Straight Lines
Inequalities and intersection with lines
Grade 11
Question:
<p>For what values of <em>α</em>, the numbers <br>
\(-3 < \alpha < \dfrac{21}{5}\) holds? (Related to a straight line/inequality problem involving intersection points at \(\left(-\dfrac{17}{3}, 0\right)\), \((-3,0)\), \(\left(\dfrac{15}{2},0\right)\), \(\left(\dfrac{21}{5},0\right)\), \((0,3)\), \(\left(0,-\dfrac{17}{5}\right)\), \(\left(0,-\dfrac{21}{4}\right)\) and the line \(y = x+3\).)</p>
<p>\(\alpha > -3\)</p>
<p>\(\alpha < \dfrac{21}{5}\)</p>
<p>\(-3 < \alpha < \dfrac{21}{5}\)</p>
<p>None of these</p>
Step-by-Step Solution
Key Concept: Use the condition for three numbers to be in AP: the middle term must equal the arithmetic mean of the outer terms, i.e., 2b = a + c. This translates to a quadratic inequality in α that must be solved carefully.
<p><strong>Step 1:</strong> For three numbers to be in AP, the condition is: 2(middle term) = first term + last term</p><p>2(sin α - 1) = -3 + sin²α</p><p><strong>Step 2:</strong> Expand and rearrange:</p><p>2sin α - 2 = -3 + sin²α</p><p>sin²α - 2sin α - 1 = 0</p><p><strong>Step 3:</strong> Let x = sin α. Then x² - 2x - 1 = 0</p><p>Using the quadratic formula: x = (2 ± √(4 + 4))/2 = (2 ± 2√2)/2 = 1 ± √2</p><p><strong>Step 4:</strong> Check validity using domain sin α ∈ [-1, 1]:</p><p>• x = 1 + √2 ≈ 2.414 (outside [-1, 1], rejected)</p><p>• x = 1 - √2 ≈ -0.414 (valid, since -1 < 1 - √2 < 0)</p><p><strong>Step 5:</strong> Solve sin α = 1 - √2:</p><p>α = arcsin(1 - √2) + 2πn or α = π - arcsin(1 - √2) + 2πn, where n ∈ ℤ</p><p>∴ Answer: B and D</p>
Correct Answer: B and D