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1.Relation and Function
hard
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