Let `f(x)=ax^2+bx+c and f(1)=3 , f(-2)=lamda , f(3)=4` then value of `lamda` for which `f(0)+f(1)+f(-2)+f(3)=14` is

Answer

To solve the problem step by step, we start with the function \( f(x) = ax^2 + bx + c \) and the given conditions.

### Step 1: Set up the equations based on given conditions
1. From \( f(1) = 3 \):
\[ a(1^2) + b(1) + c = 3 \implies a + b + c = 3 \quad \text{(Equation 1)} \]

2. From \( f(-2) = \lambda \):
\[ a(-2)^2 + b(-2) + c = \lambda \implies 4a - 2b + c = \lambda \quad \text{(Equation 2)} \]

3. From \( f(3) = 4 \):
\[ a(3^2) + b(3) + c = 4 \implies 9a + 3b + c = 4 \quad \text{(Equation 3)} \]

### Step 2: Write the equation for the sum
We need to find \( \lambda \) such that:
\[ f(0) + f(1) + f(-2) + f(3) = 14 \]
Calculating \( f(0) \):
\[ f(0) = c \]
Thus, we have:
\[ c + f(1) + f(-2) + f(3) = 14 \implies c + 3 + \lambda + 4 = 14 \]
This simplifies to:
\[ c + \lambda + 7 = 14 \implies c + \lambda = 7 \quad \text{(Equation 4)} \]

### Step 3: Solve the system of equations
Now we have four equations:
1. \( a + b + c = 3 \) (Equation 1)
2. \( 4a - 2b + c = \lambda \) (Equation 2)
3. \( 9a + 3b + c = 4 \) (Equation 3)
4. \( c + \lambda = 7 \) (Equation 4)

From Equation 4, we can express \( c \):
\[ c = 7 - \lambda \quad \text{(Equation 5)} \]

### Step 4: Substitute \( c \) into the first three equations
Substituting \( c \) from Equation 5 into Equations 1, 2, and 3:

1. From Equation 1:
\[ a + b + (7 - \lambda) = 3 \implies a + b - \lambda = -4 \implies a + b = \lambda - 4 \quad \text{(Equation 6)} \]

2. From Equation 2:
\[ 4a - 2b + (7 - \lambda) = \lambda \implies 4a - 2b + 7 - \lambda = \lambda \implies 4a - 2b = 2\lambda - 7 \quad \text{(Equation 7)} \]

3. From Equation 3:
\[ 9a + 3b + (7 - \lambda) = 4 \implies 9a + 3b + 7 - \lambda = 4 \implies 9a + 3b = \lambda - 3 \quad \text{(Equation 8)} \]

### Step 5: Solve Equations 6, 7, and 8
From Equation 6, we have \( b = \lambda - 4 - a \). Substitute \( b \) into Equations 7 and 8.

**Substituting into Equation 7:**
\[ 4a - 2(\lambda - 4 - a) = 2\lambda - 7 \]
Simplifying:
\[ 4a - 2\lambda + 8 + 2a = 2\lambda - 7 \implies 6a + 8 = 4\lambda - 7 \implies 6a = 4\lambda - 15 \quad \text{(Equation 9)} \]

**Substituting into Equation 8:**
\[ 9a + 3(\lambda - 4 - a) = \lambda - 3 \]
Simplifying:
\[ 9a + 3\lambda - 12 - 3a = \lambda - 3 \implies 6a + 3\lambda - 12 = \lambda - 3 \implies 6a + 2\lambda = 9 \quad \text{(Equation 10)} \]

### Step 6: Solve Equations 9 and 10
From Equation 9:
\[ a = \frac{4\lambda - 15}{6} \]

Substituting \( a \) into Equation 10:
\[ 6\left(\frac{4\lambda - 15}{6}\right) + 2\lambda = 9 \]
This simplifies to:
\[ 4\lambda - 15 + 2\lambda = 9 \implies 6\lambda - 15 = 9 \implies 6\lambda = 24 \implies \lambda = 4 \]

### Final Answer
Thus, the value of \( \lambda \) is:
\[ \boxed{4} \]