If 2 tan ² θ-5 sec θ=1 has exactly 7 solutions in the interval [0, (n π)/2], for the least value of n ∈ N, then Σk=1ⁿ k/2ᵏ is equal to:

Solution

Maths sir · 5 steps

Turn it into a quadratic in sec θ, get cos θ = 1/3, then count roots in order.

1. Make it quadratic. Put tan² θ = sec² θ - 1. The equation becomes 2 sec² θ - 5 sec θ - 3 = 0, which factors as (2 sec θ + 1)(sec θ - 3) = 0.
2. Keep the valid root. sec θ = -1/2 is impossible because |sec θ| ≥ 1. So sec θ = 3, that is cos θ = 1/3. Let α = cos⁻¹ 1/3, which lies in (0, π/2).
3. List the roots. The solutions are θ = 2mπ ± α. In increasing order: α, 2π - α, 2π + α, 4π - α, 4π + α, 6π - α, 6π + α. The 7th root, 6π + α, lies between 6π and 6π + π/2.
4. Least n. With n = 12 the interval ends at 6π and holds only 6 roots. With n = 13 it ends at 13π/2 = 6π + π/2, which takes in the 7th root but not an 8th, since the next root is 8π - α. So n = 13.

   Σₖ₌₁ⁿ k/2ᵏ = 2 - (n+2)/2ⁿ

5. So. This formula comes from subtracting S/2 from S. For n = 13 the sum is 2 - 15/2¹³ = 1/2¹³(2¹⁴ - 15), which is option (d).

The idea to keep. For Σ k xᵏ, write S - xS: the arithmetic part cancels and leaves a plain GP. When counting roots of cos θ = c, list them as 2mπ ± α in order before choosing the interval.

Answer from the NTA final answer key. Solution by PrepWiser.