MHTCET 17 Apr 2026 Shift 2 Paper

Concept:Here, $\lceil x\rceil$ denotes the integral part of $x$, which is the greatest integer less than or equal to $x$.So $f(x)=x-\lceil x\rceil$ is the fractional part of $x$, and it is evaluated piecewise on $[-1,0)$ and $[0,1)$.Explanation:For $-1 \le x < 0$, we have $\lceil x\rceil = -1$.Therefore, $f(x)=x-(-1)=x+1$.For $0 \le x < 1$, we have $\lceil x\rceil = 0$.Therefore, $f(x)=x-0=x$.Now split the integral over these two intervals:$$\int_{-1}^{1} f(x)\,dx = \int_{-1}^{0}(x+1)\,dx + \int_{0}^{1} x\,dx$$Evaluate the first integral:$$\int_{-1}^{0}(x+1)\,dx = \left[\frac{x^2}{2}+x\right]_{-1}^{0} = 0 - \left(\frac{1}{2}-1\right) = \frac{1}{2}$$Evaluate the second integral:$$\int_{0}^{1} x\,dx = \left[\frac{x^2}{2}\right]_{0}^{1} = \frac{1}{2}$$Adding these results:$$\frac{1}{2} + \frac{1}{2} = 1$$Answer:$$\int_{-1}^{1} f(x)\,dx = 1$$Hence, the correct option is A.