Question

Let \( f(x) \) be defined as: \[f(x) = \begin{cases} 3 - x, & x<-3 \\ 6, & -3 \leq x \leq 3 \\ 3 + x, & x>3 \end{cases}\]
Let \( \alpha \) be the number of points of discontinuity of \( f(x) \) and \( \beta \) be the number of points where \( f(x) \) is not differentiable. Then, \( \alpha + \beta \) is:

• A. 6
• B. 3
• C. 2
• D. 0

Hint:

A function is continuous at a point if the left-hand limit, right-hand limit, and function value are equal. Differentiability requires the left-hand derivative and right-hand derivative to be equal.

Answer 1

The Correct Option is C

Step 1: Continuity Verification At \( x = -3 \) \[\lim\limits_{h \to 0} f(-3 - h) = \lim\limits_{h \to 0} (3 - (-3 - h)) = 6\]\[\lim\limits_{h \to 0} f(-3 + h) = 6\]\[f(-3) = 6\]As LHL = RHL = \( f(-3) \), \( f(x) \) exhibits continuity at \( x = -3 \).At \( x = 3 \) \[\lim\limits_{h \to 0} f(3 - h) = 6\]\[\lim\limits_{h \to 0} f(3 + h) = \lim\limits_{h \to 0} (3 + (3 + h)) = 6\]\[f(3) = 6\]As LHL = RHL = \( f(3) \), \( f(x) \) exhibits continuity at \( x = 3 \). Therefore, \( \alpha = 0 \) (indicating no discontinuities).
Step 2: Differentiability Verification At \( x = -3 \) \[\lim\limits_{h \to 0} \frac{f(-3 + h) - f(-3)}{h} = \frac{6 - 6}{h} = 0\]\[\lim\limits_{h \to 0} \frac{f(-3 - h) - f(-3)}{-h} = \frac{(3 - (-3 - h)) - 6}{-h} = \frac{6 + h - 6}{-h} = -1\]Since LHD \( eq \) RHD, \( f(x) \) is not differentiable at \( x = -3 \).At \( x = 3 \) \[\lim\limits_{h \to 0} \frac{f(3 + h) - f(3)}{h} = \frac{(3 + (3 + h)) - 6}{h} = \frac{6 + h - 6}{h} = 1\]\[\lim\limits_{h \to 0} \frac{f(3 - h) - f(3)}{-h} = \frac{6 - 6}{-h} = 0\]Since LHD \( eq \) RHD, \( f(x) \) is not differentiable at \( x = 3 \). Thus, \( \beta = 2 \) (representing non-differentiability at \( x = -3 \) and \( x = 3 \)).Final Calculation: \[\alpha + \beta = 0 + 2 = 2\]