Algebra & Trigonometry
Quadratic Equations with Trigonometric Coefficients
GRB_1000_SCQ
Grade Class 11

Question:

If α and β are the roots of the equation x² − x sin 2θ + 2cos²θ = 0, θ ∈ R and the maximum value of (2 − α)(2 − β) is (a + √a), then a is equal to:
2
3
4
5

Step-by-Step Solution

Key Concept: Sum and product of roots of quadratic; trigonometric optimization
Step 1: Identify the sum and product of roots using Vieta's formulas. Since $\alpha$ and $\beta$ are roots of the equation $x^2 - x\sin 2\theta + 2\cos^2\theta = 0$, by Vieta's formulas: $$\alpha + \beta = \sin 2\theta$$ $$\alpha\beta = 2\cos^2\theta$$ Step 2: Expand the expression $(2-\alpha)(2-\beta)$. We need to find the maximum value of $(2-\alpha)(2-\beta)$. Expanding this expression: $$(2-\alpha)(2-\beta) = 4 - 2\alpha - 2\beta + \alpha\beta$$ $$= 4 - 2(\alpha + \beta) + \alpha\beta$$ Step 3: Substitute the values of $\alpha + \beta$ and $\alpha\beta$. Substituting the expressions from Step 1: $$(2-\alpha)(2-\beta) = 4 - 2\sin 2\theta + 2\cos^2\theta$$ Step 4: Simplify using the identity $\cos^2\theta = \frac{1 + \cos 2\theta}{2}$. Using the double angle formula $2\cos^2\theta = 1 + \cos 2\theta$: $$(2-\alpha)(2-\beta) = 4 - 2\sin 2\theta + 1 + \cos 2\theta$$ $$= 5 - 2\sin 2\theta + \cos 2\theta$$ Step 5: Find the maximum value of the expression $-2\sin 2\theta + \cos 2\theta$. For an expression of the form $A\sin\phi + B\cos\phi$, the maximum value is $\sqrt{A^2 + B^2}$. Here, we have $-2\sin 2\theta + \cos 2\theta$, so: $$\text{Maximum of } (-2\sin 2\theta + \cos 2\theta) = \sqrt{(-2)^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5}$$ Step 6: Determine the maximum value of $(2-\alpha)(2-\beta)$. The maximum value of $(2-\alpha)(2-\beta)$ is: $$\text{Maximum} = 5 + \sqrt{5}$$ Step 7: Match with the given form $a + \sqrt{a}$ to find $a$. We are given that the maximum value equals $a + \sqrt{a}$. Comparing: $$a + \sqrt{a} = 5 + \sqrt{5}$$ This means $a = 5$. However, let us verify: if $a = 5$, then $a + \sqrt{a} = 5 + \sqrt{5}$ ✓ But checking the options, we need to reconsider. Testing $a = 4$: $$a + \sqrt{a} = 4 + \sqrt{4} = 4 + 2 = 6 \neq 5 + \sqrt{5}$$ Testing $a = 5$: $$a + \sqrt{a} = 5 + \sqrt{5}$$ ✓ Upon careful review of the problem statement and the given correct answer, there appears to be a discrepancy. However, following the mathematical derivation strictly, the maximum value is $5 + \sqrt{5}$, which corresponds to $a = 5$. **Final Answer: $a = 5$** (Option 4)
Correct Answer: 4

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