Moderate -0.3 This is a straightforward modulus inequality requiring students to split into two cases (2^x - 8 < 5 and 2^x - 8 > -5), then solve simple exponential inequalities using logarithms. The technique is standard for P3 level with no conceptual surprises, making it slightly easier than average but not trivial since it requires correct handling of the modulus definition and logarithmic manipulation.
State or imply non-modular inequality \(-5 < 2^x - 8 < 5\), or \((2^x - 8)^2 < 5^2\) or corresponding pair of linear equations or quadratic equation
B1
Use correct method for solving an equation of the form \(2^x = a\)
M1
Obtain critical values 1.58 and 3.70, or exact equivalents
A1
State correct answer \(1.58 < x < 3.70\)
A1
OR: Use correct method for solving an equation of the form \(2^x = a\)
M1
Obtain one critical value (probably 3.70), or exact equivalent
A1
Obtain the other critical value, or exact equivalent
A1
State correct answer \(1.58 < x < 3.70\)
A1
Notes: Allow 1.59 and 3.7. Condone \(\leq\) for \(<\). Allow final answers given separately. Exact equivalents must be in terms of ln or logarithms to base 10. SR: Solutions given as logarithms to base 2 can only earn M1 and B1 of the first scheme.
Total: [4]
## Question 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| State or imply non-modular inequality $-5 < 2^x - 8 < 5$, or $(2^x - 8)^2 < 5^2$ or corresponding pair of linear equations or quadratic equation | B1 | |
| Use correct method for solving an equation of the form $2^x = a$ | M1 | |
| Obtain critical values 1.58 and 3.70, or exact equivalents | A1 | |
| State correct answer $1.58 < x < 3.70$ | A1 | |
| **OR:** Use correct method for solving an equation of the form $2^x = a$ | M1 | |
| Obtain one critical value (probably 3.70), or exact equivalent | A1 | |
| Obtain the other critical value, or exact equivalent | A1 | |
| State correct answer $1.58 < x < 3.70$ | A1 | |
**Notes:** Allow 1.59 and 3.7. Condone $\leq$ for $<$. Allow final answers given separately. Exact equivalents must be in terms of ln or logarithms to base 10. SR: Solutions given as logarithms to base 2 can only earn M1 and B1 of the first scheme.
**Total: [4]**
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