Question

If a function \( f : (-\infty, 2) \to \mathbb{R} \) is defined by \[ f(x)= \begin{cases} \dfrac{\alpha \left|x^2 - 3x + 2\right|}{x - 1}, & x < 1 \\[8pt] \dfrac{\sin([x] - x)}{x - [x]}, & x > 1 \\[8pt] \beta, & x = 1 \end{cases} \] and \( f \) is continuous at \( x = 1 \), then find \[ \frac{\alpha^2 + \beta^2}{|\alpha \beta|}. \] 

• A. \(2\)
• B. \(\dfrac{25}{12}\)
• C. \(\dfrac{5}{2}\)
• D. \(3\)

Hint:

For piecewise functions involving greatest integer functions, first determine the value of \([x]\) in the interval approaching the point. This usually converts a complicated expression into a standard trigonometric limit.

Answer 1

The Correct Option is A

Concept: For continuity at a point \(x=a\), \[ \lim_{x\to a^-}f(x) = \lim_{x\to a^+}f(x) = f(a). \] Therefore, we must calculate both one-sided limits at \(x=1\) and equate them to \(\beta\).

Step 1: Evaluate the left-hand limit. For \(x<1\), \[ f(x) = \frac{\alpha|x^2-3x+2|}{x-1}. \] Factorize: \[ x^2-3x+2=(x-1)(x-2). \] Hence, \[ f(x) = \frac{\alpha|(x-1)(x-2)|}{x-1}. \] When \(x<1\), \[ x-1<0, \qquad x-2<0. \] Therefore, \[ |(x-1)(x-2)| = (x-1)(x-2). \] Thus, \[ f(x) = \frac{\alpha(x-1)(x-2)}{x-1} = \alpha(x-2). \] Taking the limit as \(x\to1^{-}\), \[ \lim_{x\to1^-}f(x) = \alpha(1-2) = -\alpha. \]

Step 2: Evaluate the right-hand limit. For \(x>1\) and \(x<2\), \[ [x]=1. \] Hence, \[ [x]-x = 1-x = -(x-1). \] Therefore, \[ f(x) = \frac{\sin(1-x)}{x-1}. \] Using \[ \sin(1-x) = -\sin(x-1), \] we obtain \[ f(x) = -\frac{\sin(x-1)}{x-1}. \] Now, \[ \lim_{x\to1^+} f(x) = -\lim_{x\to1^+} \frac{\sin(x-1)}{x-1} = -1. \]

Step 3: Apply continuity at \(x=1\). Since \(f\) is continuous at \(x=1\), \[ -\alpha=-1=\beta. \] Therefore, \[ \alpha=1, \qquad \beta=-1. \]

Step 4: Compute the required expression. \[ \frac{\alpha^2+\beta^2}{|\alpha\beta|} = \frac{1^2+(-1)^2}{|1(-1)|} = \frac{2}{1} = 2. \] Hence, \[ \boxed{2}. \]