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  • <img alt="\mathrm{f(x)=\left\{\begin{array}{ll}\left(x^{2}+e^{\frac{1}{2-1}}\right)^{-1}, & x=2 \\ 1, & x=2\end{array}\right.}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathr

\mathrm{f(x)=\left\{\begin{array}{ll}\left(x^{2}+e^{\frac{1}{2-1}}\right)^{-1}, & x=2 \\ 1, & x=2\end{array}\right.} is continuous from right at the point \mathrm{x=2}, then \mathrm{k} equals

Option: 1

0


Option: 2

1 / 4


Option: 3

-1 / 4


Option: 4

 none of these


Answers (1)

best_answer

\mathrm{k=\lim _{x \rightarrow 2^{*}}\left(x^{2}+e^{\frac{1}{2-1}}\right)^{-1}}

\mathrm{\therefore \quad k =\lim _{x \rightarrow 0}\left[(2+h)^{2}+e^{\frac{-1}{k}}\right]^{-1}}
              \mathrm{=\left(4+c^{-}\right)^{-1}}
              \mathrm{\frac{1}{4}}

Posted by

shivangi.shekhar

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