Question Number 54273 by rahul 19 last updated on 01/Feb/19
Answered by tanmay.chaudhury50@gmail.com last updated on 01/Feb/19
![(df/dx)=f(x)+∫_0 ^1 f(x)dx (df/dx)=f(x)+a (df/(f(x)+a))=dx ((d[f(x)+a])/(f(x)+a))=dx ln[f(x)+a]=x+c f(x)+a=e^(x+c) f(0)+a=e^(0+c) 1+a=e^c f(x)+a=e^x ×e^c f(x)+a=e^x (1+a) f(x)+a=e^x +ae^x f(x)=(1+a)e^x −a ∫_0 ^1 f(x)dx=(1+a)∫_0 ^1 e^x dx−a∫_0 ^1 dx a=(1+a)(e−1)−a 2a=e−1+ae−a 3a−ae=e−1 a=((e−1)/(3−e)) f(x)=(1+a)e^x −a =(1+((e−1)/(3−e)))e^x −((e−1)/(3−e)) f(x)=(2/(3−e))e^x −((e−1)/(3−e)) ∫f(x)dx=(2/(3−e))e^x +((1−e)/(3−e))x+c so option b is answer](https://www.tinkutara.com/question/Q54277.png)
$$\frac{{df}}{{dx}}={f}\left({x}\right)+\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){dx} \\ $$$$\frac{{df}}{{dx}}={f}\left({x}\right)+{a} \\ $$$$\frac{{df}}{{f}\left({x}\right)+{a}}={dx} \\ $$$$\frac{{d}\left[{f}\left({x}\right)+{a}\right]}{{f}\left({x}\right)+{a}}={dx} \\ $$$${ln}\left[{f}\left({x}\right)+{a}\right]={x}+{c} \\ $$$${f}\left({x}\right)+{a}={e}^{{x}+{c}} \\ $$$${f}\left(\mathrm{0}\right)+{a}={e}^{\mathrm{0}+{c}} \\ $$$$\mathrm{1}+{a}={e}^{{c}} \\ $$$${f}\left({x}\right)+{a}={e}^{{x}} ×{e}^{{c}} \\ $$$${f}\left({x}\right)+{a}={e}^{{x}} \left(\mathrm{1}+{a}\right) \\ $$$${f}\left({x}\right)+{a}={e}^{{x}} +{ae}^{{x}} \\ $$$${f}\left({x}\right)=\left(\mathrm{1}+{a}\right){e}^{{x}} −{a} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){dx}=\left(\mathrm{1}+{a}\right)\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{{x}} {dx}−{a}\int_{\mathrm{0}} ^{\mathrm{1}} {dx} \\ $$$${a}=\left(\mathrm{1}+{a}\right)\left({e}−\mathrm{1}\right)−{a} \\ $$$$\mathrm{2}{a}={e}−\mathrm{1}+{ae}−{a} \\ $$$$\mathrm{3}{a}−{ae}={e}−\mathrm{1} \\ $$$${a}=\frac{{e}−\mathrm{1}}{\mathrm{3}−{e}} \\ $$$${f}\left({x}\right)=\left(\mathrm{1}+{a}\right){e}^{{x}} −{a} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\left(\mathrm{1}+\frac{{e}−\mathrm{1}}{\mathrm{3}−{e}}\right){e}^{{x}} −\frac{{e}−\mathrm{1}}{\mathrm{3}−{e}} \\ $$$${f}\left({x}\right)=\frac{\mathrm{2}}{\mathrm{3}−{e}}{e}^{{x}} −\frac{{e}−\mathrm{1}}{\mathrm{3}−{e}} \\ $$$$\int{f}\left({x}\right){dx}=\frac{\mathrm{2}}{\mathrm{3}−{e}}{e}^{{x}} +\frac{\mathrm{1}−{e}}{\mathrm{3}−{e}}{x}+{c} \\ $$$${so}\:{option}\:{b}\:{is}\:{answer} \\ $$
Commented by rahul 19 last updated on 01/Feb/19
thank you sir!
