If 2tan²θ – 5secθ = 1 has exactly 7 solutions in the interval [0, n 2 ], for the least value of n ∈ N, then _ k=1 ^n k 2^k is equal to:

JEE Main 2024 Mathematics question

Not in JEE Main syllabus

If 2tan²θ – 5secθ = 1 has exactly 7 solutions in the interval , for the least value of n ∈ N, then is equal to:

(A)
(B)Correct
(C)
(D)

Answer: (B)

Solution

2(sec²θ − 1) − 5secθ − 1 = 0 gives 2sec²θ − 5secθ − 3 = 0, i.e. (secθ − 3)(2secθ + 1) = 0. Since |secθ| ≥ 1, secθ = 3, cosθ = 1/3. The solutions are θ = 2kπ ± cos⁻¹(1/3), i.e. ≈ 1.23, 5.05,

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Question ID VK-016642