Let `f(x)=ax^(2)+bx+c`, `g(x)=ax^(2)+qx+r`, where `a`, `b`, `c`,`q`, `r in R` and `a lt 0`. If `alpha`, `beta` are the roots of `f(x)=0` and `alpha+delta`, `beta+delta` are the roots of `g(x)=0`, then

Options

A. `f_(max) gt g_(max)`
B. `f_(max) lt g_(max)`
C. `f_(max) = g_(max)`
D. cant say anything about relation between `f_(max)` and `g_(max)`

Explanation

To solve the problem, we will analyze the functions \( f(x) \) and \( g(x) \) and their maximum values.

### Step 1: Identify the maximum values of \( f(x) \) and \( g(x) \)

Given:
- \( f(x) = ax^2 + bx + c \)
- \( g(x) = ax^2 + qx + r \)

Since \( a < 0 \), both functions are concave down, meaning they have maximum values.

The maximum value of a quadratic function \( f(x) = ax^2 + bx + c \) occurs at \( x = -\frac{b}{2a} \). The maximum value can be calculated as:
\[ f_{\text{max}} = f\left(-\frac{b}{2a}\right) = a\left(-\frac{b}{2a}\right)^2 + b\left(-\frac{b}{2a}\right) + c \]

Calculating \( f_{\text{max}} \):
\[ f_{\text{max}} = a\left(\frac{b^2}{4a^2}\right) - \frac{b^2}{2a} + c = \frac{b^2}{4a} - \frac{2b^2}{4a} + c = \frac{-b^2}{4a} + c \]

### Step 2: Calculate the maximum value of \( g(x) \)

Similarly, for \( g(x) \):
\[ g_{\text{max}} = g\left(-\frac{q}{2a}\right) = a\left(-\frac{q}{2a}\right)^2 + q\left(-\frac{q}{2a}\right) + r \]

Calculating \( g_{\text{max}} \):
\[ g_{\text{max}} = a\left(\frac{q^2}{4a^2}\right) - \frac{q^2}{2a} + r = \frac{q^2}{4a} - \frac{2q^2}{4a} + r = \frac{-q^2}{4a} + r \]

### Step 3: Relate the roots of \( f(x) \) and \( g(x) \)

Given that the roots of \( f(x) = 0 \) are \( \alpha \) and \( \beta \), we have:
\[ \alpha + \beta = -\frac{b}{a} \quad \text{and} \quad \alpha \beta = \frac{c}{a} \]

For \( g(x) = 0 \), the roots are \( \alpha + \delta \) and \( \beta + \delta \):
\[ (\alpha + \delta) + (\beta + \delta) = -\frac{q}{a} \]
This implies:
\[ \alpha + \beta + 2\delta = -\frac{q}{a} \]
Substituting \( \alpha + \beta = -\frac{b}{a} \):
\[ -\frac{b}{a} + 2\delta = -\frac{q}{a} \]
Thus,
\[ 2\delta = \frac{-q + b}{a} \quad \Rightarrow \quad \delta = \frac{-q + b}{2a} \]

### Step 4: Equate the maximum values

From the calculations of maximum values:
\[ f_{\text{max}} = \frac{-b^2}{4a} + c \]
\[ g_{\text{max}} = \frac{-q^2}{4a} + r \]

To find the relationship between \( f_{\text{max}} \) and \( g_{\text{max}} \), we can analyze the expressions:
\[ f_{\text{max}} = g_{\text{max}} \implies \frac{-b^2}{4a} + c = \frac{-q^2}{4a} + r \]

This leads us to conclude that:
\[ f_{\text{max}} = g_{\text{max}} \]

### Conclusion

Thus, we conclude that \( f_{\text{max}} \) is equal to \( g_{\text{max}} \), which corresponds to option 3.