[ Let f(x)=a x^2+b x+c be such that f(1)=3, f(-2)= and ] f(3)=4.

Question

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

Accepted Answer

$f(1) = a + b + c = 3$ ..... (i)

$f(3) = 9a + 3b + c = 4$ .... (ii)

$f(0) + f(1) + f( - 2) + f(3) = 14$

OR $c + 3 + (4a - 2b + c) + 4 = 14$

OR $4a - 2b + 2c = 7$ ..... (iii)

From (i) and (ii) $8a + 2b = 1$ ..... (iv)

From (iii) $$ - (2) \times $$ (i)

$$ \Rightarrow 2a - 4b = 1$$ ..... (v)

From (iv) and (v) $$a = {1 \over 6},\,b = {{ - 1} \over 6}$$ and $c = 3$

$f( - 2) = 4a - 2b + c$

$$ = {4 \over 6} + {2 \over 6} + 3 = 4$$

Solutions · 1