Exams › JEE Advanced › Maths

It is given that the third term in the binomial expansion of (x + x^(log₁₀(x)))⁵ equals 10⁶. Find all possible values of x.

1. x = 10
2. x = 10^(-5/2)
3. x = 100
4. x = 10^(5/2)

Correct answer: x = 10

Solution

Setting T3 = 10⁶ leads to the equation 2t² + 3t - 5 = 0 where t = log₁₀(x), giving t = 1 (x = 10) or t = -5/2 (x = 10^(-5/2)).

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 →