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  • Let <img alt="f(x)= \begin{cases}(x-1)^{\frac{1}{2-x}}, & x>1, x \neq 2 \\ k & , \quad x=2\end{cases}" src="https://entrancecorner.oncodecogs.com/gif.latex?f%28x%29%3D%20%5Cbegin%7B

Let f(x)= \begin{cases}(x-1)^{\frac{1}{2-x}}, & x>1, x \neq 2 \\ k & , \quad x=2\end{cases}
The value of k for which f is continuous at x=2 is
 

Option: 1

e^{-1}


Option: 2

e
 


Option: 3

e^{-2}


Option: 4

 1


Answers (1)

best_answer

 Since, the function f(x) is continuous at x=2.
\mathrm{\begin{aligned} & \therefore \lim _{x \rightarrow 2} f(x)=f(2) \Rightarrow \lim _{x \rightarrow 2}(x-1)^{\frac{1}{2-x}}=k \\ & \Rightarrow e^{\lim _{x \rightarrow 2}[(x-1)-1] \frac{1}{2-x}}=k \quad\left[\because \text { L.H.S. is in the form of } 1^{\infty}\right] \\ & \Rightarrow e^{\lim _{x \rightarrow 2}-\left(\frac{x-2}{x-2}\right)}=k \Rightarrow e^{-1}=k \Rightarrow k=\frac{1}{e} \end{aligned} }

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Ritika Kankaria

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