<p>The number of integral values of b for which the origin and the point (1, 1) lie on the same side of straight line \(a^2x + aby + 1 = 0\) for \(a \in \mathbb{R} - \{0\}\) is.</p>
Step-by-Step Solution
Key Concept: Two points lie on the same side of a line if the line equation evaluated at both points has the same sign. We need to find when the origin (0,0) and point (1,1) give the same sign when substituted into the line equation.
<p><strong>Step 1:</strong> For two points to lie on the same side of a line, when we substitute their coordinates into the line equation, we must get values with the same sign.</p><p><strong>Step 2:</strong> At origin (0,0): Substituting into a²x + aby + 1 = 0 gives f(0,0) = 1</p><p><strong>Step 3:</strong> At point (1,1): Substituting into a²x + aby + 1 = 0 gives f(1,1) = a² + ab + 1</p><p><strong>Step 4:</strong> For same side, we need f(0,0) · f(1,1) > 0, which means: 1 · (a² + ab + 1) > 0</p><p><strong>Step 5:</strong> This requires: a² + ab + 1 > 0 for ALL a ∈ ℝ - {0}</p><p><strong>Step 6:</strong> Treating this as a quadratic in a: a² + ba + 1 > 0 for all non-zero real a. For a quadratic Aa² + Ba + C to be always positive, we need discriminant < 0 (since A = 1 > 0).</p><p><strong>Step 7:</strong> Discriminant: D = b² - 4(1)(1) = b² - 4 < 0</p><p><strong>Step 8:</strong> This gives: b² < 4, so -2 < b < 2</p><p><strong>Step 9:</strong> The integral values of b satisfying -2 < b < 2 are: b ∈ {-1, 0, 1}</p><p><strong>Step 10:</strong> We also need to verify these work at a ≠ 0. Since a² + ab + 1 > 0 for all a ∈ ℝ when -2 < b < 2, this condition is satisfied.</p><p><strong>∴ Answer:</strong> The number of integral values of b is 3</p>
Correct Answer: 1