Q145 / 240 · KCET 2026 · Mathematics
The value at $x = 2$ for $\dfrac{x^3 + 3x^2 + 3x + 1}{x^4 + 4x^3 + 6x^2 +4x + 1}$
- A$3$
- B$\dfrac{25}{61}$
- Correct answerC$\dfrac{1}{3}$
- D$\dfrac{19}{73}$
Solution
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: $\dfrac{1}{3}$
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: $\dfrac{1}{3}$
Hence, the correct answer is option: 'C'.