Mathematics · 2022
JEE Main · 28 July 2022, Shift 2 · Q64
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)=14, then λ is equal to:
Let $\displaystyle f(x)=a x^2+b x+\mathrm{c}$ be such that $\displaystyle f(1)=3, f(-2)=\lambda$ and $\displaystyle f(3)=4$. If $\displaystyle f(0)+f(1)+f(-2)+f(3)=14$, then $\displaystyle \lambda$ is equal to :
Official answer
(4)
$\displaystyle 4$
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JEE Main 2022 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.