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: Checking Continuity At \( x = -3 \) \[ \lim\limits_{h \to 0} f(-3 - h) = \lim\limits_{h \to 0} (3 - (-3 - h)) = 6 \] \[ \lim\limits_{h \to 0} f(-3 + h) = 6 \] \[ f(-3) = 6 \] Since LHL = RHL = \( f(-3) \), \( f(x) \) is continuous at \( x = -3 \). At \( x = 3 \) \[ \lim\limits_{h \to 0} f(3 - h) = 6 \] \[ \lim\limits_{h \to 0} f(3 + h) = \lim\limits_{h \to 0} (3 + (3 + h)) = 6 \] \[ f(3) = 6 \] Since LHL = RHL = \( f(3) \), \( f(x) \) is continuous at \( x = 3 \). Thus, \( \alpha = 0 \) (No discontinuities).
Step 2: Checking Differentiability 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 \( \neq \) 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 \( \neq \) RHD, \( f(x) \) is not differentiable at \( x = 3 \). Thus, \( \beta = 2 \) (Non-differentiability at \( x = -3 \) and \( x = 3 \)). Final Calculation: \[ \alpha + \beta = 0 + 2 = 2 \]