CAIE P3 2023 November — Question 1 4 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2023
SessionNovember
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicModulus function
TypeSolve |exponential| < constant
DifficultyStandard +0.3 This is a straightforward modulus inequality with an exponential. Students split into two cases (2^(x+1) - 2 < 0.5 and 2^(x+1) - 2 > -0.5), rearrange to isolate 2^(x+1), then apply logarithms. The exponential context adds slight complexity over a linear modulus inequality, but the method is standard and mechanical with no conceptual obstacles.
Spec1.02l Modulus function: notation, relations, equations and inequalities1.06g Equations with exponentials: solve a^x = b

1 Find the set of values of \(x\) satisfying the inequality \(\left| 2 ^ { x + 1 } - 2 \right| < 0.5\), giving your answer to 3 significant figures.

Question 1:
AnswerMarks Guidance
AnswerMark Guidance
State or imply non-modular inequality \(-0.5 < 2^{x+1} - 2 < 0.5\), can be in two separate statements, or \(\left(2^{x+1}-2\right)^2 < 0.5^2\), or corresponding pair of linear equations \(0.5 = 2^{x+1} - 2\) and \(-0.5 = 2^{x+1} - 2\), or quadratic equation \(\left(2^{x+1}-2\right)^2 = 0.5^2\)B1 \(-0.25 < 2^x - 1 < 0.25\), can be in two separate statements, or \(\left(2^x-1\right)^2 < 0.25^2\) or corresponding pair of linear equations \(0.25 = 2^x - 1\) and \(-0.25 = 2^x - 1\) or quadratic equation \(\left(2^x-1\right)^2 = 0.25^2\). Incorrect inequality mark recoverable by correct final answer or \(x < 0.32\) and \(x > -0.42\)
Use correct method for solving an equation or inequality of the form \(2^{x+1} = a\) or \(2^x = b\) where \(a, b > 0\)M1 Reach \((x+1)\ln 2 = \ln a\) or equivalent, do not need to reach \(x = \ldots\)
Obtain critical values \(x = 0.322\) and \(-0.415\), or awrt \(x = 0.32\) and \(-0.42\), or exact equivalentsA1 e.g. \(\frac{\ln 2.5}{\ln 2}-1\) and \(\frac{\ln 1.5}{\ln 2}-1\)
State final answer \(-0.415 < x < 0.322\) or \((-0.415, 0.322)\)A1 Need 3 significant figures. Need combined result, not \(x < 0.32\) and \(x > -0.42\). Must be strict inequalities. No working, 0/4.
Alternative method for Question 1:
AnswerMarks Guidance
AnswerMark Guidance
Use correct method for solving an equation or inequality of the form \(2^{x+1} = a\) or \(2^x = b\) where \(a, b > 0\)M1 May see \(2^{x+1} = 1.5\) and \(2^{x+1} = 2.5\). Reach \((x+1)\ln 2 = \ln a\) or equivalent, don't need to reach \(x = \ldots\)
Obtain one critical value, e.g. \(0.322\), or awrt \(x = 0.32\), or exact equivalentA1 e.g. \(\frac{\ln 2.5}{\ln 2}-1\)
Obtain the other critical value e.g. \(-0.415\) or awrt \(x = -0.42\) or exact equivalentA1 e.g. \(\frac{\ln 1.5}{\ln 2}-1\)
Question 1:
AnswerMarks Guidance
AnswerMarks Guidance
\(-0.415 < x < 0.322\) or \((-0.415, 0.322)\)A1 Need 3 significant figures. Need combined result, not \(x < 0.32\) and \(x > -0.42\). Must be strict inequalities. No working, 0/4.
4
## Question 1:

| Answer | Mark | Guidance |
|--------|------|----------|
| State or imply non-modular inequality $-0.5 < 2^{x+1} - 2 < 0.5$, can be in two separate statements, or $\left(2^{x+1}-2\right)^2 < 0.5^2$, or corresponding pair of linear equations $0.5 = 2^{x+1} - 2$ and $-0.5 = 2^{x+1} - 2$, or quadratic equation $\left(2^{x+1}-2\right)^2 = 0.5^2$ | B1 | $-0.25 < 2^x - 1 < 0.25$, can be in two separate statements, or $\left(2^x-1\right)^2 < 0.25^2$ or corresponding pair of linear equations $0.25 = 2^x - 1$ and $-0.25 = 2^x - 1$ or quadratic equation $\left(2^x-1\right)^2 = 0.25^2$. Incorrect inequality mark recoverable by correct final answer or $x < 0.32$ and $x > -0.42$ |
| Use correct method for solving an equation or inequality of the form $2^{x+1} = a$ or $2^x = b$ where $a, b > 0$ | M1 | Reach $(x+1)\ln 2 = \ln a$ or equivalent, do not need to reach $x = \ldots$ |
| Obtain critical values $x = 0.322$ and $-0.415$, or awrt $x = 0.32$ and $-0.42$, or exact equivalents | A1 | e.g. $\frac{\ln 2.5}{\ln 2}-1$ and $\frac{\ln 1.5}{\ln 2}-1$ |
| State final answer $-0.415 < x < 0.322$ or $(-0.415, 0.322)$ | A1 | Need 3 significant figures. Need combined result, not $x < 0.32$ and $x > -0.42$. Must be strict inequalities. No working, 0/4. |

**Alternative method for Question 1:**

| Answer | Mark | Guidance |
|--------|------|----------|
| Use correct method for solving an equation or inequality of the form $2^{x+1} = a$ or $2^x = b$ where $a, b > 0$ | M1 | May see $2^{x+1} = 1.5$ and $2^{x+1} = 2.5$. Reach $(x+1)\ln 2 = \ln a$ or equivalent, don't need to reach $x = \ldots$ |
| Obtain one critical value, e.g. $0.322$, or awrt $x = 0.32$, or exact equivalent | A1 | e.g. $\frac{\ln 2.5}{\ln 2}-1$ |
| Obtain the other critical value e.g. $-0.415$ or awrt $x = -0.42$ or exact equivalent | A1 | e.g. $\frac{\ln 1.5}{\ln 2}-1$ |

## Question 1:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $-0.415 < x < 0.322$ or $(-0.415, 0.322)$ | A1 | Need 3 significant figures. Need combined result, not $x < 0.32$ and $x > -0.42$. Must be strict inequalities. No working, 0/4. |
| | **4** | |

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1 Find the set of values of $x$ satisfying the inequality $\left| 2 ^ { x + 1 } - 2 \right| < 0.5$, giving your answer to 3 significant figures.\\

\hfill \mbox{\textit{CAIE P3 2023 Q1 [4]}}