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

$\begin{array}{1 1}(a)\;a=1,b=-1&(b)\;a=-1,b=1+\sqrt 2\\(c)\;a=-1,b=1&(d)\;\text{None of these}\end{array}$

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$\lim\limits_{\large x\to 1^-}f(x)=\lim\limits_{\large x\to 1}\large\frac{x^2}{a}=\frac{1}{a}$
$\Rightarrow \lim\limits_{\large x\to 1^+}f(x)$
$f(1)=a$
Hence $a^2=1\Rightarrow a=\pm 1$
Also $\lim\limits_{x\to \sqrt 2}f(x)=a$
$\lim\limits_{x\to \sqrt 2}f(x)=\large\frac{2b^2-4b}{2}$
$\therefore \large\frac{2b^2-4b}{2}$$=a$
Now if $a=-1$ then $b=1$
If $a=1,b=1\pm \sqrt{2}$
Hence (c) is the correct option.
answered Jan 4, 2014 by sreemathi.v
 

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