A) \[\frac{9}{5}\]
B) \[18\]
C) \[\frac{3}{4}\]
D) \[\frac{9}{2}\]
Correct Answer: B
Solution :
| We have, \[y=g\left( x \right)\] \[\Rightarrow f\left( x \right)={{x}^{3}}+3x+1\] |
| \[\therefore \,fog\left( x \right)=x\] \[\left[ \because g\left( x \right)is\,inverse\,of\,f\left( x \right) \right]\] |
| Area enclosed between \[y=g\left( x \right),x=1\]and \[x=15\]is |
| Area \[=\int_{1}^{15}{g\left( x \right)}dx=\int_{1}^{15}{{{f}^{-1}}\left( x \right)dx}\] |
| \[\therefore Let\,\,{{f}^{-1}}\left( x \right)=t\] |
| \[x=f\left( t \right)\] |
| \[dx=f'\left( t \right)dt\] |
| When \[x=1,t=0x=15,t=2\] |
| \[\therefore \,\,Area=\int_{0}^{2}{tf'\left( t \right)}dt=\int_{0}^{2}{t(3{{t}^{2}}+3)dt}\] \[\begin{align} & [\therefore f\left( t \right)={{t}^{3}}+3t+1 \\ & \therefore \,f'\left( t \right)=3{{t}^{2}}+3] \\ \end{align}\] |
| \[=\left[ \frac{3{{t}^{4}}}{4}+\frac{3{{t}^{2}}}{2} \right]_{0}^{2}=12+6=18\] |
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