MHTCET 17 Apr 2026 Shift 1 Paper

Concept:Use integration by parts to obtain a recurrence relation for $I_n$, then compare it with the given condition.Explanation:Given $I_{n}=\int_{1}^{e}(\log_e x)^n dx$ and $\log_e x = \ln x$.Apply integration by parts with $u = (\ln x)^n$ and $dv = dx$.Then $du = n(\ln x)^{n-1}\frac{1}{x} dx$ and $v = x$.So,$$I_n = \left[x(\ln x)^n\right]_{1}^{e} - n\int_{1}^{e} x(\ln x)^{n-1}\frac{1}{x}dx$$$$I_n = \left[x(\ln x)^n\right]_{1}^{e} - n\int_{1}^{e} (\ln x)^{n-1}dx$$Now, $\left[x(\ln x)^n\right]_{1}^{e} = e(1)^n - 1(0)^n = e$.Therefore, $I_n = e - nI_{n-1}$.Rearranging, $I_n + nI_{n-1} = e$.For index $m$, this gives $I_m + mI_{m-1} = e$.The given equation is $I_m + mI_{2026} = e$.Comparing the two, we get $I_{m-1} = I_{2026}$.For $1 < x < e$, we have $0 < \ln x < 1$, so $I_n$ decreases as $n$ increases.Thus, the indices must be equal: $m - 1 = 2026$.Hence, $m = 2027$.Answer:$m = 2027$, which is Option D.