Question

The numbers P, Q and R for which the function
$f\left(x\right)=P{e}^{2x}+Q{e}^x+Rx$ satisfies the conditions
$f\left(0\right)=-1,f^{{}^{\prime }}\left(log2\right)=31$ and ${\int }_0^{log4}\left[f\right(x)-Rx]dx=\frac{39}{2}$ are
given by

• A.
  
• B.
  
• C.
  
• D.
  

Answer 1

The correct option is D

We have $f^{{}^{\prime }}\left(x\right)=2P{e}^{2x}+Q{e}^x+R$
$\Rightarrow f^{{}^{\prime }}\left(log2\right)=8P+2Q+R$
Also, $-1=f\left(0\right)=P+Q$
$\frac{39}{2}={\int }_0^{log4}\left[f\right(x)-Rx]dx={\int }_0^{log4}(P{e}^{2x}+Q{e}^x)dx$
$\Rightarrow \frac{15P}{2}+3Q=\frac{39}{2}$
Now on solving the above equations, we get
P = 5, Q = -6 and R = 3