Original CUET-style questions on linear inequalities and linear programming, with a worked solution and the usual trap for each. Tap an option, or type a numerical answer, to check it.
Learn the chapters first: Class 11, Linear Inequalities · Class 12, Linear Programming.
Maximise \(Z = 4x + 3y\) subject to \(x + y \le 4\), \(x \le 3\), \(x, y \ge 0\). The maximum value is
Corner points: \((0, 0)\), \((3, 0)\), \((3, 1)\), \((0, 4)\). \(Z\) = 0, 12, 15, 12. Maximum 15 at \((3, 1)\).
Trap: Missing the corner \((3, 1)\), where \(x = 3\) meets \(x + y = 4\), gives 12.
The solution set of \(3x - 5 < x + 7\), \(x \in \mathbb{R}\), is
\(3x - x < 7 + 5\), so \(2x < 12\) and \(x < 6\).
The corner points of a feasible region are \((0, 3)\), \((2, 0)\), \((5, 0)\), \((5, 6)\) and \((0, 6)\). The minimum of \(Z = 3x + 2y\) occurs
\(Z\) at the corners: 6, 6, 15, 27, 12. The minimum 6 is reached at two adjacent corners, so it is reached at every point of the edge joining them (the line \(3x + 2y = 6\)).
Trap: When two adjacent corners tie, the optimum is the whole edge, not one corner.
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