Q92 / 150 · MHT CET 2026 (17 April Shift 1) · Mathematics
If $I_n = \int_1^e (\log_e x)^n\,dx$ where $n \in Z^+$ and $I_m + mI_{2026} = e$, then $m = $
- A$2024$
- B$2025$
- C$2026$
- Correct answerD$2027$
Solution
$I_n = \int_1^e (\log_e x)^n \, dx$
Using integration by parts, taking $1$ as the second function:
$I_n = \left[ x (\log_e x)^n \right]_1^e - \int_1^e x \cdot n (\log_e x)^{n-1} \cdot \dfrac{1}{x} \, dx$
$I_n = e (1)^n - 0 - n \int_1^e (\log_e x)^{n-1} \, dx$
$I_n = e - n I_{n-1}$
$I_n + n I_{n-1} = e$
Given $I_m + m I_{2026} = e$
Comparing the two equations, we get $n = m$ and $n-1 = 2026$.
$m - 1 = 2026 \Rightarrow m = 2027$
Using integration by parts, taking $1$ as the second function:
$I_n = \left[ x (\log_e x)^n \right]_1^e - \int_1^e x \cdot n (\log_e x)^{n-1} \cdot \dfrac{1}{x} \, dx$
$I_n = e (1)^n - 0 - n \int_1^e (\log_e x)^{n-1} \, dx$
$I_n = e - n I_{n-1}$
$I_n + n I_{n-1} = e$
Given $I_m + m I_{2026} = e$
Comparing the two equations, we get $n = m$ and $n-1 = 2026$.
$m - 1 = 2026 \Rightarrow m = 2027$
Hence, the correct answer is option: 'D'.