Question

Let $f\left(0\right)=0$ and ${\int }_0^2{f}^{\prime }\left(2t\right)e^{f\left(2t\right)}dt=5$.

Then the value of $f\left(4\right)$ is?

• A. $\log 2$
• B. $\log 7$
• C. $\log 11$
• D. $\log 13$

Answer 1

The correct option is C $\log 11$

We have ${\int }_0^2{f}^{\prime }\left(2t\right)e^{f\left(2t\right)}dt=5$

Substitute $e^{f\left(2t\right)}=y$

$\therefore 2f^{\prime }\left(2t\right)e^{f\left(2t\right)}dt=dy$

$\therefore \frac{1}{2}{\int }_{e^{f\left(0\right)}}^{e^{f\left(4\right)}}dy=5$

or ${\int }_{e^{f\left(0\right)}}^{e^{f\left(4\right)}}dy=10$

or $e^{f\left(4\right)}-e^{f\left(0\right)}=10$

or $e^{f\left(4\right)}=10+1=11$

or $f\left(4\right)=\log 11$