JEE Mains · Maths · STD 12 - 5. continuity and differentiation
If the function \(f\) defined on \(\left( {\frac{\pi }{6},\frac{\pi }{3}} \right)\) by \(f\,(x)\, = \,\left\{ {\begin{array}{*{20}{c}}
{\frac{{\sqrt 2 \,\cos \,x - \,1}}{{\cot \,x\, - \,1}}\,,\,x\, \ne \,\frac{\pi }{4}}\\
{k,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\, = \frac{\pi }{4}}
\end{array}} \right.\) is continuous, then \(k\) is equal to
- A \(1\)
- B \(2\)
- C \(\frac {1}{2}\)
- D \(\frac {1}{\sqrt 2}\)
Answer & Solution
Correct Answer
(C) \(\frac {1}{2}\)
Step-by-step Solution
Detailed explanation
\(\therefore \,\) function should be continuous at \(x = \frac{\pi }{4}\) \(\therefore \mathop {\lim }\limits_{x \to \frac{\pi }{4}} f\left( x \right) = f\left( {\frac{\pi }{4}} \right)\)…
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