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

ધારો કે \(a,b \in R,\left( {a \ne 0} \right)\). જો વિધેય \(f\) એ વ્યાખ્યાયિત છે કે  \(f\left( x \right)\left\{ \begin{array}{l}
\frac{{2{x^2}}}{a}\,\,\,\,\,\,\,\,\,\,\,\,,\,\,\,\,\,0 \le x < 1\,\,\,\\
a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,,\,\,\,\,\,1 \le x < \sqrt 2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\
\frac{{2{b^2} - 4b}}{{{x^3}}}\,\,\,,\,\,\,\,\,\sqrt 2  \le x < \infty 
\end{array} \right.\,\,\,\,\)  એ \(\left[ {0,\infty } \right)\) પર સતત હોય તો \((a, b)\) જોડ મેળવો.

  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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