The numbers P, Q and R for which the function | vedclass

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(D) Given $f(x) = P{e^{2x}} + Q{e^x} + Rx$. First,$f(0) = P + Q = -1$. Next,$f'(x) = 2P{e^{2x}} + Q{e^x} + R$. Given $f'(\log 2) = 31$,we have $2P(e^{

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$P = 5, Q = -6, R = 3$

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(D) Given $f(x) = P{e^{2x}} + Q{e^x} + Rx$. First,$f(0) = P + Q = -1$. Next,$f'(x) = 2P{e^{2x}} + Q{e^x} + R$. Given $f'(\log 2) = 31$,we have $2P(e^{\log 2})^2 + Q(e^{\log 2}) + R = 31$,which simplifies to $8P + 2Q + R = 31$. Now,consider the integral $\int_0^{\log 4} [f(x) - Rx] \, dx = \int_0^{\log 4} (P{e^{2x}} + Q{e^x}) \, dx = \frac{39}{2}$. Evaluating the integral: $[\frac{P}{2} e^{2x} + Q e^x]_0^{\log 4} = (\frac{P}{2} \cdot 16 + Q \cdot 4) - (\frac{P}{2} + Q) = 8P + 4Q - 0.5P - Q = 7.5P + 3Q = \frac{15P}{2} + 3Q = \frac{39}{2}$. This gives $15P + 6Q = 39$,or $5P + 2Q = 13$. We have the system: $1) P + Q = -1 \implies Q = -1 - P$ $2) 5P + 2(-1 - P) = 13 \implies 3P - 2 = 13 \implies 3P = 15 \implies P = 5$. Then $Q = -1 - 5 = -6$. Finally,substitute into $8P + 2Q + R = 31$: $8(5) + 2(-6) + R = 31 \implies 40 - 12 + R = 31 \implies 28 + R = 31 \implies R = 3$. Thus,$P = 5, Q = -6, R = 3$.

The numbers $P, Q$ and $R$ for which the function $f(x) = P{e^{2x}} + Q{e^x} + Rx$ satisfies the conditions $f(0) = -1$,$f'(\log 2) = 31$ and $\int_0^{\log 4} [f(x) - Rx] \, dx = \frac{39}{2}$ are given by

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