Question

The function  \[f\left( x \right) = \begin{cases}x^2 /a , & \text{ if } 0 \leq x < 1 \\ a , & \text{ if } 1 \leq x < \sqrt{2} \\ \frac{2 b^2 - 4b}{x^2}, & \text{ if }  \sqrt{2} \leq x < \infty\end{cases}\] is continuous on (0, ∞), then find the most suitable values of a and b.

Answer 1

Given:  f is continuous on  \[\left( 0, \infty \right)\]

 ∴  f is continuous at x = 1 and \[\sqrt{2}\]

At x = 1, we have

\[\lim_{x \to 1^-} f\left( x \right) = \lim_{h \to 0} f\left( 1 - h \right) = \lim_{h \to 0} \left[ \frac{\left( 1 - h \right)^2}{a} \right] = \frac{1}{a}\]

\[\lim_{x \to 1^+} f\left( x \right) = \lim_{h \to 0} f\left( 1 + h \right) = \lim_{h \to 0} \left( a \right) = a\]

Also, 

At x = \[\sqrt{2}\],

we have

\[\lim_{x \to \sqrt{2}^-} f\left( x \right) = \lim_{h \to 0} f\left( \sqrt{2} - h \right) = \lim_{h \to 0} \left( a \right) = a\] 

\[\lim_{x \to \sqrt{2}^+} f\left( x \right) = \lim_{h \to 0} f\left( \sqrt{2} + h \right) = \lim_{h \to 0} \left[ \frac{2 b^2 - 4b}{\left( \sqrt{2} + h \right)^2} \right] = \frac{2 b^2 - 4b}{2} = b^2 - 2b\]

 f is continuous at x = 1 and \[\sqrt{2}\]

∴  \[\lim_{x \to 1^-} f\left( x \right) = \lim_{x \to 1^+} f\left( x \right) \text{ and }  \lim_{x \to \sqrt{2}^-} f\left( x \right) = \lim_{x \to \sqrt{2}^+} f\left( x \right)\]

\[\Rightarrow \frac{1}{a} = a \text{ and } b^2 - 2b = a\]
\[ \Rightarrow a^2 = 1 \text{ and }  b^2 - 2b = a\]
\[ \Rightarrow a = \pm 1 \text{ and } b^2 - 2b = a . . . \left( 1 \right)\]

If a = 1, then

\[b^2 - 2b = 1 \left[ \text{ From eq }  . (1) \right]\]
\[ \Rightarrow b^2 - 2b - 1 = 0\]
\[ \Rightarrow b = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2}\]

If a = −1, then

\[b^2 - 2b = - 1 \left[ \text{ From eq }  . (1) \right]\]
\[ \Rightarrow b^2 - 2b + 1 = 0\]
\[ \Rightarrow \left( b - 1 \right)^2 = 0\]
\[ \Rightarrow b = 1\]

Hence, the most suitable values of a and b are

a = −1, b = 1  or a = 1,

\[b = 1 \pm \sqrt{2}\]