Moderate -0.8 This is a straightforward substitution question requiring basic index law manipulation (recognizing 2^(2x+1) = 2·2^(2x) = 2y²) and solving a quadratic equation. Part (a) is essentially guided, and part (b) requires only factorizing/using the quadratic formula then taking logarithms. This is easier than average A-level content as it's highly procedural with minimal problem-solving.
6. (a) Given \(y = 2 ^ { x }\), show that
$$2 ^ { 2 x + 1 } - 17 \left( 2 ^ { x } \right) + 8 = 0$$
can be written in the form
$$2 y ^ { 2 } - 17 y + 8 = 0$$
(b) Hence solve
$$2 ^ { 2 x + 1 } - 17 \left( 2 ^ { x } \right) + 8 = 0$$
Replaces \(2^{2x+1}\) with \(2^{2x} \times 2\), or states \(2^{2x+1} = 2^{2x} \times 2\), or states \((2^x)^2 = 2^{2x}\)
M1
Uses the addition or power law of indices on \(2^{2x}\) or \(2^{2x+1}\). E.g. \(2^x \times 2^x = 2^{2x}\) or \((2^x)^2 = 2^{2x}\) or \(2^{2x+1} = 2 \times 2^{2x}\) or \(2^{x+0.5} = 2^x \times \sqrt{2}\) or \(2^{2x+1} = (2^{x+0.5})^2\)
cso. Complete proof including explicit statements for the addition and power law of indices on \(2^{2x+1}\) with no errors. Equation needs to be as printed including "\(= 0\)". If working backwards, do not need to write down printed answer first but must end with version in \(2^x\) including '\(= 0\)'
Solves the given quadratic either in terms of \(y\) or in terms of \(2^x\). Note completing the square e.g. \(y^2 - \frac{17}{2}y + 4 = 0\) requires \(\left(y \pm \frac{17}{4}\right)^2 \pm q \pm 4 = 0 \Rightarrow y = ...\)
\((y=)\frac{1}{2}, 8\) or \((2^x =)\frac{1}{2}, 8\)
A1
Correct values
\(\Rightarrow 2^x = \frac{1}{2}, 8 \Rightarrow x = -1, 3\)
M1 A1
M1: Either finds one correct value of \(x\) for their \(2^x\) or obtains a correct numerical expression e.g. for \(k > 0\), \(2^x = k \Rightarrow x = \log_2 k\) or \(\frac{\log k}{\log 2}\). A1: \(x = -1, 3\) only. Must be values of \(x\).
Total: 4 marks (part of 6 mark question)
## Question 6:
### Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Replaces $2^{2x+1}$ with $2^{2x} \times 2$, or states $2^{2x+1} = 2^{2x} \times 2$, or states $(2^x)^2 = 2^{2x}$ | M1 | Uses the addition **or** power law of indices on $2^{2x}$ or $2^{2x+1}$. E.g. $2^x \times 2^x = 2^{2x}$ or $(2^x)^2 = 2^{2x}$ or $2^{2x+1} = 2 \times 2^{2x}$ or $2^{x+0.5} = 2^x \times \sqrt{2}$ or $2^{2x+1} = (2^{x+0.5})^2$ |
| $2^{2x+1} - 17 \times 2^x + 8 = 0 \Rightarrow 2y^2 - 17y + 8 = 0$ | A1* | cso. Complete proof including explicit statements for the addition **and** power law of indices on $2^{2x+1}$ with no errors. Equation needs to be as printed including "$= 0$". If working backwards, do not need to write down printed answer first but must end with version in $2^x$ including '$= 0$' |
# Question (b) [Logarithms/Exponentials]:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2y^2 - 17y + 8 = 0 \Rightarrow (2y-1)(y-8)(=0) \Rightarrow y = ...$ or $2(2^x)^2 - 17(2^x) + 8 = 0 \Rightarrow (2(2^x)-1)((2^x)-8)(=0) \Rightarrow 2^x = ...$ | M1 | Solves the given quadratic either in terms of $y$ or in terms of $2^x$. Note completing the square e.g. $y^2 - \frac{17}{2}y + 4 = 0$ requires $\left(y \pm \frac{17}{4}\right)^2 \pm q \pm 4 = 0 \Rightarrow y = ...$ |
| $(y=)\frac{1}{2}, 8$ or $(2^x =)\frac{1}{2}, 8$ | A1 | Correct values |
| $\Rightarrow 2^x = \frac{1}{2}, 8 \Rightarrow x = -1, 3$ | M1 A1 | M1: Either finds one correct value of $x$ for their $2^x$ or obtains a correct numerical expression e.g. for $k > 0$, $2^x = k \Rightarrow x = \log_2 k$ or $\frac{\log k}{\log 2}$. A1: $x = -1, 3$ only. Must be values of $x$. |
**Total: 4 marks (part of 6 mark question)**
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