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  • It the area enclosed by the parabolas and <img alt="P_{2}: x^{2}-y+6=0"

It the area enclosed by the parabolas P_{1}: 2 y=5 x^{2} and P_{2}: x^{2}-y+6=0 is equal to the area enclosed by P_{1} and y=\alpha x, \alpha>0, then \alpha^{3} is equal to

Option: 1

600


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer


\mathrm{y}=\frac{5 \mathrm{x}^2}{2}, \quad \mathrm{y}=\mathrm{x}^2+6
\frac{5 \mathrm{x}^2}{2}=\mathrm{x}^2+6
3 \mathrm{x}^2=12 \Rightarrow \mathrm{x}^2=4
\mathrm{x}= \pm 2
\mathrm{~A}_1=2 \int_0^2\left(\mathrm{x}^2+6-\frac{5}{2} \mathrm{x}^2\right) \mathrm{dx}
=2 \int_0^2\left(6-\frac{3 \mathrm{x}^2}{2}\right) \mathrm{dx}
=2\left[6 \mathrm{x}-\mathrm{x}^3 / 2\right]_0^2=2[12-4]
=16





y=\frac{5}{2} x^{2}, y=\alpha x(\alpha>0)
\text { area }=\frac{8}{3}\left[\mathrm{a}^{2} \mathrm{~m}^{3}\right]
=\frac{8}{3}[1 / 10]^{2} \cdot \alpha^{3}
=\frac{8}{300}-\alpha^{3}=\frac{2}{75} \alpha^{3}
\because \frac{2}{75}-\alpha^{3}=16 \quad \Rightarrow \alpha^{3}=8 \times 75
\alpha^{3}=600
 

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Riya

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