MHTCET 18 Apr 2026 Shift 2 Paper

Concept:Each new function adds one more logarithm, so the integrand is exactly the derivative of $f_{2027}(x)$.Explanation:Given $f(x) = x$, we have:$f_1(x) = f(\log x) = \log x$$f_2(x) = f_1(\log x) = \log(\log x)$Continuing this pattern, $f_n(x)$ contains $n$ nested logarithms.Hence the denominator becomes:$f(x) \cdot f_1(x) \cdot f_2(x) \cdots f_{2026}(x) = x \cdot \log x \cdot \log(\log x) \cdots f_{2026}(x)$.Now observe that $f_{2027}(x) = \log(f_{2026}(x))$.Differentiating using the chain rule:$\frac{d}{dx} f_{2027}(x) = \frac{1}{f_{2026}(x)} \cdot \frac{d}{dx} f_{2026}(x)$.Applying the chain rule repeatedly backward, each step contributes one factor from the denominator:$\frac{d}{dx} f_{2027}(x) = \frac{1}{f(x) f_1(x) f_2(x) \cdots f_{2026}(x)}$.Therefore:$\int \frac{1}{f(x) f_1(x) f_2(x) \cdots f_{2026}(x)} dx = f_{2027}(x) + c$.Answer:Option C: $f_{2027}(x) + c$