Q. If $c < 1$ and the system of equations $x + y - 1 = 0,\ 2x - y - c = 0\ and - bx + 3by - c =$0 is consistent, then the possible real values of $b$ are
A) $b \in \left( \ - 3,\frac{3}{4} \right)$
B) $b \in \left( \ - \frac{3}{4},4 \right)$
C) $b \in \left( \ - \frac{3}{4},3 \right)$
D) None of these
Explanation
(c)
Since system of equations is consistent.
$\therefore\ \left| \begin{matrix}
1 & 1 & - 1 \\
2 & - 1 & - c \\
- b & 3b & - c
\end{matrix} \right| = 0$
$\Rightarrow c + bc - 6b + b + 2c + 3bc = 0$
$\Rightarrow 3c + 4bc - 5b = 0$
$\Rightarrow c = \frac{5}{3 + 4b}$
But $c < 1 \Rightarrow \ \frac{5b}{3 + 4b} < 1$
$\Rightarrow \ \frac{b - 3}{3 + 4b} < 0$
$\Rightarrow b \in \left( \ - \frac{3}{4},\ 3 \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.