Question

Let f(x) = $\{\begin{array}{ccc}\frac{a|x^2-x-2|}{2+x-x^2}& ,& x<2\\ & & \\ b& ,& x=2\\ & & \\ \frac{x-\left[x\right]}{x-2}& ,& x>2\end{array}$ , where [.] denotes the greatest integer function.

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

$f\left(x\right)=\{\begin{array}{cc}\frac{a|x^2-x-2|}{2+x-x^2}& ,x<2\\ & \\ b& ,x=2\\ & \\ \frac{x-\left[x\right]}{x-2}& ,x>2\end{array}$

$Ix<2$

$f\left(x\right)=\frac{a|-(-x^2+x+2)|}{(-x^2+x+2)}.......\left(A\right)$

$-x^2+x+2\to Roots=-1,2$

$Graph\to$

$\therefore -x^2+x+2is+vewhenx<2$

$\left(A\right)\to f\left(x\right)=\frac{a(-x^2+x+2)}{-x^2+x+2}\Rightarrow f\left(x\right)=a,x<2-\left(1\right)$

$IIf\left(x\right)=b.x=2$

$IIIf\left(x\right)=\frac{x-\left[x\right]}{x-2},x>2$

$Letx=2+f$

$f\left(x\right)=\frac{(2+f)-[2+f]}{2+f-2}=\frac{2+f-2}{2+f-2}=1,x>2$

since f(x) is continues at x = 2

$\begin{array}{c}a=b=1\end{array}$