A) \[\frac{{{({{e}^{2}}+1)}^{3}}}{3}+\frac{{{({{e}^{2}}+1)}^{2}}}{2}\]
B) \[\frac{{{({{e}^{2}}-1)}^{3}}}{3}+\frac{{{({{e}^{2}}-1)}^{2}}}{2}\]
C) \[\frac{{{({{e}^{2}}+1)}^{3}}}{3}-\frac{{{({{e}^{2}}+1)}^{2}}}{2}\]
D) None of these
Correct Answer: B
Solution :
[b] \[\frac{dx}{dt}=t(t+1)\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=\frac{{{t}^{3}}}{3}+\frac{{{t}^{2}}}{2}+c\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=\frac{{{t}^{3}}}{3}+\frac{{{t}^{2}}}{2}\] (\[\because \,\,\,\,\,\,x=0\]at \[t=0\]) ?(1) \[\frac{dy}{dt}=\frac{1}{t+1}\] \[\Rightarrow \,\,\,\,\,\,\,\,y={{\log }_{e}}(t+1)+c\] \[\Rightarrow \,\,\,\,\,\,\,\,y={{\log }_{e}}(t+1)\] (\[\because \,\,\,\,\,\,\,y=0\] at \[t=0\]) Since the particle passes through the point \[(d,2),\] \[2={{\log }_{e}}(t+1)\] \[\Rightarrow \,\,\,\,\,\,t={{e}^{2}}-1\] Putting the value of t into (1), we get \[d=\frac{{{({{e}^{2}}-1)}^{3}}}{3}+\frac{{{({{e}^{2}}-1)}^{2}}}{2}\]You need to login to perform this action.
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