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JEE Mains · Maths · STD 12 - 5. continuity and differentiation

Let \(f\left( x \right) = \left\{ \begin{gathered}
  {\left( {x - 1} \right)^{\frac{1}{{2 - x}}}},\,\,\,x > 1,x \ne 2 \hfill \\
  k\,,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 2 \hfill \\ 
\end{gathered}  \right.\) The value of \(k\) for which \(f\) is continuous at \(x\, = 2\) is

  1. A \(e^{-2}\)
  2. B \(e\)
  3. C \(e^{-1}\)
  4. D \(1\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(e^{-1}\)

Step-by-step Solution

Detailed explanation

Since \(f(x)\) is continuous at \(x=2\). \(\therefore \mathop {\lim }\limits_{x \to 2} f\left( x \right) = f\left( 2 \right)\) \( \Rightarrow \mathop {\lim }\limits_{x \to 2} {\left( {x - 1} \right)^{\frac{1}{{2 - x}}}} = k\,\,\,\,\,\,\left( {{1^{\infty \,\,}}form} \right)\)…