The sum of the absolute maximum and absolute minimum values of the function f(x)=tan ⁻¹(sin x-cos x) in the interval [0, π] is :

Solution

Maths sir · 4 steps

Since tan⁻¹ only ever increases, find the range of sin x - cos x first, then apply tan⁻¹.

1. Inside function. Let g(x) = sin x - cos x = √2 sin(x - π/4). Because tan⁻¹ is increasing, the largest and smallest values of f come from the largest and smallest values of g.
2. Range on the interval. For x ∈ [0, π], the angle x - π/4 runs over [-π/4, 3π/4]. On this stretch the sine goes from -1/√2 up to 1, so g goes from -1 (at x = 0) up to √2 (at x = 3π/4).

   fₘₐₓ + fₘᵢₙ = tan⁻¹√2 + tan⁻¹(-1) = tan⁻¹√2 - π/4

3. Match the option. Let θ = tan⁻¹√2. Take a right triangle with opposite side √2 and adjacent side 1. Its hypotenuse is √3, so cos θ = 1/√3 and tan⁻¹√2 = cos⁻¹ 1/√3.
4. So. The sum is cos⁻¹(1/√3) - π/4, which is option (c).

The idea to keep. For f(g(x)) with f monotonic, you never need to differentiate. Find the range of g on the given interval, and remember that a restricted interval can cut off part of the full-period range.

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