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If $ f(x) = \left\{ \begin{array}{l l} \large\frac{x^3+x^2-16x+20}{(x-2)^2} & \quad when \quad x \neq 3 \\ k & \quad when \quad x = 2 \end{array} \right. $ and $f(x)$ is continuous at $x=3$ find the value of k.

$\begin{array}{1 1}(A)\;7 \\(B)\;6 \\(C)\;8 \\(D)\;0 \end{array}$

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Given $f(3)= k$----(1)
Now $\lim \limits_{x \to 3-0} f(x) =\lim \limits_{3+0} f(x) =\lim \limits_{x \to 3 } \large\frac{x^3+x^2-16 x +20}{(x-2)^2}$
=>$\lim \limits_{x \to 3} \large\frac{x^3-2x^2+3x^2-6x-10x+20}{(x-2)^2}$
=>$\lim \limits_{x \to 3} \large\frac{(x-2)(x^2+3x-10)}{(x-2)^2}$
=>$\lim \limits_{x \to 3} \large\frac{(x-2)(x-2)(x+5)}{(x-2)^2}$
=>$\lim \limits_{x \to 3} (x+5)$
=> $3+5$
=> $8$
$f(x) $ is continous at $x=3$
$\lim \limits_{x \to 3-0} f(x) =\lim \limits_{x \to 3+0} f(x)=f(3)$
So $k=8$
Hence C is the correct answer.
answered Apr 22, 2014 by meena.p
 

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