Question

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

Answer 1

$f\left(1\right)=a+b+c=3\cdots \left(i\right)$
$f\left(3\right)=9a+3b+c=4\cdots \left(ii\right)$
$f\left(0\right)+f\left(1\right)+f(-2)+f\left(3\right)=14$
OR $c+3+(4a-2b+c)+4=14$
OR $4a-2b+2c=7\cdots \left(iii\right)$
From (i) and (ii) $8a+2b=1\cdots \left(iv\right)$
From (iii) - $\left(2\right)\times \left(i\right)$
$\Rightarrow 2a-4b=1\cdots \left(v\right)$
From (iv) and (v) $a=\frac{1}{6},b=\frac{-1}{6}$ and $c=3$
$f(-2)=4a-2b+c$
$=\frac{4}{6}+\frac{2}{6}+3=4$