[JEE Advanced 1978] sketch the solution set of x 2 y 2 2x 0 3x y 12
Sketch the solution set of: $x^2+y^2-2x\ge0$, $3x-y-12\le0$, $y-x\le0$, $y\ge0$.
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· 4d ago
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Answer: The region outside/on the circle $(x-1)^2+y^2=1$, below the line $3x-y=12$, below $y=x$, and above the $x$-axis.
Each inequality is a half-plane or circle exterior; their intersection (in the first quadrant, under $y=x$ and the line $3x-y=12$, outside the unit circle centred at $(1,0)$) is the required region.
JEE Advanced 1978 · Quadratic Equations and Inequations — verified solution by the Vidaara Team.
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