Let f(x)=a x^{2}+b x+c be such that f(1)=3, f(-2) =λ and f (3)=4 . If f (0)+ f (1)+ f (-2)+ f (3)=1

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1.Relation and Function

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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

$4$

(JEE MAIN-2022)

Solution

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

$\therefore f (0)=7-\lambda= c$

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

$f (3)=9 a +3 b + c =4 \quad$…(ii)

$f (-2)=4 a -2 b + c =\lambda \quad$…(iii)

$(ii) – (iii)$

$a+b=\frac{4-\lambda}{5}$ put in equation (i)

$\frac{4-\lambda}{5}+7-\lambda=3$

$6 \lambda=24 ; \quad \lambda=4$

Standard 12

Mathematics

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