MHTCET 15 Apr 2026 Shift 2 Paper

Concept:The integral of an odd function over a symmetric interval $[-a,a]$ is zero.Explanation:Given that $f(x)$ is an even function, we have $f(-x)=f(x)$.Also, $\sin x$ is an odd function, so $\sin(-x)=-\sin x$.The product of an even function and an odd function is an odd function.Thus, $f(x)\sin x$ is an odd function.Therefore, $\int_{-2}^{2} f(x)\sin x \, dx = 0$.The given integral becomes:$\int_{-2}^{2} (|x| + f(x)\sin x) \, dx = \int_{-2}^{2} |x| \, dx + 0 = \int_{-2}^{2} |x| \, dx$.Since $|x|$ is an even function,$\int_{-2}^{2} |x| \, dx = 2\int_{0}^{2} x \, dx$.Evaluating: $2\int_{0}^{2} x \, dx = 2\left[\frac{x^2}{2}\right]_{0}^{2} = 2(2-0) = 4$.Answer:The correct option is B: $4$.