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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