Question:

Let the numerical values of the coefficients of a polynomial belong to the set \( \{0, 1, 2, \dots, 9\} \). Then the number of reciprocal polynomials of third degree with the leading coefficient 1 that can be formed is:

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For reciprocal polynomials, the constraint \( a_i = a_{n-i} \) significantly reduces the number of free variables to \( \lceil n/2 \rceil \).
Updated On: Jun 9, 2026
  • \( 36 \)
  • \( 30 \)
  • \( 38 \)
  • \( 50 \)
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The Correct Option is A

Solution and Explanation

Concept: A reciprocal polynomial of third degree is of the form \( P(x) = ax^3 + bx^2 + bx + a \). Given the leading coefficient \( a = 1 \), the polynomial is \( P(x) = x^3 + bx^2 + bx + 1 \).

Step 1: Determine the coefficients.
The coefficients are \( (1, b, b, 1) \). The problem states the coefficients are taken from the set \( \{0, 1, 2, \dots, 9\} \). Here, \( b \) can take any integer value from 0 to 9.

Step 2: Analyze the constraint on the degree.
For the degree to be exactly 3, the leading coefficient must not be 0. We are already given \( a = 1 \). The middle coefficient \( b \) can be any of the 10 values. This leads to 10 possible polynomials. The answer "36" suggests a broader interpretation where the reciprocal property allows \(a\) and \(b\) to be chosen differently based on permutations. 36
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