Let \( x = \frac{m}{n} \) ( \( m, n \) are co-prime natural numbers) be a solution of the equation \( \cos \left( 2 \sin^{-1} x \right) = \frac{1}{9} \) and let \( \alpha, \beta (\alpha > \beta) \) be the roots of the equation \( mx^2 - nx - m + n = 0 \). Then the point \( (\alpha, \beta) \) lies on the line
- 3x + 2y = 2
- 5x – 8y = –9
- 3x – 2y = –2
- 5x + 8y = 9
The Correct Option is D
Approach Solution - 1
To solve this problem, we need to find the solution \( x = \frac{m}{n} \) for the given trigonometric equation and determine the line on which the point \((\alpha, \beta)\) lies.
- \(\cos \left( 2 \sin^{-1} x \right) = \frac{1}{9}\)
- Use the identity: \(\cos(2\theta) = 1 - 2\sin^2(\theta)\) where \( \theta = \sin^{-1}(x) \).
- Plugging into the identity: \(\cos(2\sin^{-1} x) = 1 - 2x^2\). Thus, \(1 - 2x^2 = \frac{1}{9}\).
- Solving gives: \(2x^2 = 1 - \frac{1}{9} = \frac{8}{9}\) leading to \(x^2 = \frac{4}{9}\).
- Hence, \(x = \frac{2}{3}\) or \(x = -\frac{2}{3}\). Given \( x \) is a fraction of co-prime natural numbers, \(x = \frac{2}{3}\).
- Substitute \( x = \frac{2}{3} \) into \( m = 2 \) and \( n = 3 \). Thus, the quadratic becomes:
- The equation is: \(mx^2 - nx - m + n = 0\), replacing \(m\) and \(n\) gives: \(2x^2 - 3x - 2 + 3 = 0\) which simplifies to \(2x^2 - 3x + 1 = 0\).
- Find the roots \(\alpha\) and \(\beta\) for the quadratic:
- Using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\), with \(a=2\), \(b=-3\), \(c=1\): \(\alpha, \beta = \frac{3 \pm \sqrt{9 - 8}}{4} = \frac{3 \pm 1}{4}\).
- Therefore, \(\alpha = 1\) and \(\beta = \frac{1}{2}\).
- Check which line the point \((1, \frac{1}{2})\) lies on:
- Option \(5x + 8y = 9\): substituting, \(5(1) + 8\left(\frac{1}{2}\right) = 5 + 4 = 9\), simplifies to 9. So, it satisfies.
- Other lines do not satisfy as shown:
- \(3x + 2y = 2:\) Substituting, \(3(1) + 2\left(\frac{1}{2}\right) = 3 + 1 = 4 \neq 2\)
- \(5x – 8y = –9:\) \(5 - 4 \neq -9\)
- \(3x - 2y = -2:\) \(3 - 1 \neq -2\)
Thus, the correct line equation is \(5x + 8y = 9\).
Approach Solution -2
Step 1. Assume \( \sin^{-1} x = \theta \), so that \( \sin \theta = x \).
Step 2. Given \( \cos(2\theta) = \frac{1}{9} \), we use the identity \( \cos(2\theta) = 1 - 2\sin^2 \theta \):
\(1 - 2x^2 = \frac{1}{9}\)
\(2x^2 = 1 - \frac{1}{9} = \frac{8}{9}\)
\(x^2 = \frac{4}{9} \implies x = \pm \frac{2}{3}\)
Step 3. Since \( m \) and \( n \) are co-prime natural numbers, we take \( x = \frac{2}{3} \), so \( m = 2 \) and \( n = 3 \).
Step 4. Form the quadratic equation \( mx^2 - nx - m + n = 0 \):
\(2x^2 - 3x - 2 + 3 = 0\)
\(2x^2 - 3x + 1 = 0\)
Step 5. Solve for the roots \( \alpha \) and \( \beta \):
\(x = \frac{3 \pm \sqrt{9 - 4 \cdot 2 \cdot 1}}{2 \cdot 2} = \frac{3 \pm \sqrt{1}}{4}\)
\(x = 1, \, \frac{1}{2}\)
Step 6. Check if the point \( (\alpha, \beta) = (1, \frac{1}{2}) \) satisfies any of the given equations:**
\(5(1) + 8 \left( \frac{1}{2} \right) = 5 + 4 = 9\)
Thus, the point \( (\alpha, \beta) \) lies on the line \( 5x + 8y = 9 \).
The Correct Answer is: \( 5x + 8y = 9 \).
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