Question

The function 

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

 

Options

•  a = 1, b = −1

•  a = −1, b = 1 + \[\sqrt{2}\]

   

• a = −1, b = 1

• none of these

Answer 1

a = -1, b = 1 

Given: 

\[f\left( x \right)\]  is continuous for 0 ≤ x < ∞.

This means that 

\[f\left( x \right)\]  is continuous for

\[x = 1, \sqrt{2}\]

Now,

If  \[f\left( x \right)\]  is continuous at x = 1, then

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

If  \[f\left( x \right)\]  is continuous at x = \[\sqrt{2}\], then​

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

∴ For a = 1, we have 

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

Also,
For a = −1, we have

\[b^2 - 2b + 1 = 0\]
\[ \Rightarrow \left( b - 1 \right)^2 = 0\]
\[ \Rightarrow b = 1\]

Thus, 

\[a = - 1 \text{ and } b = 1\]