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

यदि फलन \(f(x)=\left\{\begin{array}{ll}-x, & x<1 \\ a+\cos ^{-1}(x+b), & 1 \leq x \leq 2\end{array}\right.\) \(x=1\) पर अवकलनीय है, तो \(\frac{a}{b}\) का मान है 

  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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