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  •  If <img alt="\mathrm{f(x)}= \begin{cases}\mathrm{\frac{x^3+x^2-16 x+20}{(x-2)^2}} & , \mathrm{x \neq 2} \\ \mathrm{k} & \mathrm{x=2}\end{cases}" src="https://entrancecorner.oncodecogs

 If \mathrm{f(x)}= \begin{cases}\mathrm{\frac{x^3+x^2-16 x+20}{(x-2)^2}} & , \mathrm{x \neq 2} \\ \mathrm{k} & \mathrm{x=2}\end{cases}

\mathrm{f(x)} is continuous at \mathrm{x=2} then find the value of \mathrm{k}.

Option: 1

7


Option: 2

8


Option: 3

9


Option: 4

1


Answers (1)

best_answer

\mathrm{ \lim _{x \rightarrow 2} f(x)=f(2) }
Because \mathrm{ f(x) } is continuous at \mathrm{ x=2 }.

\begin{aligned} & \mathrm{=\lim _{x \rightarrow 2} \frac{x^3+x^2-16 x+20}{(x-2)^2}} \\ & \mathrm{=\lim _{x \rightarrow 2} \frac{x^2(x-2)+3 x(x-2)-10(x-2)}{(x-2)^2} }\\ & \mathrm{=\lim _{x \rightarrow 2} \frac{(x-2)\left(x^2+3 x-10\right)}{(x-2)^2} }\\ & \mathrm{=\lim _{x \rightarrow 2} \frac{(x-2)(x+5)(x-2)}{(x-2)^2}=\lim _{x \rightarrow 2}(x+5)=7}\\ &\mathrm{Thus,~ k=7 }\end{aligned}

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