Mathematics · 2023
JEE Main · 24 January 2023, Shift 1 · Q66
Let α be a root of the equation (a-c) x^2+(b-a) x+(c-b)=0 where a, b, c are distinct real numbers such that the matrix [α^2, α, 1; 1, 1, 1; a, b, c]…
Let $\displaystyle \alpha$ be a root of the equation $\displaystyle (a-c) x^2+(b-a) x+(c-b)=0$
where $\displaystyle \mathrm{a}, \mathrm{b}, \mathrm{c}$ are distinct real numbers such that the matrix $\displaystyle \left[\begin{array}{ccc}\alpha^2 & \alpha & 1 \\ 1 & 1 & 1 \\ a & b & c\end{array}\right]$
is singular. Then, the value of $\displaystyle \frac{(a-c)^2}{(b-a)(c-b)}+\frac{(b-a)^2}{(a-c)(c-b)}+\frac{(c-b)^2}{(a-c)(b-a)}$ is
Official answer
From NTA’s final answer key for this paper.
(2)
$\displaystyle 3$
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JEE Main 2023 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.