Q90 / 150 · MHT CET 2026 (18 April Shift 2) · Mathematics
Let $f(x) = x$, $f_1(x) = f(\log x)$, $f_2(x) = f_1(\log x)$, $f_3(x) = f_2(\log x)$, $\ldots$ and so on. Then $\int \dfrac{1}{f(x)\,f_1(x)\,f_2(x)\,\ldots f_{2026}(x)}\,dx = \ldots$
- A$f_{2025}(x) + c$
- B$2025 f_{2025}(x) + c$
- Correct answerC$f_{2027}(x) + c$
- D$2027 f_{2027}(x) + c$
Solution
Given $f(x) = x$
$f_1(x) = f(\log x) = \log x$
$f_2(x) = f_1(\log x) = \log(\log x)$
In general, $f_n(x) = \log(f_{n-1}(x))$.
Differentiating $f_n(x)$ with respect to $x$ using the chain rule:
$f_1'(x) = \dfrac{1}{x} = \dfrac{1}{f(x)}$
$f_2'(x) = \dfrac{d}{dx}(\log(f_1(x))) = \dfrac{1}{f_1(x)} f_1'(x) = \dfrac{1}{f_1(x) f(x)}$
$f_3'(x) = \dfrac{d}{dx}(\log(f_2(x))) = \dfrac{1}{f_2(x)} f_2'(x) = \dfrac{1}{f_2(x) f_1(x) f(x)}$
By induction, $f_n'(x) = \dfrac{1}{f_{n-1}(x) f_{n-2}(x) \ldots f_1(x) f(x)}$.
For $n = 2027$, we have:
$f_{2027}'(x) = \dfrac{1}{f_{2026}(x) f_{2025}(x) \ldots f_1(x) f(x)}$
The given integral is:
$I = \int \dfrac{1}{f(x) f_1(x) f_2(x) \ldots f_{2026}(x)} \, dx$
$I = \int f_{2027}'(x) \, dx$
$I = f_{2027}(x) + c$
Answer: $f_{2027}(x) + c$
$f_1(x) = f(\log x) = \log x$
$f_2(x) = f_1(\log x) = \log(\log x)$
In general, $f_n(x) = \log(f_{n-1}(x))$.
Differentiating $f_n(x)$ with respect to $x$ using the chain rule:
$f_1'(x) = \dfrac{1}{x} = \dfrac{1}{f(x)}$
$f_2'(x) = \dfrac{d}{dx}(\log(f_1(x))) = \dfrac{1}{f_1(x)} f_1'(x) = \dfrac{1}{f_1(x) f(x)}$
$f_3'(x) = \dfrac{d}{dx}(\log(f_2(x))) = \dfrac{1}{f_2(x)} f_2'(x) = \dfrac{1}{f_2(x) f_1(x) f(x)}$
By induction, $f_n'(x) = \dfrac{1}{f_{n-1}(x) f_{n-2}(x) \ldots f_1(x) f(x)}$.
For $n = 2027$, we have:
$f_{2027}'(x) = \dfrac{1}{f_{2026}(x) f_{2025}(x) \ldots f_1(x) f(x)}$
The given integral is:
$I = \int \dfrac{1}{f(x) f_1(x) f_2(x) \ldots f_{2026}(x)} \, dx$
$I = \int f_{2027}'(x) \, dx$
$I = f_{2027}(x) + c$
Answer: $f_{2027}(x) + c$
Hence, the correct answer is option: 'C'.