Question

The function $f\left(x\right)=\{\begin{array}{cc}\frac{x^2}{a}& 0\le x<1\\ a& 1\le x<\sqrt{2}\\ (2b^2-4b)/x^2& \sqrt{2}\le x<\infty \end{array}$ is continuous for $0\le x<\infty$, then the most suitable values of $a$ and $b$ are

• A. $a=1,b=1$
• B. $a=-1,b=1+\sqrt{2}$
• C. $a=-1,b=1$
• D. none of these

Answer 1

Answer: (C)

$a=-1,b=1$

The function is continuous
At $x=1$,
$\Rightarrow \frac{x^2}{a}{|}_{x=1}=a{|}_{x=1}$
$\Rightarrow 1=a^2\Rightarrow a=\pm 1$
At $x=\sqrt{2},$
$\Rightarrow \frac{1}{a}{|}_{x=\sqrt{2}}=\frac{2b^2-4b}{x^2}{|}_{x=\sqrt{2}}$
$\Rightarrow 2a=2b^2-4b$
$\Rightarrow b^2-2b\mp 1=0$
For $a=-1,b=1$