Moderate -0.8 This is a straightforward exponential equation requiring a standard substitution technique. The substitution is explicitly given, leading to a simple quadratic equation. The algebraic manipulation is routine and the question only requires numerical solving, making it easier than average for A-level.
1 Using the substitution \(u = \mathrm { e } ^ { x }\), or otherwise, solve the equation
$$\mathrm { e } ^ { x } = 1 + 6 \mathrm { e } ^ { - x }$$
giving your answer correct to 3 significant figures.
Rearrange as \(e^{2x} - e^x - 6 = 0\), or \(u^2 - u - 6 = 0\), or equivalent
B1
Solve a 3-term quadratic for \(e^x\) or for \(u\)
M1
Obtain simplified solution \(e^x = 3\) or \(u = 3\)
A1
Obtain final answer \(x = 1.10\) and no other
A1
[4]
Rearrange as $e^{2x} - e^x - 6 = 0$, or $u^2 - u - 6 = 0$, or equivalent | B1 |
Solve a 3-term quadratic for $e^x$ or for $u$ | M1 |
Obtain simplified solution $e^x = 3$ or $u = 3$ | A1 |
Obtain final answer $x = 1.10$ and no other | A1 | [4]
1 Using the substitution $u = \mathrm { e } ^ { x }$, or otherwise, solve the equation
$$\mathrm { e } ^ { x } = 1 + 6 \mathrm { e } ^ { - x }$$
giving your answer correct to 3 significant figures.
\hfill \mbox{\textit{CAIE P3 2011 Q1 [4]}}