Rational Term
Within the expansion of (\sqrt{3} + \sqrt[3]{2})^{6}, the number of terms that turn out to be rational, free of any surd, is counted to be which value?
Select the correct option:
Solution
2
A term is rational only when every fractional exponent simplifies to an integer, so locating rational terms means imposing simultaneous integrality conditions on the general term, a precise JEE Advanced surd analysis. The general term is T_{r+1} = C(6, r) (3^{1/2})^{6-r} (2^{1/3})^{r} = C(6, r) 3^{(6-r)/2} 2^{r/3}. For rationality we need (6 - r)/2 to be an integer and r/3 to be an integer simultaneously, with r ranging from 0 to 6. The condition r/3 ∈ Z forces r ∈ {0, 3, 6}. Among these we test (6 - r)/2: for r = 0 it is 3 (integer, valid); for r = 3 it is 3/2 (not an integer, rejected); for r = 6 it is 0 (integer, valid). Thus exactly two values, r = 0 and r = 6, satisfy both conditions, giving two rational terms. At r = 0 the term is C(6,0) 3^3 = 27, and at r = 6 it is C(6,6) 2^2 = 4, both free of any surd. Hence the count is 2. Option 1 undercounts by dropping one of the two valid values of r. Option 3 wrongly accepts r = 3, which leaves the half-integer exponent 3/2 on 3 and so retains a square-root factor. Option 4 grossly overestimates. Plausibility check: r/3 ∈ Z gives candidates {0, 3, 6}, and the extra parity requirement 2 | (6 - r) discards only r = 3, leaving exactly two rational terms.
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About This Question
- Subject
- mathematics
- Chapter
- binomial theorem and its simple applications
- Topic
- rational term
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
2
A term is rational only when every fractional exponent simplifies to an integer, so locating rational terms means imposing simultaneous integrality conditions on the general term, a precise JEE Advanced surd analysis. The general term is T_{r+1} = C(6, r) (3^{1/2})^{6-r} (2^{1/3})^{r} = C(6, r) 3^{(6-r)/2} 2^{r/3}. For rationality we need (6 - r)/2 to be an integer and r/3 to be an integer simultaneously, with r ranging from 0 to 6. The condition r/3 ∈ Z forces r ∈ {0, 3, 6}. Among these we test (6 - r)/2: for r = 0 it is 3 (integer, valid); for r = 3 it is 3/2 (not an integer, rejected); for r = 6 it is 0 (integer, valid). Thus exactly two values, r = 0 and r = 6, satisfy both conditions, giving two rational terms. At r = 0 the term is C(6,0) 3^3 = 27, and at r = 6 it is C(6,6) 2^2 = 4, both free of any surd. Hence the count is 2. Option 1 undercounts by dropping one of the two valid values of r. Option 3 wrongly accepts r = 3, which leaves the half-integer exponent 3/2 on 3 and so retains a square-root factor. Option 4 grossly overestimates. Plausibility check: r/3 ∈ Z gives candidates {0, 3, 6}, and the extra parity requirement 2 | (6 - r) discards only r = 3, leaving exactly two rational terms.
This hard difficulty mathematics question is from the chapter binomial theorem and its simple applications, covering the topic of rational term. It appeared in the 2025 exam.
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