Define: `f (x) =|x^(2)-4x +3| ln x+2 (x-2)^(1//3) , x gt 0`
`h (x)= {{:(x-1"," , x in Q),(x ^(2) -x-2"," , x cancel (in)Q):}`
h (x) is discontinous at `x =……`

Options

A. `1+ sqrt2`
B. `tan ""(3pi)/(8)`
C. `tan ""(7pi)/(8)`
D. `sqrt2-1`

Answer

To determine where the function \( h(x) \) is discontinuous, we will analyze the function defined as follows:

\[ h(x) = \begin{cases} x - 1 & \text{if } x \in \mathbb{Q} \\ x^2 - x - 2 & \text{if } x \notin \mathbb{Q} \end{cases} \]

### Step 1: Identify the points of discontinuity
To find the points of discontinuity, we need to check where the left-hand limit and right-hand limit do not equal the function value at a point \( \alpha \).

### Step 2: Set up the limits
Let’s denote \( \alpha \) as a point where we want to check for continuity. We need to find:

1. \( h(\alpha) \)
2. \( \lim_{x \to \alpha^-} h(x) \)
3. \( \lim_{x \to \alpha^+} h(x) \)

### Step 3: Calculate \( h(\alpha) \)
- If \( \alpha \in \mathbb{Q} \), then \( h(\alpha) = \alpha - 1 \).
- If \( \alpha \notin \mathbb{Q} \), then \( h(\alpha) = \alpha^2 - \alpha - 2 \).

### Step 4: Calculate the left-hand limit
For \( x \) approaching \( \alpha \) from the left (\( x \to \alpha^- \)):
- If \( x \in \mathbb{Q} \), then \( h(x) = x - 1 \).
- If \( x \notin \mathbb{Q} \), then \( h(x) = x^2 - x - 2 \).

Thus, we have:
\[ \lim_{x \to \alpha^-} h(x) = \begin{cases} \alpha - 1 & \text{if } \alpha \in \mathbb{Q} \\ \alpha^2 - \alpha - 2 & \text{if } \alpha \notin \mathbb{Q} \end{cases} \]

### Step 5: Calculate the right-hand limit
For \( x \) approaching \( \alpha \) from the right (\( x \to \alpha^+ \)):
- The analysis is the same as for the left-hand limit.

Thus, we have:
\[ \lim_{x \to \alpha^+} h(x) = \begin{cases} \alpha - 1 & \text{if } \alpha \in \mathbb{Q} \\ \alpha^2 - \alpha - 2 & \text{if } \alpha \notin \mathbb{Q} \end{cases} \]

### Step 6: Set the limits equal for continuity
For \( h(x) \) to be continuous at \( \alpha \):
\[ \lim_{x \to \alpha^-} h(x) = \lim_{x \to \alpha^+} h(x) = h(\alpha) \]

### Step 7: Solve the equation
We need to solve the equation:
\[ \alpha - 1 = \alpha^2 - \alpha - 2 \]
Rearranging gives:
\[ \alpha^2 - 2\alpha - 1 = 0 \]
Using the quadratic formula:
\[ \alpha = \frac{2 \pm \sqrt{(2)^2 - 4(1)(-1)}}{2(1)} = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2} \]

### Step 8: Identify discontinuous points
The points where \( h(x) \) is discontinuous are \( 1 + \sqrt{2} \) and \( 1 - \sqrt{2} \).

### Conclusion
Thus, \( h(x) \) is discontinuous at \( x = 1 + \sqrt{2} \) and \( x = 1 - \sqrt{2} \).