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If-f-x-a-e-2x-b-e-x-cx-satisfies-the-condition-f-0-1-f-log-2-31-0-log-4-f-x-cx-dx-39-2-then-

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Question Number 31538 by akhilesh2684894@gmail.com last updated on 09/Mar/18
If f(x)= a e^(2x) +b e^x +cx satisfies the condition f(0)= −1, f ′(log 2)=31, ∫_( 0) ^(log 4) (f(x)−cx) dx = ((39)/2), then
$$\mathrm{If}\:{f}\left({x}\right)=\:{a}\:{e}^{\mathrm{2}{x}} +{b}\:{e}^{{x}} +{cx}\:\mathrm{satisfies}\:\mathrm{the} \\ $$$$\mathrm{condition}\:{f}\left(\mathrm{0}\right)=\:−\mathrm{1},\:{f}\:'\left(\mathrm{log}\:\mathrm{2}\right)=\mathrm{31}, \\ $$$$\:\underset{\:\mathrm{0}} {\overset{\mathrm{log}\:\mathrm{4}} {\int}}\left({f}\left({x}\right)−{cx}\right)\:{dx}\:=\:\frac{\mathrm{39}}{\mathrm{2}},\:\mathrm{then} \\ $$
Answered by MJS last updated on 09/Mar/18
I hope log=ln, not log_(10) ... f(0)=1 ⇒ a+b=−1 f′(log 2)=31 ⇒ 8a+2b+c=31 ∫_0 ^(log 4) ae^(2x) +be^x dx=((39)/2) ⇒ ⇒ ((ae^(2x) )/2)+be^x ∣_0 ^(log 4) =((39)/2) ⇒ ⇒ (8a+4b)−((a/2)+b)=((39)/2) ⇒ ⇒ ((15a)/2)+3b=((39)/2) (I) a+b=−1 (II) 8a+2b+c=31 (III) ((15a)/2)+3b=((39)/2) −3×(I)+(III) ((9a)/2)=((45)/2) ⇒ a=5 (I) 5+b=−1 ⇒ b=−6 (II) 40−12+c=31 ⇒ ⇒ c=3 f(x)=5e^(2x) −6e^x +3x
$$\mathrm{I}\:\mathrm{hope}\:\mathrm{log}=\mathrm{ln},\:\mathrm{not}\:\mathrm{log}_{\mathrm{10}} … \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{1}\:\Rightarrow\:{a}+{b}=−\mathrm{1} \\ $$$${f}'\left(\mathrm{log}\:\mathrm{2}\right)=\mathrm{31}\:\Rightarrow\:\mathrm{8}{a}+\mathrm{2}{b}+{c}=\mathrm{31} \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{log}\:\mathrm{4}} {\int}}{ae}^{\mathrm{2}{x}} +{be}^{{x}} \:{dx}=\frac{\mathrm{39}}{\mathrm{2}}\:\Rightarrow \\ $$$$\Rightarrow\:\frac{{ae}^{\mathrm{2}{x}} }{\mathrm{2}}+{be}^{{x}} \mid_{\mathrm{0}} ^{\mathrm{log}\:\mathrm{4}} =\frac{\mathrm{39}}{\mathrm{2}}\:\Rightarrow \\ $$$$\Rightarrow\:\left(\mathrm{8}{a}+\mathrm{4}{b}\right)−\left(\frac{{a}}{\mathrm{2}}+{b}\right)=\frac{\mathrm{39}}{\mathrm{2}}\:\Rightarrow \\ $$$$\Rightarrow\:\frac{\mathrm{15}{a}}{\mathrm{2}}+\mathrm{3}{b}=\frac{\mathrm{39}}{\mathrm{2}} \\ $$$$\left({I}\right)\:{a}+{b}=−\mathrm{1} \\ $$$$\left({II}\right)\:\mathrm{8}{a}+\mathrm{2}{b}+{c}=\mathrm{31} \\ $$$$\left({III}\right)\:\frac{\mathrm{15}{a}}{\mathrm{2}}+\mathrm{3}{b}=\frac{\mathrm{39}}{\mathrm{2}} \\ $$$$−\mathrm{3}×\left({I}\right)+\left({III}\right)\:\frac{\mathrm{9}{a}}{\mathrm{2}}=\frac{\mathrm{45}}{\mathrm{2}}\:\Rightarrow\:{a}=\mathrm{5} \\ $$$$\left({I}\right)\:\mathrm{5}+{b}=−\mathrm{1}\:\Rightarrow\:{b}=−\mathrm{6} \\ $$$$\left({II}\right)\:\mathrm{40}−\mathrm{12}+{c}=\mathrm{31}\:\Rightarrow \\ $$$$\Rightarrow\:{c}=\mathrm{3} \\ $$$${f}\left({x}\right)=\mathrm{5}{e}^{\mathrm{2}{x}} −\mathrm{6}{e}^{{x}} +\mathrm{3}{x} \\ $$

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