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

જો વિધેય  \(f(x)\, = \left\{ {\begin{array}{*{20}{c}}{ - x,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x < 1\,\,\,\,}\\{a + {{\cos }^{ - 1}}(x + b),\,\,\,\,\,\,\,\,\,1 \le x \le 2} \end{array}} \right.\)  એ  \(x = 1\) આગળ વિકલનીય હોય તો \(\frac {a}{b}\) મેળવો.

  1. A \(\frac {\pi + 2}{2}\)
  2. B \(\frac {\pi - 2}{2}\)
  3. C \(\frac {-\pi - 2}{2}\)
  4. D \(-1-cos^{-1}\,(2)\)
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Answer & Solution

Correct Answer

(A) \(\frac {\pi + 2}{2}\)

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Detailed explanation

\(f\left( x \right) = \left\{ \begin{array}{l} - x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x < 1\\ a + {\cos ^{ - 1}}\left( {x + b} \right)\,\,\,1 \le x \le 2 \end{array} \right.\) \(f(x)\) is continuous…
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