JEE Mains · Maths · STD 12 - 5. continuity and differentiation
Let \(a,b \in R,\left( {a \ne 0} \right)\). if the function \(f\) defined as \(f\left( x \right)\left\{ \begin{array}{l}
\frac{{2{x^2}}}{a}\,\,\,\,\,\,\,\,\,\,\,\,,\,\,\,\,\,0 \le x < 1\,\,\,\\
a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,,\,\,\,\,\,1 \le x < \sqrt 2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\
\frac{{2{b^2} - 4b}}{{{x^3}}}\,\,\,,\,\,\,\,\,\sqrt 2 \le x < \infty
\end{array} \right.\,\,\,\,\) is continuous in the interval \(\left[ {0,\infty } \right)\) , then an ordered pair \((a, b)\) is
- A \(\left( { - \sqrt 2 ,1 - \sqrt 3 } \right)\)
- B \(\left( {\sqrt 2 , - 1 + \sqrt 3 } \right)\)
- C \(\left( {\sqrt 2 ,1 - \sqrt 3 } \right)\)
- D \(\left( { - \sqrt 2 ,1 + \sqrt 3 } \right)\)
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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