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f(x) = \dfrac{e^x - 1}{e - 1} \ for \ x \ \geq 0[br] \[br]Let y = f(x)[br]y = \dfrac{e^x - 1}{e - 1} \[br]y(e - 1) = e^x - 1[br]y(e - 1) +1 = e^x[br]x = \ln (y(e - 1) + 1)[br]g(x) = \ln (x(e - 1) + 1)[br] \[br]Drawing the graph...[br] \[br]y = \dfrac{e^x - 1}{e - 1}
\int^{\frac{1}{2}}_0 f(x) \ dx[br]= \int^{\frac{1}{2}}_0 \dfrac{e^x - 1}{e - 1} \ dx[br]= \dfrac{1}{e - 1} \int^{\frac{1}{2}}_0e^x - 1 \ dx[br]= \dfrac{1}{e - 1} \left[ e^x - x\right]^{\frac{1}{2}}_0[br]= \dfrac{1}{e - 1} \left[ \sqrt e - \frac{3}{2} \right] = \dfrac{2 \sqrt e - 3}{2(e - 1)}[br]\[br]\int^k_0 g(x) \ dx = \int^k_0 \ln (x(e - 1) + 1) \ dx [br]Let \ u = x(e - 1) + 1 = ex - x + 1[br]\dfrac{du}{dx} = e - 1[br]\dfrac{1}{e - 1} du = dx
Spoiler
Introduction
Part (i)
Position of P
Velocity of P
Total kinetic energy
\angleP_1OP_2=90^{\circ},\mathrm{\ so\ } \angleP_1CP_2=90^{\circ}\mathrm{\ and\ }P_1OCP_2 \mathrm{\ is\ cyclic\ }
\mathrm{if\ }ON\mathrm{\ is\ vertical\ then\ we\ have}\angleCGO=2\lambda
\angleGON=2\lambda \mathrm{\ alternate\ angles\ }\angleNOP_1=\alpha \mathrm{\ alternate\ angles\ and\ }\angleGOP_1=\frac{\pi}{4}
\mathrm{if\ }\theta\mathrm{\ is\ the\ angle\ of\ the\ rod\ to \ the\ horizontal\ then\ }\angleCGO=\theta\mathrm{\ hence\ }\theta=2\lambda
2\lambda+\alpha=\frac{\pi}{4} \Rightarrow 0<\lambda<22\frac{1}{2}^\circ}
-kmc+m(-v+u\cos\theta)=0\mathrm{\ and\ }u=at\mathrm_\ so}
at\cos\theta-v=at\sin\theta \Rightarrowv=at(\cos\theta-\sin\theta)
\Rightarrow \sin\theta= \frac{k\cos\theta}{k+1} \RIghtarrow\tan\theta= \frac{k}{k+1} \mathrm{\ as\ required}
\mathrm{Resolving\ along\ the\ slope\ }mg\sintheta-\muN=m\left(-\frac{dv}{dt}\cos\theta+a\Right) \Rightarrow3mg-5\muN=5m\left(-\frac{4a}{25}+a\right)
\mathrm{Resolving\ perpendicular\ to\ slope\ }mg\cosTheta-N=m\frac{dv}{dt}\sin\theta \Rightarrow4mg-5N=\frac{3am}{5}
=\frac{1}{2n-3}-\left(\frac{1}{2n-3}\righ)^2=\frac{2(n-2)}{(2n-3)^2} \mathrm{\ etc.}
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