JEE Main 2020 · 4 September · evening · Single correct · +4 / −1
The minimum value of 2ˢⁱⁿˣ + 2ᶜᵒˢˣ is :
- (a) 2^(-1 + √2)
- (b) 2^(1 - 1/√2)
- (c) 2^(1 - √2)
- (d) 2^(-1 + 1/√2)
Answer and solution
Answer (b) 2^(1 - 1/√2)
Solution
Maths sir · 4 steps
Use AM-GM to turn the sum of powers into one power, then push sin x + cos x to its lowest value.
- Apply AM-GM. Both 2ˢⁱⁿˣ and 2ᶜᵒˢˣ are positive, so their sum is at least twice their geometric mean. Multiplying the two powers adds the exponents.
2ˢⁱⁿˣ + 2ᶜᵒˢˣ ≥ 2√2ˢⁱⁿˣ⁺ᶜᵒˢˣ = 2^(1 + (sin x + cos x)/2)
- Lowest exponent. Since sin x + cos x = √2 sin(x + π/4), its least value is -√2. The exponent is then 1 - √2/2 = 1 - 1/√2.
- Check equality. AM-GM is an equality when sin x = cos x, and sin x + cos x = -√2 needs x = 5π/4. At x = 5π/4, sin x = cos x = -1/√2, so both conditions hold together and the bound is reached.
- So. The minimum value is 2^(1 - 1/√2), which is option (b).
The idea to keep. For a sum like aᶠ + aᵍ, AM-GM turns it into 2a^((f+g)/2). Then you only need the extreme value of f + g, and you must confirm both equality conditions can hold at the same x.
Answer from the NTA final answer key. Solution by PrepWiser.