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 :

  1. 9/29

  2. 91/213

  3. 7/27

  4. 11/212

Show Answer

Answer :

91/213

Solution :

2tan2θ – 5secθ – 1 = 0
⇒ 2(sec2θ – 1) – 5secθ – 1 = 0
⇒ 2sec2θ – 5secθ – 3 = 0
⇒ 2sec2θ – 6secθ + secθ – 3 = 0
⇒ (2secθ + 1)(secθ – 3) = 0
secθ = 3
⇒ cosθ = 1/3
2 solutions in [0, 2π]
2 solutions in [2π, 4π]
2 solutions in [4π, 6π]
1 solution in [6π, 13π/2]
⇒ n = 13

Report
More Similar Tests

Related Tests