Question

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

• A. 3x + 2y = 2
• B. 5x – 8y = –9
• C. 3x – 2y = –2
• D. 5x + 8y = 9

Answer 1

Answer: (D)

5x + 8y = 9

Answer 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: $5x + 8y = 9$.