Question

Let \[ f(x) = \begin{cases} e^{x-1}, & x < 0 \\ x^2 - 5x + 6, & x \ge 0 \end{cases} \] and \( g(x) = f(|x|) + |f(x)| \). If \( \alpha \) = number of points of discontinuity of \( g(x) \) and \( \beta \) = number of points of non-differentiability of \( g(x) \), then \( \alpha + \beta = \)

• A. \(2\)
• B. \(4\)
• C. \(3\)
• D. \(5\)

Hint:

Absolute value functions create non-differentiable points where the inner function becomes zero.

Answer 1

The Correct Option is B

To solve for \( \alpha + \beta \), where \( \alpha \) is the number of points of discontinuity of \( g(x) \) and \( \beta \) is the number of points of non-differentiability of \( g(x) \), we need to analyze the function \( g(x) = f(|x|) + |f(x)| \).

The function \( f(x) \) is given as a piecewise function:

• For \( x < 0 \), \( f(x) = e^{x-1} \).
• For \( x \ge 0 \), \( f(x) = x^2 - 5x + 6 \).

Let us explore \( g(x) \) under these conditions:

1. For \( x \ge 0 \):
   • \( f(|x|) = f(x) = x^2 - 5x + 6 \).
   • \( |f(x)| = |x^2 - 5x + 6| \).
   • Therefore, \( g(x) = (x^2 - 5x + 6) + |x^2 - 5x + 6| \).
2. For \( x < 0 \):
   • \( f(|x|) = f(-x) = (-x)^2 - 5(-x) + 6 = x^2 + 5x + 6 \).
   • \( f(x) = e^{x-1} \), so \( |f(x)| = e^{x-1} \) because the exponential is positive.
   • Therefore, \( g(x) = (x^2 + 5x + 6) + e^{x-1} \).

Now, examine the points of discontinuity and non-differentiability:

1. Discontinuity: Analyze around \( x=0 \) as the transition between piecewise conditions often causes discontinuity.
   • Both functions \( f(|x|) \) and \( |f(x)| \) are continuous. Therefore, \( g(x) \) itself is continuous at \( x=0 \).
   • Thus, \( \alpha = 0 \).
2. Non-Differentiability: Consider \( x=0 \) and \( x=2, x=3, \text{and} x=1 \) as points where the modulus function and quadratic may create cusps.
   • \( x=0 \) might be non-differentiable because symmetric function behavior at 0, \( f(|x|) \) and \( |f(x)| \) near 0 can cause \( g(x) \) to have sharp changes.
   • Quadratic \( |f(x)| \) changes slope abruptly at its roots \( x=2 \) and \( x=3 \).
   • Additionally, the vertex \( x=1 \) adds another point due to derivative shift.
   • Thus, \( \beta = 4 \) (at \( x=0, x=1, x=2, x=3 \)).

Summing them, \( \alpha + \beta = 0 + 4 = 4 \).

The correct answer is \( 4 \).