Question:
If \(\alpha\) and \(\beta\) are the roots of the equation \(3x^2 + 2x + 1 = 0\), then the equation whose roots are \(\alpha + \beta^{-1}\) and \(\beta + \alpha^{-1}\) is:
यदि \(\alpha\) और \(\beta\) समीकरण \(3x^2 + 2x + 1 = 0\) के मूल हैं, तो वह समीकरण जिसके मूल \(\alpha + \beta^{-1}\) और \(\beta + \alpha^{-1}\) हैं:
Options:
- \(3x^2 + 8x + 16 = 0\)
- \(3x^2 – 8x + 16 = 0\)
- \(3x^2 + 8x – 16 = 0\)
- \(x^2 + 8x + 16 = 0\)
- \(3x^2 + 8x + 16 = 0\)
- \(3x^2 – 8x + 16 = 0\)
- \(3x^2 + 8x – 16 = 0\)
- \(x^2 + 8x + 16 = 0\)
Correct Answer: a)
Explanation:
From \(3x^2 + 2x + 1 = 0\), \(\alpha + \beta = -\frac{2}{3}\) and \(\alpha\beta = \frac{1}{3}\).
The new roots are \(r_1 = \alpha + \frac{1}{\beta} = \frac{\alpha\beta + 1}{\beta} = \frac{1/3 + 1}{\beta} = \frac{4/3}{\beta}\) and \(r_2 = \frac{4/3}{\alpha}\).
Sum of new roots:
\[S = \frac{4}{3}\left(\frac{1}{\alpha} + \frac{1}{\beta}\right) = \frac{4}{3}\left(\frac{\alpha+\beta}{\alpha\beta}\right) = \frac{4}{3}\left(\frac{-2/3}{1/3}\right) = \frac{4}{3}(-2) = -\frac{8}{3}\]
Product of new roots:
\[P = \left(\frac{4}{3}\right)^2 \frac{1}{\alpha\beta} = \frac{16}{9} \times 3 = \frac{16}{3}\]
The required quadratic equation is:
\[x^2 – Sx + P = 0 \implies x^2 – \left(-\frac{8}{3}\right)x + \frac{16}{3} = 0 \implies 3x^2 + 8x + 16 = 0\]
Hence, the correct option is (a).
\(3x^2 + 2x + 1 = 0\) से, \(\alpha + \beta = -\frac{2}{3}\) तथा \(\alpha\beta = \frac{1}{3}\)।
नए मूल: \(r_1 = \alpha + \frac{1}{\beta} = \frac{4/3}{\beta}\) और \(r_2 = \frac{4/3}{\alpha}\)।
नए मूलों का योगफल:
\[S = \frac{4}{3}\left(\frac{\alpha+\beta}{\alpha\beta}\right) = \frac{4}{3}(-2) = -\frac{8}{3}\]
नए मूलों का गुणनफल:
\[P = \frac{16}{9} \times \frac{1}{1/3} = \frac{16}{3}\]
अभीष्ट द्विघात समीकरण:
\[x^2 – Sx + P = 0 \implies 3x^2 + 8x + 16 = 0\]
अतः सही विकल्प (a) है।