Question:medium

If \(U_n (n = 1, 2)\) denotes the \(n\)-th derivative (\(n = 1, 2\)) of \(U(x) = \frac{Lx + M}{x^2 - 2Bx + C}\) (\(L, M, B, C\) are constants), then \(PU_2 + QU_1 + RU = 0\) holds for:

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When dealing with derivatives of rational functions, use the quotient rule, and when solving for related constants, match the degree of terms on both sides of the equation.
Updated On: Nov 28, 2025
  • \(P = x^2 - 2B, Q = 2x, R = 3x\)
  • \(P = x^2 - 2Bx + C, Q = 4(x - B), R = 2\)
  • \(P = 2x, Q = 2B, R = 2\)
  • \(P = x, Q = x, R = 3\)
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The Correct Option is B

Solution and Explanation

Step 1: The objective is to determine the relationships between \( P, Q, \) and \( R \) in the equation \( PU_2 + QU_1 + RU = 0 \). Here, \( U(x) = \frac{Lx + M}{x^2 - 2Bx + C} \), and \( U_1 \) and \( U_2 \) represent the first and second derivatives of \( U(x) \).

Step 2: Calculate the first and second derivatives of \( U(x) \). Employ the quotient rule to find the first derivative:

\[ U_1(x) = \frac{(x^2 - 2Bx + C)(L) - (Lx + M)(2x - 2B)}{(x^2 - 2Bx + C)^2} \]

Subsequently, compute the second derivative \( U_2(x) \).

Step 3: Substitute \( U_1(x) \) and \( U_2(x) \) into the equation \( PU_2 + QU_1 + RU = 0 \).

Step 4: Solving for \( P, Q, \) and \( R \) yields:

\[ P = x^2 - 2Bx + C, \quad Q = 4(x - B), \quad R = 2 \]

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