Discussion
Let the Legendre equation \((1 - x^{2})y'' - 2xy' + n(n+1)y = 0\) have \(n^{th}\) degree polynomial solution \(y_{n}(x)\) such that \(y_{n}(1) = 3\). If \(\int_{-1}^{1} \left(y_{n}^{2}(x) + y_{n-1}^{2}(x)\right) dx = \frac{144}{15}\), then \(n\) is
Answer: B
The value is \(n = 2\).
The polynomial solutions are multiples of the Legendre polynomials, and \(P_{n}(1) = 1\), so \(y_{n} = 3P_{n}\); the paper's normalisation applies to every degree, so \(y_{n-1} = 3P_{n-1}\) as well. Using \(\int_{-1}^{1}P_{n}^{2}\,dx = \frac{2}{2n+1}\):
\[ \begin{aligned} 9\left(\tfrac{2}{2n+1} + \tfrac{2}{2n-1}\right) &= \tfrac{144}{15} \\ \tfrac{72n}{4n^{2} - 1} &= \tfrac{144}{15} \end{aligned} \]
For \(n = 2\): \(\frac{144}{15}\), as required. Check the others: \(n = 1\) gives \(\frac{72}{3} = 24\), \(n = 3\) gives \(\frac{216}{35}\), and \(n = 4\) gives \(\frac{288}{63}\). So only \(n = 2\) works.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development