Let f(x) = ax² + bx + c be such that f(1) = 3, f(-2) = λ and f(3) = 4. If f(0) + f(1) + f(-2) + f(3)

Q: Why can't we just solve for $a, b, c$ as numbers immediately?

That is a great question! Notice that $\lambda$ is an unknown parameter. Because $\lambda$ is embedded in the value of $f(-2)$, we cannot treat it as a constant yet. We must treat $\lambda$ as an algebraic entity throughout the process, solving for $a, b,$ and $c$ in terms of $\lambda$ first, and only then using the final condition to 'lock' the value of $\lambda$.

Q: Is there a faster way than solving the system of equations?

You could use Lagrange Interpolation or Newton's Divided Differences, but for a simple quadratic, those methods are often overkill and prone to arithmetic errors. The system of equations method is the most robust and 'JEE-proof' way to handle this. It keeps your logic linear and easy to verify.

Q: I keep getting a different answer for $\lambda$. Where do I usually go wrong?

The most common trap is the sign convention when subtracting equations. When you subtract $(9a + 3b) - (3a + 3b)$, ensure you distribute the negative sign to every term on the right side of the equation. Also, double-check your substitution of $c = 7 - \lambda$ into the $f(-2)$ equation; it is very easy to drop a negative sign there!

Question

Let $f(x) = ax^2 + bx + c$ be such that $f(1) = 3, f(-2) = \lambda$ and $f(3) = 4$. If $f(0) + f(1) + f(-2) + f(3) = 14$, then $\lambda$ is equal to

Options:

(A) $-4$

(B) $\frac{13}{2}$

(C) $\frac{23}{2}$

(D) 4

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are exploring a quadratic function defined by $f(x) = ax^2 + bx + c$. Our goal is to determine the unknown coefficients by utilizing the provided constraints.

We begin by evaluating the function at $x = 0$. Since the terms $ax^2$ and $bx$ vanish, we find:
$$f(0) = c$$
This value represents the $y$-intercept of the parabola.

### The Summation Constraint

The problem provides the following summation constraint:
$$f(0) + f(1) + f(-2) + f(3) = 14$$

Given the values $f(1) = 3$, $f(3) = 4$, and $f(-2) = \lambda$, we substitute these into the equation:
$$c + 3 + \lambda + 4 = 14$$

Simplifying this expression, we obtain:
$$c + \lambda + 7 = 14 \Rightarrow c = 7 - \lambda$$
We have successfully reduced our number of unknowns from three to two.

### Building the System

Next, we translate the known points into a system of linear equations. For $f(1) = 3$:
$$a(1)^2 + b(1) + c = 3 \Rightarrow a + b + c = 3$$
Substituting $c = 7 - \lambda$, we get:
$$a + b = \lambda - 4$$

For $f(3) = 4$:
$$a(3)^2 + b(3) + c = 4 \Rightarrow 9a + 3b + c = 4$$
Substituting $c = 7 - \lambda$, we get:
$$9a + 3b = \lambda - 3$$

To solve for $a$ and $b$, we multiply the first equation by $3$:
$$3a + 3b = 3\lambda - 12$$

Subtracting this from the second equation ($9a + 3b = \lambda - 3$):
$$6a = 9 - 2\lambda \Rightarrow a = \frac{9 - 2\lambda}{6}$$

Substituting $a$ back into the first equation, we solve for $b$:
$$b = \frac{8\lambda - 33}{6}$$

### The Final Convergence

We now use the final constraint $f(-2) = \lambda$ to solve for $\lambda$. Substituting $x = -2$ into the quadratic form $4a - 2b + c = \lambda$:
$$4\left(\frac{9 - 2\lambda}{6}\right) - 2\left(\frac{8\lambda - 33}{6}\right) + (7 - \lambda) = \lambda$$

Multiplying the entire equation by $6$ to clear the denominators:
$$4(9 - 2\lambda) - 2(8\lambda - 33) + 6(7 - \lambda) = 6\lambda$$
$$36 - 8\lambda - 16\lambda + 66 + 42 - 6\lambda = 6\lambda$$

Grouping the terms involving $\lambda$:
$$144 - 30\lambda = 6\lambda$$
$$144 = 36\lambda$$

Dividing both sides by $36$, we arrive at the final result:
**$\lambda = 4$**

Animated Solution for Mathematics - Quadratic Equations: Let $f (x) = a x^{2} + b x + c$ be such that $f (1) = 3, f (- 2) = λ$ and $f (3) = 4$ . If $f (0) + f (1) + f (- 2) + f (3) = 14$ , then $λ$ is equal to

Select Answer:

Visualized Solution

Defining the Function $f (x)$

Using the Summation Condition

Substituting Known Values

Expressing $c$ in terms of $λ$

Equation from $f (1) = 3$

Equation from $f (3) = 4$

Eliminating $b$ to find $a$

Solving for $a$ in terms of $λ$

Solving for $b$ in terms of $λ$

Final Constraint: $f (- 2) = λ$

Substituting $a, b, c$ into the Equation

Solving the Linear Equation

Finding the Final Value of $λ$

The Sigma Insight: Solution of Quadratic Equations

Analyzing the Setup

The Summation Constraint

Building the System

The Final Convergence

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Animated Solution for Mathematics - Quadratic Equations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=ax2+bx+c be such that f(1)=3,f(−2)=λ and f(3)=4. If f(0)+f(1)+f(−2)+f(3)=14, then λ is equal to

Select Answer:

Visualized Solution

Defining the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Using the Summation Condition

Substituting Known Values

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Equation from .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(1)=3

Equation from .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(3)=4

Eliminating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b to find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a in terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b in terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ

Final Constraint: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(−2)=λ

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a,b,c into the Equation