A) \[\sqrt{{{\log }_{e}}(3y-2)}\]
B) \[{{\left( lo{{g}_{e}}\left( 3y-2 \right) \right)}^{1/3}}\]
C) \[{{\left( lo{{g}_{e}}\left( 2-\text{ }3y \right) \right)}^{1/3}}\]
D) None of these
Correct Answer: B
Solution :
[b] \[y=\int{{{x}^{2}}{{e}^{{{x}^{3}}}}dx\,{{x}^{3}}=t}\] \[{{x}^{2}}dx=\frac{1}{3}dt\] \[y=\frac{1}{3}{{e}^{{{x}^{3}}}}+c\] ... (1) passes through (0, 1) \[1=\frac{1}{3}+c\Rightarrow c=\frac{2}{3}\] From (1), we have \[y=\frac{1}{3}{{e}^{{{x}^{3}}}}+\frac{2}{3}\] \[\Rightarrow {{e}^{{{x}^{3}}}}=3y-2\Rightarrow {{x}^{3}}=\ell {{n}_{e}}(3y-2)\] \[\Rightarrow x={{(\ell {{n}_{e}}(3y-2))}^{1/3}}\]
You need to login to perform this action.
You will be redirected in
3 sec
