JEE MAIN - Mathematics (2020 - 5th September Morning Slot - No. 6)
If
$$\int {\left( {{e^{2x}} + 2{e^x} - {e^{ - x}} - 1} \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx} $$ = $$g\left( x \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}} + c$$
where c is a constant of integration, then g(0) is equal to :
$$\int {\left( {{e^{2x}} + 2{e^x} - {e^{ - x}} - 1} \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx} $$ = $$g\left( x \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}} + c$$
where c is a constant of integration, then g(0) is equal to :
1
2
e
e2
Explanation
I = $$\int {\left( {{e^{2x}} + 2{e^x} - {e^{ - x}} - 1} \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx} $$
= $$\int {\left( {\left( {{e^{2x}} + {e^x} - 1} \right) + \left( {{e^x} - {e^{ - x}}} \right)} \right)} {e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx$$
= $$\int {\left( {{e^{2x}} + {e^x} - 1} \right)} {e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx$$
+ $$\int {{e^{\left( {{e^x} + {e^{ - x}}} \right)}}\left( {{e^x} - {e^{ - x}}} \right)dx} $$
= $$\int {{{\left( {{e^{2x}} + {e^x} - 1} \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}}} \over {{e^x}}}.} {e^x}dx$$
$$ + \int {{e^{\left( {{e^x} + {e^{ - x}}} \right)}}\left( {{e^x} - {e^{ - x}}} \right)dx} $$
= $$\int {\left( {{e^x} - {e^{ - x}} + 1} \right)} {e^{\left( {{e^x} + {e^{ - x}} + x} \right)}}dx$$
+ $$\int {{e^{\left( {{e^x} + {e^{ - x}}} \right)}}\left( {{e^x} - {e^{ - x}}} \right)dx} $$
Let $${{e^x} + {e^{ - x}} + x}$$ = t $$ \Rightarrow $$ $${\left( {{e^x} - {e^{ - x}} + 1} \right)}$$dx = dt
and let $${{e^x} + {e^{ - x}}}$$ = u $$ \Rightarrow $$ $${\left( {{e^x} - {e^{ - x}}} \right)dx}$$ = du
$$ \therefore $$ I = $$\int {{e^t}} dt + \int {{e^u}} du$$
= $${{e^t}}$$ + $${{e^u}}$$ + C
= $${e^{\left( {{e^x} + {e^{ - x}} + x} \right)}}$$ + $${{e^{\left( {{e^x} + {e^{ - x}}} \right)}}}$$ + C
= $${e^{\left( {{e^x} + {e^{ - x}}} \right)}}\left( {{e^x} + 1} \right)$$ + C
$$ \therefore $$ g(x) = ex + 1
$$ \Rightarrow $$ g(0) = 2
= $$\int {\left( {\left( {{e^{2x}} + {e^x} - 1} \right) + \left( {{e^x} - {e^{ - x}}} \right)} \right)} {e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx$$
= $$\int {\left( {{e^{2x}} + {e^x} - 1} \right)} {e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx$$
+ $$\int {{e^{\left( {{e^x} + {e^{ - x}}} \right)}}\left( {{e^x} - {e^{ - x}}} \right)dx} $$
= $$\int {{{\left( {{e^{2x}} + {e^x} - 1} \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}}} \over {{e^x}}}.} {e^x}dx$$
$$ + \int {{e^{\left( {{e^x} + {e^{ - x}}} \right)}}\left( {{e^x} - {e^{ - x}}} \right)dx} $$
= $$\int {\left( {{e^x} - {e^{ - x}} + 1} \right)} {e^{\left( {{e^x} + {e^{ - x}} + x} \right)}}dx$$
+ $$\int {{e^{\left( {{e^x} + {e^{ - x}}} \right)}}\left( {{e^x} - {e^{ - x}}} \right)dx} $$
Let $${{e^x} + {e^{ - x}} + x}$$ = t $$ \Rightarrow $$ $${\left( {{e^x} - {e^{ - x}} + 1} \right)}$$dx = dt
and let $${{e^x} + {e^{ - x}}}$$ = u $$ \Rightarrow $$ $${\left( {{e^x} - {e^{ - x}}} \right)dx}$$ = du
$$ \therefore $$ I = $$\int {{e^t}} dt + \int {{e^u}} du$$
= $${{e^t}}$$ + $${{e^u}}$$ + C
= $${e^{\left( {{e^x} + {e^{ - x}} + x} \right)}}$$ + $${{e^{\left( {{e^x} + {e^{ - x}}} \right)}}}$$ + C
= $${e^{\left( {{e^x} + {e^{ - x}}} \right)}}\left( {{e^x} + 1} \right)$$ + C
$$ \therefore $$ g(x) = ex + 1
$$ \Rightarrow $$ g(0) = 2
Comments (0)
