Jee Main 2024 27 01 2024 Shift 2 - Question15
Question 15
The area bounded by $0 \leq y \leq \min {2 x, 6 x-x^{2} }$ and $x$-axis is $A$. Then $12 A$ is
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Answer: (304)
Solution:
$\min {6 x-x^{2}, y=2 x }$
$$ \begin{alignedat} Area =\frac{1}{2} \times 4 \times 8+\int _{4}^{6}\left(6 x-x^{2}\right) d x \ & =16+\left[3 x^{2}-\frac{x^{3}}{3}\right] _4^{6} \end{aligned} $$
$$ \begin{gathered} A=16+\frac{28}{3} \\ 12 A=(12 \times 16+28 \times 4) \text{ V} \ 304 \text { square units } \end{gathered} $$
BETA
