A) \[\frac{115}{4}\]
B) \[\frac{125}{2}\]
C) \[\frac{121}{2}\]
D) \[\frac{121}{4}\]
Correct Answer: D
Solution :
(d)| The given equations of parabola are | ||
| \[y=4x-{{x}^{2}}\operatorname{or}{{\left( x-2 \right)}^{2}}=-\left( y-4 \right)\] | …(1) | |
| and \[y={{x}^{2}}-x\] | ||
| Or \[{{\left( x-\frac{1}{2} \right)}^{2}}=\left( y+\frac{1}{4} \right)\] | …(2) | |
| Solving the equation of two parabolas we get their points of intersection as \[O\left( 0,0 \right),A\left( \frac{5}{2},\frac{15}{4} \right)\] | ||
| Here the area below x-axis, | ||
 |
||
| \[{{A}_{1}}=\int_{0}^{1}{\left( -{{y}_{2}} \right)dx=\int_{0}^{1}{\left( x-{{x}^{2}} \right)dx}}\]\[=\left( \frac{{{x}^{2}}}{2}-\frac{{{x}^{3}}}{3} \right)_{0}^{1}=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\operatorname{sq}.units\] |
| Area above x-axis, |
| \[{{A}_{2}}=\int_{0}^{5/2}{{{y}_{1}}dx-\int_{1}^{5/2}{{{y}^{2}}dx}}\]\[=\int_{0}^{5/2}{\left( 4x-{{x}^{2}} \right)dx-\int_{1}^{5/2}{\left( {{x}^{2}}-x \right)dx}}\]\[=\left( 2{{x}^{2}}-\frac{{{x}^{3}}}{3} \right)_{0}^{5/2}-\left( \frac{{{x}^{3}}}{3}-\frac{{{x}^{2}}}{2} \right)_{1}^{5/2}\]\[=\left( \frac{25}{2}-\frac{125}{24} \right)-\left[ \left( \frac{125}{24}-\frac{25}{8} \right)-\left( \frac{1}{3}-\frac{1}{2} \right) \right]\]\[=\frac{25}{2}-\frac{125}{12}+\frac{25}{8}-\frac{1}{6}\]\[=\frac{300-250+75-4}{24}=\frac{121}{24}\] |
| \[\therefore \]Ratio of area above \[x-\operatorname{axis}\]and below \[x-\operatorname{axis}\] |
| \[{{A}_{2}}:{{A}_{1}}=\frac{121}{24}:\frac{1}{6}=\frac{121}{4}=121:4\]. |
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