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  • Let <img alt="f(x)= \begin{cases}\frac{x^3+x^2-16 x+20}{(x-2)^2}, & x \neq 2 \\ k & , x=2\end{cases}" src="https://entrancecorner.oncodecogs.com/gif.latex?f%28x%29%3D%20%5Cbegin%7Bcases%7D%

Let f(x)= \begin{cases}\frac{x^3+x^2-16 x+20}{(x-2)^2}, & x \neq 2 \\ k & , x=2\end{cases} If f(x) is continuous for all x, then k=

Option: 1

7


Option: 2

4


Option: 3

3


Option: 4

6


Answers (1)

best_answer

\mathrm{\because f(x) } is continuous
\mathrm{k=f(2)=\lim _{x \rightarrow 2} \frac{x^3+x^2-16 x+20}{(x-2)^2} }
\mathrm{\begin{aligned} & =\lim _{x \rightarrow 2} \frac{3 x^2+2 x-16}{2(x-2)} \\ & =\lim _{x \rightarrow 2} \frac{6 x+2}{2}=7 . \end{aligned} }

[By L.H. Rule]

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Rakesh

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