Question

Consider a function of the form $f\left(x\right)=\alpha e^{2x}+\beta e^x-\gamma x,$ where $\alpha ,\beta ,\gamma$ are independent of $x$ and $f\left(x\right)$ satisfies the following conditions : $f\left(0\right)=-1,$ $f^{\prime }\left(\ln 2\right)=30$ and $\underset{0}{\overset{\ln 4}{\int }}(f\left(x\right)+\gamma x)dx=24.$ Then the value of $(\alpha +\beta +\gamma )$ is

Answer 1

$f\left(x\right)=\alpha e^{2x}+\beta e^x-\gamma x$
$f\left(0\right)=\alpha +\beta$
$\Rightarrow \alpha +\beta =-1\cdots \left(1\right)$
and $f^{\prime }\left(x\right)=2\alpha e^{2x}+\beta e^x-\gamma$
$\Rightarrow f^{\prime }\left(\ln 2\right)=2\alpha e^{2\ln 2}+\beta e^{\ln 2}-\gamma$
$\Rightarrow 8\alpha +2\beta -\gamma =30\cdots \left(2\right)$

Also, $\underset{0}{\overset{\ln 4}{\int }}\left(f\right(x)+\gamma x)dx$
$\Rightarrow \underset{0}{\overset{\ln 4}{\int }}(\alpha e^{2x}+\beta e^x)dx=24$
$\Rightarrow 5\alpha +2\beta =16\cdots \left(3\right)$

From $\left(1\right),\left(2\right)$ and $\left(3\right),$ we get
$\alpha =6,\beta =-7,\gamma =4$
Hence, $\alpha +\beta +\gamma =6-7+4=3$