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

Question

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 $c = 3$

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

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

"

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

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

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