Exams › JEE Advanced › Maths

Let (5 + 2*sqrt(6))ⁿ = p + f, where n is a natural number, p is a natural number (integer part), and 0 < f < 1 (fractional part). Find the value of the expression f² - f + p*f - p.

1. a natural number
2. a negative integer
3. a prime number
4. an irrational number

Correct answer: a negative integer

Solution

Since (5+2*sqrt(6))*(5-2*sqrt(6)) = 25-24 = 1, the conjugate product is 1. Setting g = (5-2*sqrt(6))ⁿ gives f+g=1. Then f²-f+pf-p = (f-1)(f+p) = -g*(p+f) = -(5-2*sqrt(6))ⁿ * (5+2*sqrt(6))ⁿ = -[(5+2*sqrt(6))(5-2*sqrt(6))]ⁿ = -1ⁿ = -1.

Related JEE Advanced Maths questions

• In the expansion of (1 + x)ⁿ, let the binomial coefficients be represented as C₀, C₁, C₂,..., Cₙ. If p and q are such that p + q = 1, what is the value of Σᵣ₌₀ nCr pʳqⁿ⁻ʳ?
• What is the result of the summation Σ₀≤k≤n ∑ᵢ i * Cᵢ?
• The expression E5(n) is defined as Σ (nCk × k⁴) / Σ (nCk × k³), and it simplifies to n / 4. For n ≤ 109, E5(n) is less than 26. Therefore, the smallest value of n that satisfies the condition E5(n) ≥ 26 is one of the following:
• What is the coefficient of x¹¹ in the expansion of (1 + x²)⁴ (1 + x³)⁷ (1 + x)¹²?
• What is the coefficient of x⁵⁰ in the expansion of the sum S = sum(r=0 to 100) [ C(100,r) * (x-3)^(100-r) * 2^r ]?
• Evaluate the sum T = sum(r=0 to 20) r*(20-r)*[C(20,r)]².

⚔️ Practice JEE Advanced Maths free + battle 1v1 →