Moderate -0.3 This is a standard quadratic-in-exponential problem requiring substitution of y = e^x, multiplying through by e^x to get 3y² - 14y + 8 = 0, then solving the quadratic and taking logarithms. It's a routine technique taught in P2/C3 with clear steps, making it slightly easier than average, though not trivial since it requires recognizing the substitution and handling both positive roots correctly.
3 Given that \(3 \mathrm { e } ^ { x } + 8 \mathrm { e } ^ { - x } = 14\), find the possible values of \(\mathrm { e } ^ { x }\) and hence solve the equation \(3 \mathrm { e } ^ { x } + 8 \mathrm { e } ^ { - x } = 14\) correct to 3 significant figures.
Rearrange to \(3e^{2x} - 14e^x + 8 = 0\) or equivalent involving substitution
B1
Solve quadratic equation in \(e^x\) to find two values of \(e^x\)
*M1
Obtain \(\frac{2}{3}\) and 4
A1
Use natural logarithms to solve equation of form \(e^x = k\) where \(k > 0\) dep on
DM1
Allow M mark if left in exact form
M1
Obtain −0.405
A1
Obtain 1.39
A1
[6]
Rearrange to $3e^{2x} - 14e^x + 8 = 0$ or equivalent involving substitution | B1 |
Solve quadratic equation in $e^x$ to find two values of $e^x$ | *M1 |
Obtain $\frac{2}{3}$ and 4 | A1 |
Use natural logarithms to solve equation of form $e^x = k$ where $k > 0$ dep on | DM1 |
Allow M mark if left in exact form | M1 |
Obtain −0.405 | A1 |
Obtain 1.39 | A1 | [6]
3 Given that $3 \mathrm { e } ^ { x } + 8 \mathrm { e } ^ { - x } = 14$, find the possible values of $\mathrm { e } ^ { x }$ and hence solve the equation $3 \mathrm { e } ^ { x } + 8 \mathrm { e } ^ { - x } = 14$ correct to 3 significant figures.
\hfill \mbox{\textit{CAIE P2 2016 Q3 [6]}}