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Let $f(x)=\left\{\small\begin{array}{1 1} \large\frac{(x^3+x^2-16x+20)}{(x-2)^2}&if\;x\neq 2\\K&if\;x=2\end{array}\right.$. If $f(x)$ is continuous for all $x$ then $K$ =

$(a)\;6\qquad(b)\;7\qquad(c)\;8\qquad(d)\;9$

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It will be continuous at $x=2$ if $\lim\limits_{x\to 2}f(x)=f(2)$
$\Rightarrow \lim\limits_{x\to 2}\large\frac{x^3+x^2-16x+20}{(x-2)^2}$$=K$
$\Rightarrow K=\lim\limits_{x\to 2}\large\frac{(x-2)^2(x+5)}{(x-2)^2}$
$\quad\quad=\lim\limits_{x\to 2}(x+5)=7$
$K=7$
Hence (b) is the correct answer.
answered Dec 20, 2013 by sreemathi.v
 

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