A) 2e
B) \[lo{{g}_{e}}2\]
C) \[2+lo{{g}_{e}}2\]
D) \[1+lo{{g}_{e}}2\]
Correct Answer: D
Solution :
| [d] |
| \[{{e}^{y}}=t\] |
| \[{{e}^{y}}\frac{dy}{dx}=\frac{dt}{dx}\] |
| \[\Rightarrow \frac{dt}{dx}-t={{e}^{x}}\] |
| \[I.F.={{e}^{\int{-1dx}}}={{e}^{-x}}\] |
| \[\Rightarrow t.{{e}^{-x}}=x+c\] |
| \[{{e}^{y-x}}=x+c\] |
| \[c=1\] |
| \[\Rightarrow f(1)=1+\ln \,\,2\] |
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