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

વિધેય \(f: R \rightarrow R\) એ નીચે મુજબ વ્યાખ્યાયિત છે . \(f(x)=\sin x-e^{x} \,\,\,\, \text { if } x \leq 0\) \(\quad\quad\quad a+[-x] \,\,\,\, \text { if } 0\,<\,x\,<\,1\) \(\quad\quad\quad 2 x-b \,\,\,\,\,\,\,\, \text { if } \geq 1\) કે જ્યાં \([\mathrm{x}]\) એ \(\mathrm{x}\) નું  મહતમ પૃણાંક વિધેય છે. જો \(\mathrm{f}\) એ  \(\mathrm{R}\) પર સતત હોય તો \((\mathrm{a}+\mathrm{b})\) ની કિમંત મેળવો.

  1. A \(5\)
  2. B \(3\)
  3. C \(2\)
  4. D \(4\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(3\)

Step-by-step Solution

Detailed explanation

Continuous at \(x=0\) \(f\left(0^{+}\right)=f^{-} \Rightarrow a-1=0-e^{0}\) \(\Rightarrow a=0\) Continuous at \(\mathrm{x}=1\) \(f\left(1^{+}\right)=f(1^{-})\) \(\Rightarrow 2(1)-b=a+(-1)\) \(\Rightarrow b=2-a+1 \Rightarrow b=3\) \(\therefore a+b=3\)
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