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

Concept:
Points of discontinuity occur where left and right limits differ.
Points of non-differentiability occur where the derivative is not defined (corners, cusps, or discontinuities).

Step 1: Analyse \(f(x)\)


\[ f(x) = \begin{cases} e^{x-1}, & x < 0 \\ x^2 - 5x + 6, & x \ge 0 \end{cases} \] For \(x \ge 0\): \[ x^2 - 5x + 6 = (x-2)(x-3) \] Zeros occur at \[ x = 2,\, 3 \] Step 2: Graph of \(f(|x|)\)


Because of \(|x|\), the right side of the graph reflects symmetrically to the left.
Thus \(f(|x|)\) is an even function.
At \(x=0\), left and right derivatives differ, hence \[ f(|x|) \text{ is non-differentiable at } x=0 \] Step 3: Graph of \(|f(x)|\)


Absolute value reflects negative parts above the \(x\)-axis.
Hence cusps occur where \(f(x)=0\): \[ x = 0,\, 2,\, 3 \] Thus \(|f(x)|\) is non-differentiable at \[ x = 0,\, 2,\, 3 \] Also, \(|f(x)|\) is discontinuous at \(x = 0\). Step 4: Analyse \(g(x)\)
\[ g(x) = f(|x|) + |f(x)| \] Since \(|f(x)|\) is discontinuous at \(x = 0\), \[ \alpha = 1 \] Points of non-differentiability occur at \[ x = 0,\, 2,\, 3 \] Thus \[ \beta = 3 \] Step 5: Final result
\[ \alpha + \beta = 1 + 3 = 4 \]