Let f be a real-valued function defined on the interval (−∞, 1) such that e⁻ˣf(x) = 2 + ∫₀ˣ (4t + 1)dt, for all x ∈ (−∞, 1) and let f⁻¹ be the inverse function of f. Then (f⁻¹)'(2) is equal to 1, 1/e, 1/3, 1/2?

e⁻ˣf(x) = 2 + ∫₀ˣ (4t + 1)dt …(i)
f(f⁻¹(x)) = x …(ii)
Differentiating (ii) with respect to x, we get
f'(f⁻¹(x))(f⁻¹(x))' = 1
⇒ (f⁻¹(2))' = 1/f'(f⁻¹(2))
Putting x = 0 in equation (i), we get
f(0) = 2 …(iii)
From (ii) and (iii), f⁻¹(2) = 0
⇒ (f⁻¹(2))' = 1/f'(0)
Now e⁻ˣ(f'(x) - f(x)) = 4x + 1 (differentiating equation (i))
Put x = 0 ⇒ f'(0) - f(0) = 1
⇒ f'(0) - 2 = 1
⇒ f'(0) = 3
⇒ (f⁻¹(2))' = 1/3