First slide

Continuity

Question

$The function f (x) = \{\begin{cases} x^{2} / a, 0 \leq x < 1 \\ a, 1 \leq x < \sqrt{2} \\ (2 b^{2} - 4 b) / x^{2}, \sqrt{2} \leq x < \infty \end{cases}$ $Is continuous for 0 \leq x < \infty, then the most suitable values of a and b are$

Moderate

A

$a = 1, b = - 1$
B

$a = - 1, b = 1 + \sqrt{2}$
C

$a = - 1, b = 1$
D

none of these

Solution

$\begin{array}{l} f (x) is continuous at x = 1. \\ ∴ f (1 - 0) = f (1) = f (1 + 0) \\ \Rightarrow 1 / a = a \Rightarrow a = \pm 1. \\ f (x) i s c o n t i n u o u s a t x = \sqrt{2} . \\ ∴ f (\sqrt{2} - 0) = f (\sqrt{2}) = f (\sqrt{2} + 0) \\ \Rightarrow a = (2 b^{2} - 4 b) / 2 \Rightarrow b^{2} - 2 b = 2 \\ When a= - 1, b^{2} - 2 b = - 1 \Rightarrow {(b - 1)}^{2} = 0 \Rightarrow b = 1. \\ When a = 1, b^{2} - 2 b = 1 \Rightarrow b = 1 \pm \sqrt{2} \\ Which are not given . \\ Hence (3)is the correct answer. \end{array}$

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