f(x)=x3+x2−16x+20(x−2)2 if x≠2 =k if x=2f(x)is continuous at x=2then

f(x)=x3+x2−16x+20(x−2)2=(x−2)2(x+5)(x−2)2=x+5f(2)=Ltx→2f(x)⇒k=Ltx→2(x+5)=7⇒k=7

f(x)=x3+x2−16x+20(x−2)2 if x≠2 =k if x=2f(x)is continuous at x=2then

f(x)=x3+x2−16x+20(x−2)2=(x−2)2(x+5)(x−2)2=x+5f(2)=Ltx→2f(x)⇒k=Ltx→2(x+5)=7⇒k=7

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$\{\begin{array}{l} f (x) = \frac{x^{3} + x^{2} - 16 x + 20}{{(x - 2)}^{2}} if x \neq 2 \\ = k if x = 2 \end{array}\}$ $f (x)$ is continuous at $x = 2$ then

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$k = 3$

$k = 5$

$k = 7$

$k = 9$

answer is C.

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Detailed Solution

$f (x) = \frac{x^{3} + x^{2} - 16 x + 20}{{(x - 2)}^{2}} = \frac{{(x - 2)}^{2} (x + 5)}{{(x - 2)}^{2}} = x + 5$

$f (2) = \underset{x \to 2}{L t} f (x)$

$\Rightarrow k = \underset{x \to 2}{L t} (x + 5) = 7$

$\Rightarrow k = 7$

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f(x)=x3+x2−16x+20(x−2)2 if x≠2 =k if x=2f(x)is continuous at x=2then

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$\{\begin{array}{l} f (x) = \frac{x^{3} + x^{2} - 16 x + 20}{{(x - 2)}^{2}} if x \neq 2 \\ = k if x = 2 \end{array}\}$ $f (x)$ is continuous at $x = 2$ then