[JEE Main 2013] The number of values of $k$ for which the system $(k+1)x+8y=4k$, $kx+(k+3)y=3k-1$ has no solution is
The number of values of $k$ for which the system $(k+1)x+8y=4k$, $kx+(k+3)y=3k-1$ has no solution is
(a) infinite
(b) $1$
(c) $2$
(d) $3$
1 Answer
VAVidaara Admin
✓ Vidaara Team
✓ Accepted
· 1mo ago
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Correct answer: (b) $1$
Determinant $(k-1)(k-3)=0$. At $k=1$ the equations are proportional (infinite solutions); at $k=3$ they are inconsistent (no solution). So exactly $1$ value.
JEE Main 2013 · Quadratic Equations and Inequations — verified solution by the Vidaara Team.
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