JEE Main 2022 · 28 June · evening · Single correct · +4 / −1

Let f : R → R be a differentiable function such that f( π/4 ) = √2 , f( π/2 ) = 0 and f'( π/2 ) = 1 and
let g(x) = ∫ₓ^(π /4) (f'(t)sec t + tan t sec t f(t)) dt for x ∈ [ π/4,π/2 ). Then lim _(x → ( π/2 ) -) g(x) is equal to :

  1. (a) 2
  2. (b) 3
  3. (c) 4
  4. (d) -3
Answer and solution

Answer (b) 3

Solution

Maths sir · 4 steps

That integrand is the derivative of f(t)sec t, so integrate it in one line and then take the limit.

  1. Spot the product rule. The derivative of sec t is sec t tan t. So f'(t)sec t + f(t)sec t tan t is exactly d/dt(f(t)sec t).

    g(x) = [f(t)sec t]ₓ^(π/4) = f(π/4)sec π/4 - f(x)sec x = 2 - f(x)/(cos x)

  2. The tricky term. Here f(π/4)sec π/4 = √2 × √2 = 2. As x → π/2⁻, both f(x) → 0 and cos x → 0, so f(x)/(cos x) is a 0/0 form.
  3. Use L'Hospital. lim f(x)/(cos x) = lim f'(x)/(-sin x) = 1/-1 = -1, using f'(π/2) = 1.
  4. So. lim(x → π/2⁻) g(x) = 2 - (-1) = 3, which is option (b).

The idea to keep. When an integrand looks like f'u + fu', write it as (fu)' straight away. If a limit then gives 0/0, that is a cue for L'Hospital, not a reason to call it zero.

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

Study Definite Integration

Asked about 230 times in JEE Main since 2020. This question is from 2022.