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Question Number 31538 by akhilesh2684894@gmail.com last updated on 09/Mar/18

$$\mathrm{If}f\left(x\right)=a{e}^{2x}+b{e}^x+cx\mathrm{satisfies}\mathrm{the}$$$$\mathrm{condition}f\left(0\right)=-1,f{}^{\prime }\left(\log 2\right)=31,$$$$\underset{0}{\stackrel{\log 4}{\int }}(f\left(x\right)-cx)dx=\frac{39}{2},\mathrm{then}$$

Answered by MJS last updated on 09/Mar/18

$$\mathrm{I}\mathrm{hope}\log =\ln ,\mathrm{not}{\log }_{10}...$$$$f\left(0\right)=1\Rightarrow a+b=-1$$$$f^{\prime }\left(\log 2\right)=31\Rightarrow 8a+2b+c=31$$$$\underset{0}{\stackrel{\log 4}{\int }}{ae}^{2x}+{be}^{x}dx=\frac{39}{2}\Rightarrow$$$$\Rightarrow \frac{{ae}^{2x}}{2}+{be}^x{\mid }_0^{\log 4}=\frac{39}{2}\Rightarrow$$$$\Rightarrow (8a+4b)-(\frac{a}{2}+b)=\frac{39}{2}\Rightarrow$$$$\Rightarrow \frac{15a}{2}+3b=\frac{39}{2}$$$$\left(I\right)a+b=-1$$$$\left(II\right)8a+2b+c=31$$$$\left(III\right)\frac{15a}{2}+3b=\frac{39}{2}$$$$-3\times \left(I\right)+\left(III\right)\frac{9a}{2}=\frac{45}{2}\Rightarrow a=5$$$$\left(I\right)5+b=-1\Rightarrow b=-6$$$$\left(II\right)40-12+c=31\Rightarrow$$$$\Rightarrow c=3$$$$f\left(x\right)=5e^{2x}-6e^x+3x$$

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