Question

lf $f\left(x\right)=a{e}^{2x}+b{e}^x+cx$, satisfies the conditions $f\left(0\right)=-1$, $f^{\prime }(\log 2)=31$,and ${\int }_0^{\log 4}\left(f\right(x)-cx)dx=\frac{39}{2}$, then

• A. $a=5,b=6,c=3$
• B. $a=5,b=-6,c=3$
• C. $a=-5,b=6,c=3$
• D. $a=1,b=2,c=3$

Answer 1

Answer: (B)

$a=5,b=-6,c=3$

$f\left(x\right)=a{e}^{2x}+b{e}^x+cx$
$f\left(0\right)=-1$
$\Rightarrow a+b=-\mathrm{1....}\left(1\right)$
Now differentiating both sides of the given equation we get
$f^{\prime }\left(x\right)=2a{e}^{2x}+b{e}^x+cx$
Using $f^{\prime }\left(\log 2\right)=31$, we get
$f^{\prime }\left(\log 2\right)=2a\left(4\right)+2b+c=31$
$\Rightarrow 8a+2b+c=\mathrm{31....}\left(2\right)$
${\int }_0^{\log 4}\left(f\right(x)-cx)dx=\frac{39}{2}$,
$\Rightarrow {\int }_0^{\log 4}(a{e}^{2x}+b{e}^x)dx=\frac{a}{2}\times e^{2x}{|}_0^{\log 4}+b\times e^x{|}_0^{\log 4}=39/2$
$=a/2[16-1]+b[4-1]=39/2$
$=a/2\left[15\right]+3b=39/2$
$\Rightarrow 15a+6b=\mathrm{39....}\left(3\right)$
Solving $\left(1\right)$, and $\left(3\right)$ simultaneously, we get $a=5$ and $b=-6$
Using the values of $a$ and $b$ in $\left(2\right)$, we get $c=3$.