A) \[\frac{2}{3}\]
B) \[\frac{4}{3}\]
C) \[\frac{5}{3}\]
D) \[\frac{1}{3}\]
Correct Answer: B
Solution :
\[x+2{{y}^{2}}=0\Rightarrow {{y}^{2}}=-\frac{x}{2}\] parabola \[x+3{{y}^{2}}=1\Rightarrow {{y}^{2}}=-\frac{1}{3}\left( x-1 \right)\] parabola Solving equation of two parabolas simultaneously, we get \[x=-2;\,\,y=\pm 1\] Area of the region ABCA \[=\left| \int\limits_{0}^{1}{\left( -2{{y}^{2}}-1+3{{y}^{2}} \right)dy} \right|=\left| \int\limits_{0}^{1}{\left( {{y}^{2}}-1 \right)dy} \right|\]\[=\left| \,\left| \frac{{{y}^{3}}}{3}-y \right. \right|_{0}^{1}\left| =\left| \frac{1}{3}-1 \right|=\frac{2}{3} \right.\] Hence area of region bounded by given curves is equal to \[\frac{4}{3}\].You need to login to perform this action.
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