MHTCET 20 Apr 2026 Shift 2 Paper

Concept:The absolute value $\mid \log x \mid$ changes its sign at $x = 1$, so the integral must be split there.Explanation:For $1/e \le x \le 1$, we have $\log x \le 0$, so $\mid \log x \mid = -\log x$.For $1 \le x \le e$, we have $\log x \ge 0$, so $\mid \log x \mid = \log x$.Therefore,$$\int_{1/e}^{e} \frac{\mid \log x \mid}{x^2} dx = \int_{1/e}^{1} \frac{-\log x}{x^2} dx + \int_{1}^{e} \frac{\log x}{x^2} dx.$$Now integrate $\frac{\log x}{x^2}$ by parts, taking $u = \log x$ and $dv = \frac{1}{x^2}dx$.We get$$\int \frac{\log x}{x^2} dx = -\frac{\log x + 1}{x} + C.$$Evaluate the first part:$$\int_{1/e}^{1} \frac{-\log x}{x^2} dx = -\left[ -\frac{\log x + 1}{x} \right]_{1/e}^{1}.$$At $x = 1$, the bracket value is $-1$, and at $x = 1/e$, it is $0$.So the first part equals $1$.Evaluate the second part:$$\int_{1}^{e} \frac{\log x}{x^2} dx = \left[ -\frac{\log x + 1}{x} \right]_{1}^{e} = -\frac{2}{e} - (-1) = 1 - \frac{2}{e}.$$Adding both parts,$$1 + \left(1 - \frac{2}{e}\right) = 2 - \frac{2}{e}.$$Answer:$$2 - \frac{2}{e}$$Hence, the correct option is C.