If `p(x)=ax^(2)+bx` and `q(x)=lx^(2)+mx+n` with `p(1)=q(1), p(2)-q(2)=1`, and `p(3)-q(3)=4`, then `p(4)-q(4)` is equal to

Options

A. 7
B. 16
C. 9
D. none of these

Answer

To solve the problem, we need to find \( p(4) - q(4) \) given the equations \( p(x) = ax^2 + bx \) and \( q(x) = lx^2 + mx + n \) along with the conditions provided.

### Step-by-Step Solution:

1. **Define the function \( f(x) \)**:
\[ f(x) = p(x) - q(x) = (ax^2 + bx) - (lx^2 + mx + n) = (a - l)x^2 + (b - m)x - n \]

2. **Use the first condition \( p(1) = q(1) \)**:
\[ f(1) = p(1) - q(1) = 0 \]
This gives us:
\[ (a - l) \cdot 1^2 + (b - m) \cdot 1 - n = 0 \implies a - l + b - m - n = 0 \quad \text{(Equation 1)} \]

3. **Use the second condition \( p(2) - q(2) = 1 \)**:
\[ f(2) = p(2) - q(2) = 1 \]
This gives us:
\[ (a - l) \cdot 2^2 + (b - m) \cdot 2 - n = 1 \implies 4(a - l) + 2(b - m) - n = 1 \quad \text{(Equation 2)} \]

4. **Use the third condition \( p(3) - q(3) = 4 \)**:
\[ f(3) = p(3) - q(3) = 4 \]
This gives us:
\[ (a - l) \cdot 3^2 + (b - m) \cdot 3 - n = 4 \implies 9(a - l) + 3(b - m) - n = 4 \quad \text{(Equation 3)} \]

5. **Now we have a system of equations**:
- From Equation 1: \( a - l + b - m - n = 0 \)
- From Equation 2: \( 4(a - l) + 2(b - m) - n = 1 \)
- From Equation 3: \( 9(a - l) + 3(b - m) - n = 4 \)

6. **Let \( A = a - l \) and \( B = b - m \)**. Rewrite the equations:
- \( A + B - n = 0 \) (Equation 1)
- \( 4A + 2B - n = 1 \) (Equation 2)
- \( 9A + 3B - n = 4 \) (Equation 3)

7. **From Equation 1, express \( n \)**:
\[ n = A + B \]

8. **Substituting \( n \) into Equations 2 and 3**:
- From Equation 2:
\[ 4A + 2B - (A + B) = 1 \implies 3A + B = 1 \quad \text{(Equation 4)} \]
- From Equation 3:
\[ 9A + 3B - (A + B) = 4 \implies 8A + 2B = 4 \quad \text{(Equation 5)} \]

9. **Now solve Equations 4 and 5**:
- From Equation 4: \( B = 1 - 3A \)
- Substitute \( B \) into Equation 5:
\[ 8A + 2(1 - 3A) = 4 \implies 8A + 2 - 6A = 4 \implies 2A = 2 \implies A = 1 \]
- Substitute \( A = 1 \) back into Equation 4:
\[ 3(1) + B = 1 \implies 3 + B = 1 \implies B = -2 \]

10. **Now substitute \( A \) and \( B \) back to find \( n \)**:
\[ n = A + B = 1 - 2 = -1 \]

11. **Now we have \( A = 1 \), \( B = -2 \), and \( n = -1 \)**:
- Hence, \( a - l = 1 \) and \( b - m = -2 \).

12. **Now we can find \( p(4) - q(4) \)**:
\[ f(4) = (a - l) \cdot 4^2 + (b - m) \cdot 4 - n \]
\[ f(4) = 1 \cdot 16 - 2 \cdot 4 - (-1) = 16 - 8 + 1 = 9 \]

### Final Answer:
\[ p(4) - q(4) = 9 \]