If ƒ'(x) = tan –1 (secx + tanx), -π/2<x<π/2,
and ƒ(0) = 0, then ƒ(1) is equal to :

Solution

Maths sir · 4 steps

Turn sec x + tan x into a single tangent, and the inverse tangent opens up into a straight line in x.

1. Half-angle rewrite. Write sec x + tan x = (1 + sin x)/(cos x). Use 1 + sin x = (cos x/2 + sin x/2)² and cos x = cos² x/2 - sin² x/2, then cancel one factor.

   sec x + tan x = (cos x/2 + sin x/2)/(cos x/2 - sin x/2) = (1 + tan x/2)/(1 - tan x/2) = tan(π/4 + x/2)

2. Check the range. For -π/2 < x < π/2, the angle π/4 + x/2 lies between 0 and π/2. That is inside the principal range of tan⁻¹, so f'(x) = π/4 + x/2.
3. Integrate. f(x) = (π x)/4 + x²/4 + C. Putting f(0) = 0 gives C = 0.
4. So. f(1) = π/4 + 1/4 = (π + 1)/4, which is option (c).

The idea to keep. Whenever you see tan⁻¹ of a trig expression, first convert the expression into one tangent of an angle in x. Then confirm that angle lies in (-π/2, π/2) before you cancel.

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