Mathematics · 2024
JEE Main · 8 April 2024, Shift 2 · Q4
If α ≠ a, β ≠ b, γ ≠ c and |α, b, c; a, β, c; a, b, γ| =0, then a/(α-a)+b/(β-b)+(γ)/(γ-c) is equal to:
If $\displaystyle \alpha \neq a, \beta \neq b, \gamma \neq c$ and $\displaystyle \left|\begin{array}{lll}\alpha & b & c \\ a & \beta & c \\ a & b & \gamma\end{array}\right|=0$, then $\displaystyle \frac{a}{\alpha-a}+\frac{b}{\beta-b}+\frac{\gamma}{\gamma-c}$ is equal to :
Official answer
From NTA’s final answer key for this paper.
(2)
$\displaystyle 0$
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JEE Main 2024 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.