Let f(x) = a x ^ 2 + bx + c be such that f(1) = 3, f(- 2) = λ and...

Given $f\left(x\right) = a{x}^{2} + bx + c$ be such that $f\left(1\right) = 3, f\left(- 2\right) = \lambda$ and $f\left(3\right) = 4$. If $f\left(0\right) + f\left(1\right) + f\left(- 2\right) + f\left(3\right) = 14$, then $\lambda$ is equal to

Now given $f\left(0\right) + f\left(1\right) + f\left(- 2\right) + f\left(3\right) = 14$

$⇒ f\left(0\right) + 3 + \lambda + 4 = 14$

$⇒ f\left(0\right) = 7 - \lambda = c$

Now $f\left(1\right) = a + b + c = 3 . . . \left(i\right)$

$f\left(3\right) = 9a + 3b + c = 4 . . . \left(ii\right)</p><p>$f\left(- 2\right)$= 4a - 2b + c = \lambda . . . \left(iii\right)$

Now subtracting equation \left(ii\right) - \left(iii\right)

We get, $a + b =$\dfrac{4 - \lambda}{5}, now putting$in equation\left(i\right)</p><p>We get, \dfrac{4 - \lambda}{5} +7 - \lambda = 3$

$⇒ 6\lambda = 24$

$⇒ \lambda = 4$