Moderate -0.8 This is a straightforward exponential equation requiring students to equate the two expressions, convert to the same base (both powers of 2), equate exponents, and solve a simple linear equation. The 'show that' format removes any challenge of finding the answer independently. This is easier than average A-level work as it's purely procedural with no problem-solving element.
2. The curves \(C _ { 1 }\) and \(C _ { 2 }\) have equations
$$\begin{aligned}
& C _ { 1 } : \quad y = 2 ^ { 3 x + 2 } \\
& C _ { 2 } : \quad y = 4 ^ { - x }
\end{aligned}$$
Show that the \(x\)-coordinate of the point where \(C _ { 1 }\) and \(C _ { 2 }\) intersect is \(\frac { - 2 } { 5 }\).
Writes \(4^{-x}\) as a power of 2 or equivalent, e.g. \(4^{-x} = 2^{-2x}\) or alternatively writes \(2^{-2x}\) as a power of 4, e.g. \(2^{3x+2} = \left(4^z\right)^{(3x+2)}\)
\(-2x = 3x + 2 \Rightarrow x = -\frac{2}{5}\)
dM1
Equates the indices and attempts to find \(x = \ldots\)
A1 cso
\(x = -\frac{2}{5}\)
(3) marks total for Question 2
**Show that the x-coordinate of the point where $C_1$ and $C_2$ intersect is $x = -\frac{2}{5}$**
| $4^{-x} = 2^{3x+2} \Rightarrow 2^{-2x} = 2^{3x+2}$ | M1 | Writes $4^{-x}$ as a power of 2 or equivalent, e.g. $4^{-x} = 2^{-2x}$ or alternatively writes $2^{-2x}$ as a power of 4, e.g. $2^{3x+2} = \left(4^z\right)^{(3x+2)}$ |
| --- | --- | --- |
| $-2x = 3x + 2 \Rightarrow x = -\frac{2}{5}$ | dM1 | Equates the indices and attempts to find $x = \ldots$ |
| | A1 cso | $x = -\frac{2}{5}$ |
**(3) marks total for Question 2**
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2. The curves $C _ { 1 }$ and $C _ { 2 }$ have equations
$$\begin{aligned}
& C _ { 1 } : \quad y = 2 ^ { 3 x + 2 } \\
& C _ { 2 } : \quad y = 4 ^ { - x }
\end{aligned}$$
Show that the $x$-coordinate of the point where $C _ { 1 }$ and $C _ { 2 }$ intersect is $\frac { - 2 } { 5 }$.\\
\hfill \mbox{\textit{Edexcel PMT Mocks Q2 [3]}}