JEE MAIN - Mathematics (2025 - 7th April Evening Shift - No. 12)
If the area of the region $ \{(x, y) : 1 + x^2 \leq y \leq \min \{x+7, 11-3x\}\} $ is $ A $, then $ 3A $ is equal to :
50
46
49
47
Explanation
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$$\begin{aligned} & A=\int_{-2}^1\left(x+7-x^2-1\right) d x+\int_1^2\left(11+3 x-x^2-1\right) d x \\ & =\left[\frac{x^2}{2}+6 x-\frac{x^3}{3}\right]_{-2}^1+\left[10 x-\frac{3 x^2}{2}-\frac{x^3}{3}\right]_1^2 \\ & =\frac{50}{3} \Rightarrow 3 A=50 \quad \text { Option (1) } \end{aligned}$$
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