MHT CET202618 April 2026Evening ShiftMathematicsIndefinite IntegrationActual
Let f(x) = x , f₁(x) = f( x) , f₂(x) = f₁( x) , f₃(x) = f₂( x) , and so on. Then 1 f(x) ,f₁(x) ,f₂(x) , f₂₀₂₆(x) ,dx =
Options
- Af₂₀₂₅(x) + c
- B2025 f₂₀₂₅(x) + c
- Cf₂₀₂₇(x) + c
- D2027 f₂₀₂₇(x) + c
Correct answer
C. f₂₀₂₇(x) + c
Step-by-step solution
Given f(x) = x f₁(x) = f( x) = x f₂(x) = f₁( x) = ( x) In general, f_n(x) = (f_ n-1 (x)) . Differentiating f_n(x) with respect to x using the chain rule: f₁'(x) = 1 x = 1 f(x) f₂'(x) = d dx ( (f₁(x))) = 1 f₁(x) f₁'(x) = 1 f₁(x) f(x) f₃'(x) = d dx ( (f₂(x))) = 1 f₂(x) f₂'(x) = 1 f₂(x) f₁(x) f(x) By induction, f_n'(x) = 1 f_ n-1 (x) f_ n-2 (x) f₁(x) f(x) . For n = 2027 , we have: f₂₀₂₇'(x) = 1 f₂₀₂₆(x) f₂₀₂₅(x) f₁(x) f(x) The given integral is: I = 1 f(x) f₁(x) f₂(x) f₂₀₂₆(x) , dx I = f₂₀₂₇'(x) , dx I = f₂₀₂₇(x) + c