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KCET2026MathematicsBinomial Theorem

The value at x = 2 for x^3 + 3x^2 + 3x + 1 x^4 + 4x^3 + 6x^2 +4x + 1

Options

  1. A3
  2. B25 61
  3. C1 3
  4. D19 73

Correct answer

C. 1 3

Step-by-step solution

The given expression is x^3 + 3x^2 + 3x + 1 x^4 + 4x^3 + 6x^2 + 4x + 1 Using the binomial expansion, the numerator can be written as (x + 1)^3 and the denominator can be written as (x + 1)^4 . The expression simplifies to (x + 1)^3 (x + 1)^4 = 1 x + 1 Substituting x = 2 , we get 1 2 + 1 = 1 3 Answer: 1 3

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