Let `f(x) = x2 + b_1x + c_1. g(x) = x^2 + b_2x + c_2`. Real roots of `f(x) = 0` be `alpha, beta` and real roots of `g(x) = 0` be `alpha+gamma, beta+gamma`. Least values of `f(x)` be `- 1/4`Least value of `g(x)` occurs at `x=7/2`

Options

A. `-1`
B. `-1/2`
C. `-1/3`
D. `-1/4`

Explanation

To solve the problem, we start with the given functions and analyze the information provided:

1. **Identify the functions**:
- Let \( f(x) = x^2 + b_1 x + c_1 \)
- Let \( g(x) = x^2 + b_2 x + c_2 \)

2. **Roots of the functions**:
- The roots of \( f(x) = 0 \) are \( \alpha \) and \( \beta \).
- The roots of \( g(x) = 0 \) are \( \alpha + \gamma \) and \( \beta + \gamma \).

3. **Given conditions**:
- The least value of \( f(x) \) is \( -\frac{1}{4} \).
- The least value of \( g(x) \) occurs at \( x = \frac{7}{2} \).

4. **Finding the least value of \( f(x) \)**:
- The least value of a quadratic function \( ax^2 + bx + c \) occurs at \( x = -\frac{b}{2a} \).
- For \( f(x) \), since \( a = 1 \), the least value occurs at \( x = -\frac{b_1}{2} \).
- The least value of \( f(x) \) is given as \( -\frac{1}{4} \).
- Therefore, we can write:
\[ f\left(-\frac{b_1}{2}\right) = -\frac{1}{4} \]
- Substituting into \( f(x) \):
\[ f\left(-\frac{b_1}{2}\right) = \left(-\frac{b_1}{2}\right)^2 + b_1\left(-\frac{b_1}{2}\right) + c_1 = -\frac{1}{4} \]
- Simplifying:
\[ \frac{b_1^2}{4} - \frac{b_1^2}{2} + c_1 = -\frac{1}{4} \]
\[ -\frac{b_1^2}{4} + c_1 = -\frac{1}{4} \]
\[ c_1 = \frac{b_1^2}{4} - \frac{1}{4} \]

5. **Finding the least value of \( g(x) \)**:
- The least value of \( g(x) \) occurs at \( x = \frac{7}{2} \).
- Therefore, we can write:
\[ g\left(\frac{7}{2}\right) = \left(\frac{7}{2}\right)^2 + b_2\left(\frac{7}{2}\right) + c_2 \]
- We need to find the least value of \( g(x) \).

6. **Using the relationship between \( f(x) \) and \( g(x) \)**:
- The roots of \( g(x) \) are shifted by \( \gamma \) from the roots of \( f(x) \).
- The least value of \( g(x) \) will also be \( -\frac{1}{4} \) because shifting the roots does not change the minimum value of the quadratic function.

7. **Conclusion**:
- Therefore, the least value of \( g(x) \) is also \( -\frac{1}{4} \).

### Final Answer:
The least value of \( g(x) \) is \( -\frac{1}{4} \).