Let `f(x)=ax^(2)+bx+c`, `a ne 0`, `a`, `b`, `c in I`. Suppose that `f(1)=0`, `50 lt f(7) lt 60 ` and `70 lt f(8) lt 80`.
The least value of `f(x)` is

Options

A. `3//4`
B. `9//2`
C. `-9//8`
D. `3//4`

Explanation

To solve the problem step by step, we will follow the given conditions and derive the necessary equations.

### Step 1: Write the function and the first condition
Given the function:
\[ f(x) = ax^2 + bx + c \]
We know that:
\[ f(1) = 0 \]
This implies:
\[ a(1)^2 + b(1) + c = 0 \]
Thus, we have:
\[ a + b + c = 0 \quad \text{(Equation 1)} \]

### Step 2: Use the second condition
We know that:
\[ 50 < f(7) < 60 \]
Calculating \( f(7) \):
\[ f(7) = a(7^2) + b(7) + c = 49a + 7b + c \]
From Equation 1, we can substitute \( c \):
\[ f(7) = 49a + 7b - (a + b) = 48a + 6b \]
Thus, we have:
\[ 50 < 48a + 6b < 60 \quad \text{(Equation 2)} \]

### Step 3: Simplify Equation 2
Dividing the entire inequality by 6:
\[ \frac{50}{6} < 8a + b < \frac{60}{6} \]
This simplifies to:
\[ 8.33 < 8a + b < 10 \]
Since \( a \) and \( b \) are integers, the only integer value for \( 8a + b \) is:
\[ 8a + b = 9 \quad \text{(Equation 3)} \]

### Step 4: Use the third condition
Now, we know:
\[ 70 < f(8) < 80 \]
Calculating \( f(8) \):
\[ f(8) = a(8^2) + b(8) + c = 64a + 8b + c \]
Substituting \( c \) from Equation 1:
\[ f(8) = 64a + 8b - (a + b) = 63a + 7b \]
Thus, we have:
\[ 70 < 63a + 7b < 80 \quad \text{(Equation 4)} \]

### Step 5: Simplify Equation 4
Dividing the entire inequality by 7:
\[ \frac{70}{7} < 9a + b < \frac{80}{7} \]
This simplifies to:
\[ 10 < 9a + b < 11.43 \]
The only integer value for \( 9a + b \) is:
\[ 9a + b = 11 \quad \text{(Equation 5)} \]

### Step 6: Solve Equations 3 and 5
Now we have two equations:
1. \( 8a + b = 9 \) (Equation 3)
2. \( 9a + b = 11 \) (Equation 5)

Subtract Equation 3 from Equation 5:
\[ (9a + b) - (8a + b) = 11 - 9 \]
This simplifies to:
\[ a = 2 \]

Substituting \( a = 2 \) back into Equation 3:
\[ 8(2) + b = 9 \]
\[ 16 + b = 9 \]
Thus:
\[ b = 9 - 16 = -7 \]

### Step 7: Find \( c \)
Using Equation 1:
\[ a + b + c = 0 \]
Substituting \( a = 2 \) and \( b = -7 \):
\[ 2 - 7 + c = 0 \]
Thus:
\[ c = 5 \]

### Step 8: Write the function
Now we have:
\[ f(x) = 2x^2 - 7x + 5 \]

### Step 9: Find the least value of \( f(x) \)
To find the minimum value of a quadratic function \( ax^2 + bx + c \), we use the vertex formula:
\[ x = -\frac{b}{2a} = -\frac{-7}{2 \cdot 2} = \frac{7}{4} \]

Now substituting \( x = \frac{7}{4} \) into \( f(x) \):
\[ f\left(\frac{7}{4}\right) = 2\left(\frac{7}{4}\right)^2 - 7\left(\frac{7}{4}\right) + 5 \]
Calculating:
\[ = 2 \cdot \frac{49}{16} - \frac{49}{4} + 5 \]
\[ = \frac{98}{16} - \frac{196}{16} + \frac{80}{16} \]
\[ = \frac{98 - 196 + 80}{16} = \frac{-18}{16} = -\frac{9}{8} \]

Thus, the least value of \( f(x) \) is:
\[ \boxed{-\frac{9}{8}} \]