Let the range of the function f(x)=1/(2+sin 3 x+cos 3 x), x ∈ ℝ be [a, b]. If α and β ar respectively the A.M. and the G.M. of a and b, then α/β is equal to

Solution

Maths sir · 4 steps

Squeeze sin 3x + cos 3x between -√2 and √2 first, then flip for the reciprocal.

1. Bound the sum. For any angle, sin θ + cos θ = √2 sin(θ + π/4), so it runs over [-√2, √2]. With θ = 3x, the denominator 2 + sin 3x + cos 3x runs over [2-√2, 2+√2].
2. Take reciprocals. The denominator is always positive, so taking reciprocals reverses the order. The largest denominator gives the smallest value of f.

   a = 1/(2+√2) = (2-√2)/2, b = 1/(2-√2) = (2+√2)/2

3. AM and GM. α = (a+b)/2 = 1/2×4/2 = 1. Also ab = 1/(2+√2)(2-√2) = 1/(4-2) = 1/2, so β = √ab = 1/√2.
4. So. α/β = 1/(1/√2) = √2, which is option (c).

The idea to keep. For p sin θ + q cos θ the range is [-√(p²+q²), √(p²+q²)]. For a reciprocal, check that the denominator never hits zero, then swap the ends.

Answer from the NTA final answer key. Solution by PrepWiser.