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

माना \(a, b \in R ,(a \neq 0)\) । यदि फलन \(f\) जो, निम्न द्वारा परिभाषित है \(f(x)=\left\{\begin{array}{ll}\frac{2 x^{2}}{a}, & 0 \leq x<1 \\ a, & 1 \leq x<\sqrt{2} \\ \frac{2 b^{2}-4 b}{x^{3}}, & \sqrt{2} \leq x<\infty\end{array}\right.\) अंतराल \([0, \infty)\) में सतत है, तो एक क्रमित युग्म \((a, b)\) है 

  1. A \(\left( { - \sqrt 2 ,1 - \sqrt 3 } \right)\)
  2. B \(\left( {\sqrt 2 , - 1 + \sqrt 3 } \right)\)
  3. C \(\left( {\sqrt 2 ,1 - \sqrt 3 } \right)\)
  4. D \(\left( { - \sqrt 2 ,1 + \sqrt 3 } \right)\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\left( {\sqrt 2 ,1 - \sqrt 3 } \right)\)

Step-by-step Solution

Detailed explanation

Continuity at \(x=1\) \(\frac{2}{a} = a \Rightarrow a = \pm \sqrt 2 \) Continuity at \(x = \sqrt 2 \,a = \sqrt 2 \) \(a = \frac{{2{b^2} - 4b}}{{2\sqrt 2 }}\) Put \(a = \sqrt 2 \) \(2 = {b^2} - 2b\,\,\,\,\,\,\,\, \Rightarrow {b^2} - 2b - 2 = 0\)…
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