To determine which statements about the function \( g(x) = \frac{1}{f(x)} \) are not true given that \( f: \mathbb{R} \to \mathbb{R} \), we will analyze each statement one by one.
### Step-by-Step Solution:
1. **Statement (a): g is onto if f is onto.**
- A function \( f \) is onto if for every \( y \in \mathbb{R} \), there exists an \( x \in \mathbb{R} \) such that \( f(x) = y \).
- For \( g(x) = \frac{1}{f(x)} \) to be onto, for every \( z \in \mathbb{R} \), there must exist an \( x \) such that \( g(x) = z \), or \( \frac{1}{f(x)} = z \) which implies \( f(x) = \frac{1}{z} \).
- If \( f \) is onto, it can take all real values, including zero. However, if \( f(x) = 0 \) for some \( x \), then \( g(x) \) is undefined. Therefore, \( g \) cannot be onto if \( f \) is onto.
- **Conclusion:** Statement (a) is **not true**.
2. **Statement (b): g is one-one if f is onto.**
- A function \( g \) is one-one (injective) if \( g(x_1) = g(x_2) \) implies \( x_1 = x_2 \).
- If \( f \) is onto and \( f(x_1) = f(x_2) \), then \( g(x_1) = g(x_2) \) implies \( \frac{1}{f(x_1)} = \frac{1}{f(x_2)} \), which leads to \( f(x_1) = f(x_2) \).
- Thus, if \( f \) is onto and injective, \( g \) will also be one-one.
- **Conclusion:** Statement (b) is **true**.
3. **Statement (c): g is continuous if f is continuous.**
- A function \( g \) is continuous if the limit of \( g(x) \) as \( x \) approaches \( a \) equals \( g(a) \).
- If \( f(x) \) is continuous and does not equal zero in an interval, then \( g(x) = \frac{1}{f(x)} \) will also be continuous in that interval.
- However, if \( f(x) \) approaches zero, \( g(x) \) will approach infinity, which means \( g \) can be discontinuous.
- **Conclusion:** Statement (c) is **not true**.
4. **Statement (d): g is differentiable if f is differentiable.**
- A function \( g \) is differentiable if it can be differentiated at a point.
- If \( f \) is differentiable and \( f(x) \neq 0 \), then \( g(x) = \frac{1}{f(x)} \) is differentiable using the quotient rule.
- However, if \( f(x) = 0 \) at some point, \( g(x) \) is not defined and hence not differentiable.
- **Conclusion:** Statement (d) is **not true**.
### Final Conclusion:
The statements that are **not true** are:
- (a) g is onto if f is onto.
- (c) g is continuous if f is continuous.
- (d) g is differentiable if f is differentiable.
