Question

$Let\left(x\right)=\{\begin{array}{ccc}\frac{a|x^2-x-2|}{2+x-x^2},& x<2& \\ b,& x=2& \left(\right[.\left]denotesthegreatestintegerfunction\right)\\ \frac{x-\left[x\right]}{x-2},& x>2& \end{array}$
If f(x) is continuous at x = 2, then

• A. a = 1, b = 2
• B. a = 1, b = 1
• C. a = 0, b = 1
• D. a = 2, b = 1

Answer 1

The correct option is B a = 1, b = 1

$f\left(x\right)=\{\begin{array}{cc}\frac{a|x^2-x-2|}{2+x-x^2},& x<2\\ \frac{x-\left[x\right]}{x-2},& x=2\end{array}\{\begin{array}{cc}-\frac{a(x^2-x-2)}{2+x-x^2},& x<2\\ \frac{x-\left[x\right]}{x2},& x>2\end{array}f\left(x\right)=\{\begin{array}{cc}a,& x<2\\ b,& x=2\\ \frac{x-\left[x\right]}{x-2},& x>2\end{array}$
Check at x = 2
$LHL=\underset{x\to 2^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(2-h)=\underset{h\to 0}{lim}a=aRHL=\underset{x\to 2^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(2-h)=\underset{h\to 0}{lim}\frac{2+h-[2+h]}{2+h-2}=\underset{h\to 0}{lim}\frac{2+h-2}{2+h-2}=1$