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KCET · Maths · Binomial Theorem

The value at \(x = 2\) for \(\dfrac{x^3 + 3x^2 + 3x + 1}{x^4 + 4x^3 + 6x^2 +4x + 1}\)

  1. A \(3\)
  2. B \(\dfrac{25}{61}\)
  3. C \(\dfrac{1}{3}\)
  4. D \(\dfrac{19}{73}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\dfrac{1}{3}\)

Step-by-step Solution

Detailed explanation

The given expression is \(\dfrac{x^3 + 3x^2 + 3x + 1}{x^4 + 4x^3 + 6x^2 + 4x + 1}\)

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 \(\dfrac{(x + 1)^3}{(x + 1)^4} = \dfrac{1}{x + 1}\)

Substituting \(x = 2\), we get \(\dfrac{1}{2 + 1} = \dfrac{1}{3}\)

Answer: \(\dfrac{1}{3}\)