Integral Calculus
Integral Calculus
star_batch_jee_advanced_2025
Grade None
Question:
A function $f(x)$ which satisfies the relation $f(x) = e^x + \int_0^x e^t f(t) dt$, then:
f(x) = e^x(1 + x)
f(x) = e^{2x}
f(x) = e^x(1 + \frac{x^2}{2})
f(x) = e^x + xe^x
Step-by-Step Solution
Key Concept: Compare derivatives and use monotonicity along with the constraint that $f'(x) > \frac{1}{1+x^2}$ to establish the integral inequality.
Given $f(x) = e^x + \int_0^1 e^t f(t) dt = e^x + ke^x$ where $k = \int_0^1 f(t) dt$. Computing the integral: $k = \int_0^1 (e^t + ke^t) dt = e + ke - 1 - k$, which gives $k = \frac{e-1}{2-e}$. Therefore $f(x) = e^x\left(1 + \frac{e-1}{2-e}\right) = \frac{e^x(e-1)}{2-e}$. We verify $f(0) = \frac{e-1}{2-e} 0$ for $x > 0$, we have $f''(x) = \frac{3(1+x^2) - 3x(2x)}{(1+x^2)^2} = \frac{3(1-x^2)}{(1+x^2)^2} > 0$ for $x \tan^{-1}x - \tan^{-1}1 \Rightarrow f(x) > \frac{\pi}{4} - \tan^{-1}x$ for $x \geq 1$.
Correct Answer: 3,4