JEE Main 2021 · 26 February · evening · Single correct · +4 / −1
Let f(x) = ∫₀ˣ eᵗf(t)dt + eˣ be a differentiable function for all x∈R. Then f(x) equals :
- (a) e^((eˣ⁻¹))
- (b) 2e^(eˣ) - 1
- (c) 2e^(eˣ - 1) - 1
- (d) e^(eˣ) - 1
Answer and solution
Answer (c) 2e^(eˣ - 1) - 1
Solution
Maths sir · 4 steps
Differentiate both sides to get a differential equation, then let x = 0 fix the constant.
- Differentiate. By the Leibniz rule, d/dx∫₀ˣ eᵗ f(t) dt = eˣ f(x). So f'(x) = eˣ f(x) + eˣ = eˣ(f(x) + 1).
- Starting value. Put x = 0 in the original equation. The integral from 0 to 0 is zero, so f(0) = e⁰ = 1.
- Separate and integrate. Write it as f'(x)/(f(x)+1) = eˣ. Integrating both sides gives ln(f(x)+1) = eˣ + C.
ln 2 = 1 + C ⇒ f(x) + 1 = e^(eˣ + ln 2 - 1) = 2e^(eˣ - 1)
- So. f(x) = 2e^(eˣ - 1) - 1. Check: f(0) = 2e⁰ - 1 = 1. The answer is (c).
The idea to keep. When f sits both outside and inside an integral with a variable limit, differentiate to get a differential equation. Then put the variable limit equal to the lower limit to find the constant.
Answer from the NTA final answer key. Solution by PrepWiser.