JEE Main 2020 · 5 September · morning · Single correct · +4 / −1
If
∫ (e²ˣ+2eˣ-e⁻ˣ-1)e^((eˣ+e⁻ˣ))dx = g(x)e^((eˣ+e⁻ˣ))+c
where c is a constant of integration, then g(0) is equal to :
- (a) 1
- (b) 2
- (c) e
- (d) e²
Answer and solution
Answer (b) 2
Solution
Maths sir · 4 steps
Split the bracket so the integrand becomes the derivative of (eˣ+1) e^(eˣ+e⁻ˣ).
- Find the pattern. Let u = eˣ + e⁻ˣ, so u' = eˣ - e⁻ˣ. By the product rule, d/dx[g eᵘ] = (g' + g u')eᵘ. Write the bracket in the form g' + g u'.
- Split the bracket. Expand (eˣ+1)(eˣ-e⁻ˣ) = e²ˣ + eˣ - 1 - e⁻ˣ. Adding eˣ gives exactly e²ˣ + 2eˣ - e⁻ˣ - 1, the given bracket.
(e²ˣ+2eˣ-e⁻ˣ-1)eᵘ = [eˣ + (eˣ+1)(eˣ-e⁻ˣ)]eᵘ = d/dx[(eˣ+1) eᵘ]
- Read off g. So the integral is (eˣ+1) e^(eˣ+e⁻ˣ) + c, which gives g(x) = eˣ + 1.
- So. g(0) = e⁰ + 1 = 1 + 1 = 2, which is option (b).
The idea to keep. When the answer has the form g(x)eᶠ⁽ˣ⁾, the integrand must be (g' + g f')eᶠ. Rewrite the bracket in that shape and you can read g straight off.
Answer from the NTA final answer key. Solution by PrepWiser.