Q77 / 150 · MHT CET 2026 (15 April Shift 2) · Mathematics
If $f(x)$ is an even function, then $\int\limits_{-2}^{2} (|x| + f(x)\sin x)\, \text{d}x$ is
- A$2$
- Correct answerB$4$
- C$6$
- D$8$
Solution
Let $I = \int_{-2}^{2} (|x| + f(x)\sin x)\, dx$
$I = \int_{-2}^{2} |x|\, dx + \int_{-2}^{2} f(x)\sin x\, dx$
Since $f(x)$ is an even function, $f(-x) = f(x)$.
Let $g(x) = f(x)\sin x$. Then $g(-x) = f(-x)\sin(-x) = -f(x)\sin x = -g(x)$.
Thus, $g(x)$ is an odd function, which gives $\int_{-2}^{2} f(x)\sin x\, dx = 0$.
The function $|x|$ is an even function, so $\int_{-2}^{2} |x|\, dx = 2\int_{0}^{2} x\, dx$.
$I = 2 \left[ \dfrac{x^2}{2} \right]_{0}^{2} + 0$
$I = 4$
Answer: $4$
$I = \int_{-2}^{2} |x|\, dx + \int_{-2}^{2} f(x)\sin x\, dx$
Since $f(x)$ is an even function, $f(-x) = f(x)$.
Let $g(x) = f(x)\sin x$. Then $g(-x) = f(-x)\sin(-x) = -f(x)\sin x = -g(x)$.
Thus, $g(x)$ is an odd function, which gives $\int_{-2}^{2} f(x)\sin x\, dx = 0$.
The function $|x|$ is an even function, so $\int_{-2}^{2} |x|\, dx = 2\int_{0}^{2} x\, dx$.
$I = 2 \left[ \dfrac{x^2}{2} \right]_{0}^{2} + 0$
$I = 4$
Answer: $4$
Hence, the correct answer is option: 'B'.