Rightarrow \lim\limits_{x\to 2^-}f(x)=\lim\limits_{h\to 0}f(2-h)$
$\Rightarrow \lim\limits_{h\to 0}k(2-h)^2$
Step 2:
RHL at $x=2$
$\Rightarrow \lim\limits_{x\to 2^+}f(x)=\lim\limits_{x\to 2^+}3$
Since the function is continuous.
$\therefore$ LHL =RHL
$\lim\limits_{h\to 0}k(2-h)^2=\lim\limits_{x\to 2^+}3$
$\Rightarrow \lim\limits_{h\to 0}k[4-4h+h^2]$
$\Rightarrow 3$
(i.e) $4k=3$
$\Rightarrow k=\large\frac{3}{4}$
answered Sep 30, 2013 by sreemathi.v

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