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Suppose $f(x)=\left\{\begin{array}{1 1}a+bx,&x<1\\4,&x=1 \\b-ax,&x>1\end{array}\right.$ and if $ \lim\limits_{x \to 1} f(x) = f(1) $ what are possible values of $a$ and $b$?

1 Answer

The given function is
$f(x)=\left\{\begin{array}{1 1}a+bx,&x<1\\4,&x=1 \\b-ax,&x>1\end{array}\right.$
$ \lim\limits_{x \to 1^-}f(x) = \lim\limits_{x \to 1}(a+bx)=(a+b)$
$ \lim\limits_{x \to 1^+}f(x) = \lim\limits_{x \to 1}(b-ax)=(b-a)$
$f(1)=4$
It is given that $ \lim\limits_{x \to 1}f(x) = f(1)$
$ \therefore \lim\limits_{x \to 1^-} f(x) = \lim\limits_{x \to 1^+} f(x)= \lim\limits_{x \to 1} f(x) = f(1)$
$ \Rightarrow a+b = 4 \: and\: b-a = 4$
On solving these two equations, we obtain $ a = 0 \: and \: b = 4$
Thus, the respective possible values of $a$ and $b$ are $0$ and $4$
answered Apr 7, 2014 by thanvigandhi_1
 

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