Question

Let $f(x)=a x^2+b x+c$ 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

• A. $-4$
• B. $\frac{13}{2}$
• C. $\frac{23}{2}$
• D. 4

Answer 1

Answer: (D)

$ f (0)+3+\lambda+4=14 $
$ \therefore f (0)=7-\lambda= c $
$ f (1)= a + b + c =3 $.....(i)
$ f (3)=9 a +3 b + c =4 $.....(ii)
$ f (-2)=4 a -2 b + c =\lambda$.....(iii)
(ii) - (iii)$ a + b =\frac{4-\lambda}{5} $ put in equation (ii)
$ \frac{4-\lambda}{5}+7-\lambda=3 $
$6 \lambda=24 ; \lambda=4$