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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
- A \(e^{-2}\)
- B \(e\)
- C \(e^{-1}\)
- D \(1\)
Answer & Solution
Correct Answer
(C) \(e^{-1}\)
Step-by-step Solution
Detailed explanation
Since \(f(x)\) is continuous at \(x=2\). \(\therefore \mathop {\lim }\limits_{x \to 2} f\left( x \right) = f\left( 2 \right)\) \( \Rightarrow \mathop {\lim }\limits_{x \to 2} {\left( {x - 1} \right)^{\frac{1}{{2 - x}}}} = k\,\,\,\,\,\,\left( {{1^{\infty \,\,}}form} \right)\)…
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