Commented by tanmay.chaudhury50@gmail.com last updated on 01/Feb/19
$${most}\:{welcome}… \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 01/Feb/19
$${is}\:{there}\:{any}\:{short}\:{trick}\:{to}\:{get}\:{answer}…{then} \\ $$$${how}\:{these}\:{type}\:{of}\:{question}\:{can}\:{be}\:{solved}\:{in}\:{short}\:{time}.. \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 02/Feb/19
$${thank}\:{you} \\ $$
Commented by rahul 19 last updated on 02/Feb/19
$${sir},\:{i}'{ve}\:{posted}\:{ans}.\:{of}\:{those}\:{integrals}. \\ $$$${do}\:{check}! \\ $$
Commented by rahul 19 last updated on 03/Feb/19
$${Q}.\:{No}.=\:\mathrm{54248} \\ $$
Answered by ajfour last updated on 01/Feb/19
$${f}\:''\left({x}\right)={f}\:'\left({x}\right) \\ $$$$\Rightarrow\:{f}\:'\left({x}\right)={Ae}^{{x}} \\ $$$$\:\:{f}\left({x}\right)={Ae}^{{x}} +{b} \\ $$$$\:\:{f}\left(\mathrm{0}\right)={A}+{b}\:=\mathrm{1}\:\:\Rightarrow\:{b}=\mathrm{1}−{A} \\ $$$$\:\:{f}\:'\left({x}\right)=\:{f}\left({x}\right)+\int_{\mathrm{0}} ^{\:\:\mathrm{1}} {f}\left({x}\right){dx}\:\:\Rightarrow \\ $$$$\:\:{Ae}^{{x}} ={Ae}^{{x}} +{b}+{A}\left({e}−\mathrm{1}\right)+{b} \\ $$$${but}\:{b}=\mathrm{1}−{A} \\ $$$$\Rightarrow\:\:\mathrm{2}−\mathrm{2}{A}+{Ae}−{A}=\mathrm{0} \\ $$$$\Rightarrow\:\:\boldsymbol{{A}}=\:\frac{\mathrm{2}}{\mathrm{3}−{e}}\:\:\:,\:{hence} \\ $$$$\Rightarrow\:\:{f}\left({x}\right)=\frac{\mathrm{2}{e}^{{x}} }{\mathrm{3}−{e}}+\mathrm{1}−\frac{\mathrm{2}}{\mathrm{3}−{e}} \\ $$$${f}\left({x}\right)=\:\frac{\mathrm{2}{e}^{{x}} +\mathrm{1}−{e}}{\mathrm{3}−{e}}\:\: \\ $$
Commented by ajfour last updated on 02/Feb/19
$${I}\:{must}\:{have}\:{made}\:{mistake},\:{please} \\ $$$${help}..\:\:\:\:\:\:\:\:\:\:\: \\ $$
Commented by rahul 19 last updated on 02/Feb/19
$${Sir},\:{how}\:{can}\:{you}\:{say}\:{f}\:^{'} \left({x}\right)={Ae}^{{x}} \:{is}\:{the} \\ $$$${only}\:{possibility}? \\ $$
Commented by ajfour last updated on 02/Feb/19
$${f}\:''\left({x}\right)={f}\:'\left({x}\right) \\ $$$$\Rightarrow\:\:\int\frac{{f}\:''\left({x}\right){dx}}{{f}\:'\left({x}\right)}=\int{dx} \\ $$$$\Rightarrow\:\mathrm{ln}\:{f}\:'\left({x}\right)={x}+{c} \\ $$$$\Rightarrow\:{f}\:'\left({x}\right)={e}^{{x}} {e}^{{c}} \:=\:{Ae}^{{x}} \:. \\ $$
Commented by rahul 19 last updated on 03/Feb/19
$${thank}\:{you}\:{sir}. \\ $$