Question
If 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 ∑ k = 1 n k 2 n is equal to
Options :
9/29
91/213
7/27
11/212
Answer :
Solution :
2tan2θ – 5secθ – 1 = 0⇒ 2(sec2θ – 1) – 5secθ – 1 = 0⇒ 2sec2θ – 5secθ – 3 = 0⇒ 2sec2θ – 6secθ + secθ – 3 = 0⇒ (2secθ + 1)(secθ – 3) = 0secθ = 3⇒ cosθ = 1/32 solutions in [0, 2π]2 solutions in [2π, 4π]2 solutions in [4π, 6π]1 solution in [6π, 13π/2]⇒ n = 13
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