JEE Mains · Maths · STD 12 - 5. continuity and differentiation
If the function \(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}
{\frac{{\sqrt {2 + \cos \,x} - 1}}{{\left( {\pi - {x^2}} \right)}},}&{x \ne \pi } \\
{k\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,,}&{x = \pi }
\end{array}} \right.\) is continuous at \(x\, =\pi \) , then \(k\) equals
- A \(0\)
- B \(\frac{1}{2}\)
- C \(2\)
- D \(0.25\)
Answer & Solution
Correct Answer
(D) \(0.25\)
Step-by-step Solution
Detailed explanation
Since \(f\left( x \right) = \frac{{\sqrt {2 + \cos x }- 1 }}{{{{\left( {\pi - x} \right)}^2}}}\) is Continuos at \(x = \pi \) \(\therefore L.H.L = R.H.L = f\left( \pi \right)\) Let \(\left( {\pi - x} \right) = \theta ,\theta \to 0\) when \(x \to \pi \)…
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