For let `h(x)={1/q if x=p/q and 0` if x is irrational where `p & q > 0` are relatively prime integers 0 then which one does not hold good?

Options

A. (a)h(x) is discontinuous for all x in `(0, oo)`
B. (b)h(x) is continuous for each irrational in `(0, oo)`
C. (c)h(x) is discontinuous for each rational in `(0, oo)`
D. (d)h(x) is not derivable for all x in `(0, oo)`

Answer

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

\[ h(x) = \begin{cases} \frac{1}{q} & \text{if } x = \frac{p}{q} \text{ (where } p \text{ and } q \text{ are relatively prime integers and } q > 0) \\ 0 & \text{if } x \text{ is irrational} \end{cases} \]

We want to determine which statement about this function does not hold true.

### Step 1: Understanding the function's behavior

1. **For rational numbers**: If \( x \) is a rational number in the form \( \frac{p}{q} \), then \( h(x) = \frac{1}{q} \). This value depends on the denominator \( q \).
2. **For irrational numbers**: If \( x \) is irrational, then \( h(x) = 0 \).

### Step 2: Check continuity at rational points

- Let \( x \) be a rational number \( \frac{p}{q} \).
- The limit of \( h(x) \) as \( x \) approaches \( \frac{p}{q} \) from both sides involves both rational and irrational numbers.
- As we approach \( \frac{p}{q} \) through irrational numbers, \( h(x) \) approaches \( 0 \).
- However, \( h\left(\frac{p}{q}\right) = \frac{1}{q} \), which is not equal to \( 0 \) (since \( q > 0 \)).
- Therefore, \( h(x) \) is discontinuous at rational points.

### Step 3: Check continuity at irrational points

- Let \( x \) be an irrational number.
- As we approach \( x \) through rational numbers, \( h(x) \) will take values of the form \( \frac{1}{q} \) for various \( q \).
- As we approach \( x \) through irrational numbers, \( h(x) \) approaches \( 0 \).
- Since \( h(x) = 0 \) for irrational \( x \), but the limit from rational numbers does not equal \( 0 \), \( h(x) \) is also discontinuous at irrational points.

### Step 4: Conclusion about differentiability

Since \( h(x) \) is discontinuous at both rational and irrational points, it cannot be differentiable at any point.

### Step 5: Identify which statement does not hold

Given the analysis, we can summarize:

- \( h(x) \) is discontinuous at all rational points.
- \( h(x) \) is discontinuous at all irrational points.
- Since a function must be continuous to be differentiable, \( h(x) \) is not differentiable at any point.

### Final Answer

The statement that does not hold true is: **"h(x) is continuous for each irrational."**