Let `f(x)={(0 , x "is irrational"),(2/(2q^(3)-q^(2)+q+sin^(2)q+5) , if x=p/q ("rational")):}` (where HCF `(p,q)=1,p,q,gt0`) and `f(x)` is defined `AAxgt0` then which of the following is/are incorrect?

Options

A. `f(x)` is continuous at each irrational in `(0,oo)`
B. `f(x)` is continuous at each rational in `(0,oo)`
C. `f(x)` is discontinuous at each rational in `(0,oo)`
D. `f(x)` is discontinuous for all `x` in `(0,oo)`

Explanation

To solve the problem, we need to analyze the function \( f(x) \) defined as follows:

\[ f(x) = \begin{cases} 0 & \text{if } x \text{ is irrational} \\ \frac{2}{2q^3 - q^2 + q + \sin^2 q + 5} & \text{if } x = \frac{p}{q} \text{ (rational, where HCF(p, q) = 1, } p, q > 0) \end{cases} \]

We need to determine which of the following statements about the continuity of \( f(x) \) are incorrect:

1. \( f(x) \) is continuous at each irrational value.
2. \( f(x) \) is continuous at each rational value.
3. \( f(x) \) is continuous at all \( x \) values.

### Step-by-step Solution:

**Step 1: Analyze continuity at irrational points.**

For any irrational number \( x \), \( f(x) = 0 \). To check continuity at an irrational point \( c \), we need to see if:

\[ \lim_{x \to c} f(x) = f(c) = 0 \]

Since \( c \) is irrational, we consider rational numbers approaching \( c \). For rational numbers \( x = \frac{p}{q} \), as \( q \) increases, the value of \( f(x) \) becomes:

\[ f\left(\frac{p}{q}\right) = \frac{2}{2q^3 - q^2 + q + \sin^2 q + 5} \]

As \( q \to \infty \), the denominator \( 2q^3 - q^2 + q + \sin^2 q + 5 \) approaches infinity, leading \( f\left(\frac{p}{q}\right) \) to approach \( 0 \). Thus:

\[ \lim_{x \to c} f(x) = 0 = f(c) \]

**Conclusion:** \( f(x) \) is continuous at each irrational point.

**Step 2: Analyze continuity at rational points.**

For a rational number \( r = \frac{p}{q} \), \( f(r) = \frac{2}{2q^3 - q^2 + q + \sin^2 q + 5} \). To check continuity at \( r \), we need to see if:

\[ \lim_{x \to r} f(x) = f(r) \]

As \( x \) approaches \( r \) through irrational numbers, \( f(x) \) approaches \( 0 \). However, \( f(r) \) is a positive value (since \( q > 0 \)). Therefore:

\[ \lim_{x \to r} f(x) \neq f(r) \]

**Conclusion:** \( f(x) \) is not continuous at rational points.

**Step 3: Analyze continuity at all \( x \) values.**

Since \( f(x) \) is continuous at irrational points but not at rational points, it cannot be continuous at all \( x \) values.

### Final Conclusion:

- **Statement 1:** Correct (Continuous at each irrational point)
- **Statement 2:** Incorrect (Not continuous at each rational point)
- **Statement 3:** Incorrect (Not continuous at all \( x \) values)

### Summary of Incorrect Statements:

The incorrect statements are:
- \( f(x) \) is continuous at each rational value.
- \( f(x) \) is continuous at all \( x \) values.