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

જો વિધેય \(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}
  {\frac{{\sqrt {2  + \cos \,x} - 1}}{{\left( {\pi  - {x^2}} \right)}},}&{x \ne \pi } \\ 
  {k\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,,}&{x = \pi } 
\end{array}} \right.\) એ \(x\, =\pi \) આગળ સતત હોય તો  \(k\) મેળવો.

  1. A \(0\)
  2. B \(\frac{1}{2}\)
  3. C \(2\)
  4. D \(0.25\)
Verified Solution

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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