MHTCET 15 Apr 2026 Shift 2 Paper

Concept:The area under the curve can be found by integrating the function between the given limits.Since $y=|x-4|$ changes its definition at $x=4$, the integral must be split into two parts.Explanation:The curve $y=|x-4|$ lies above the X-axis for all $x$ from $3$ to $5$.Thus, the required area is the area under this curve from $x=3$ to $x=5$.For $3 \le x \le 4$, we have:$$|x-4| = 4-x$$For $4 \le x \le 5$, we have:$$|x-4| = x-4$$So the total area is given by:$$\text{Area} = \int_{3}^{4}(4-x)\,dx + \int_{4}^{5}(x-4)\,dx$$Evaluate the first integral:$$\int_{3}^{4}(4-x)\,dx = \left[4x - \frac{x^2}{2}\right]_{3}^{4}$$$$= \left(16 - 8\right) - \left(12 - \frac{9}{2}\right) = 8 - \frac{15}{2} = \frac{1}{2}$$Evaluate the second integral:$$\int_{4}^{5}(x-4)\,dx = \left[\frac{x^2}{2} - 4x\right]_{4}^{5}$$$$= \left(\frac{25}{2} - 20\right) - \left(\frac{16}{2} - 16\right) = \left(-\frac{15}{2}\right) - \left(-8\right) = \frac{1}{2}$$Therefore, the total area is:$$\frac{1}{2} + \frac{1}{2} = 1$$So the area of the required region is $1$ square unit.Answer:The area of the region is $1$ sq. unit.Correct option: D.