JEE Main 2021 · 31 August · evening · Single correct · +4 / −1
The number of solutions of the equation 32^(tan ²x) + 32^(sec ²x) = 81, 0 ≤ x ≤ π/4 is :
- (a) 3
- (b) 1
- (c) 0
- (d) 2
Answer and solution
Answer (b) 1
Solution
Maths sir · 3 steps
Write sec² x as 1 + tan² x, and both terms become multiples of one power of 32.
- Same base, one unknown. Since sec² x = 1 + tan² x, we get 32^(sec² x) = 32^(1+tan² x) = 32 × 32^(tan² x). Put t = 32^(tan² x).
t + 32t = 81 ⇒ 33t = 81 ⇒ t = 27/11
- Check the range. For 0 ≤ x ≤ π/4, tan x rises from 0 to 1, so tan² x rises from 0 to 1. So t = 32^(tan² x) rises steadily from 1 to 32, and it takes each value in between exactly once.
- So. 27/11 ≈ 2.45 lies between 1 and 32, so exactly one x in the interval works. The answer is 1, option (b).
The idea to keep. When tan² x and sec² x sit in exponents, use sec² x = 1 + tan² x to get one base. Then count solutions by checking whether the value fits in the range on the given interval.
Answer from the NTA final answer key. Solution by PrepWiser.