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 the number of points where $g$ is not continuous and is not differentiable are $\alpha$ and $\beta$ respectively, then $\alpha + \beta$ is equal to _______.

Answer 1

Correct Answer: 4

Step 1: Understanding the Concept:
We must build the piecewise function $g(x) = f(|x|) + |f(x)|$ by evaluating the absolute values based on the intervals of $x$. Then, analyze $g(x)$ for continuity and differentiability at critical "joint" points and roots of the quadratic.

Step 2: Key Formula or Approach:
For $|x|$: $|x| = x$ for $x \ge 0$, and $|x| = -x$ for $x<0$.
Check continuity at $x=a$: $\lim_{x \to a^-} g(x) = \lim_{x \to a^+} g(x) = g(a)$.
Check differentiability at $x=a$: $LHD = RHD$.

Step 3: Detailed Explanation:
Let's first define $f(|x|)$:
If $x<0$, $|x| = -x>0$. So $f(|x|) = (-x)^2 - 5(-x) + 6 = x^2 + 5x + 6$.
If $x \ge 0$, $|x| = x \ge 0$. So $f(|x|) = x^2 - 5x + 6$.
Now define $|f(x)|$:
For $x<0$, $f(x) = e^{x-1}$. Since $e^{x-1}>0$ always, $|f(x)| = e^{x-1}$.
For $x \ge 0$, $f(x) = x^2 - 5x + 6 = (x-2)(x-3)$.
$|f(x)| = \begin{cases} x^2 - 5x + 6, & 0 \le x \le 2 \text{ or } x \ge 3
-(x^2 - 5x + 6), & 2<x<3 \end{cases}$.
Construct $g(x) = f(|x|) + |f(x)|$:
For $x<0$: $g(x) = (x^2 + 5x + 6) + e^{x-1}$. (Continuous and smooth polynomial + exponential)
For $0 \le x \le 2$: $g(x) = (x^2 - 5x + 6) + (x^2 - 5x + 6) = 2(x^2 - 5x + 6)$.
For $2<x<3$: $g(x) = (x^2 - 5x + 6) - (x^2 - 5x + 6) = 0$.
For $x \ge 3$: $g(x) = (x^2 - 5x + 6) + (x^2 - 5x + 6) = 2(x^2 - 5x + 6)$.
1. Check Continuity at $x=0$:
$g(0^-) = \lim_{x \to 0^-} (x^2 + 5x + 6 + e^{x-1}) = 6 + 1/e$.
$g(0^+) = g(0) = 2(0 - 0 + 6) = 12$.
Since $g(0^-) \neq g(0^+)$, $g(x)$ is discontinuous at $x = 0$. Thus, $\alpha = 1$.
(Any point of discontinuity is strictly non-differentiable too, so $x=0$ contributes to $\beta$).
2. Check Differentiability at $x=2$:
$g'(x)$ for $x \in (0, 2) = 2(2x - 5)$. LHD at $x=2$ is $2(4 - 5) = -2$.
$g'(x)$ for $x \in (2, 3) = 0$. RHD at $x=2$ is $0$.
Since $LHD \neq RHD$, $g$ is non-differentiable at $x = 2$.
3. Check Differentiability at $x=3$:
$g'(x)$ for $x \in (2, 3) = 0$. LHD at $x=3$ is $0$.
$g'(x)$ for $x \in (3, \infty) = 2(2x - 5)$. RHD at $x=3$ is $2(6 - 5) = 2$.
Since $LHD \neq RHD$, $g$ is non-differentiable at $x = 3$.
The points of non-differentiability are $x = 0, 2, 3$. Thus, $\beta = 3$.
Finally, $\alpha + \beta = 1 + 3 = 4$.

Step 4: Final Answer:
The value of $\alpha + \beta$ is 4.