| Exam Board | OCR MEI |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2020 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Inequalities |
| Type | Exponential or logarithmic inequality |
| Difficulty | Standard +0.3 This is a straightforward multi-part question on inequalities and logarithms. Part (a) tests understanding of domain restrictions (x≥0 for square root), part (b) is routine exponential inequality solving, and part (c) involves standard logarithm manipulation and showing uniqueness. All parts use well-practiced techniques with no novel insight required, making it slightly easier than average. |
| Spec | 1.02a Indices: laws of indices for rational exponents1.06c Logarithm definition: log_a(x) as inverse of a^x1.06g Equations with exponentials: solve a^x = b1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| The argument is not correct. \(x < 16\) includes negative values of \(x\) for which \(x^{\frac{1}{2}}\) does not exist, so the statement does not imply that \(x^{\frac{1}{2}} < 4\) | E1 [1] | Allow that \(x\) must be positive. Allow correct solution \(0 \leq x < 16\) or \(0 < x < 16\) without further explanation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Take logs of both sides: \(x\log\left(\frac{1}{2}\right) < \log 4\) | M1 | Use of laws of logs must be seen. Allow equivalent with natural logs |
| \(x > \dfrac{\log 4}{\log\left(\frac{1}{2}\right)}\) since \(\log\left(\frac{1}{2}\right)\) is negative | B1 | Award for boundary value even if only evaluated. Correct inequality |
| \(x > -2\) | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Solve \(\left(\frac{1}{2}\right)^x = 4\) by taking logs base \(\frac{1}{2}\) | M1 | Using log base \(\frac{1}{2}\) |
| \(\log_{\frac{1}{2}}(4) = -2\) | B1 | Award for boundary value even if only seen as part of an equation or incorrect inequality |
| Test value e.g. \(x=0\): \(\left(\frac{1}{2}\right)^0 = 1 < 4\), so \(x > -2\) | A1 [3] | Correct inequality |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Using laws of logs: \(\log_2(x+8)^2 - \log_2(x+6) = 3\) | M1 | At least one correct use of laws of logs |
| \(\log_2 \dfrac{(x+8)^2}{(x+6)} = 3\) | ||
| \(\dfrac{(x+8)^2}{(x+6)} = 2^3\) | M1 | Clearing logs to obtain \(2^3\) or \(8\) seen in an equation |
| \((x+8)^2 = 8(x+6)\) | A1 | Correct quadratic |
| \(x^2 + 8x + 16 = 0\); Discriminant is \(8^2 - 4 \times 1 \times 16 = 0\) | M1 | Attempt to find discriminant of their quadratic (allow one slip). Allow M1 for attempt to solve their quadratic |
| So there is only one solution | A1 [5] | Correct argument from zero discriminant or repeated root \(x = -4\) found |
## Question 11:
**Part (a):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| The argument is not correct. $x < 16$ includes negative values of $x$ for which $x^{\frac{1}{2}}$ does not exist, so the statement does not imply that $x^{\frac{1}{2}} < 4$ | E1 [1] | Allow that $x$ must be positive. Allow correct solution $0 \leq x < 16$ or $0 < x < 16$ without further explanation |
**Part (b):**
**EITHER method:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Take logs of both sides: $x\log\left(\frac{1}{2}\right) < \log 4$ | M1 | Use of laws of logs must be seen. Allow equivalent with natural logs |
| $x > \dfrac{\log 4}{\log\left(\frac{1}{2}\right)}$ since $\log\left(\frac{1}{2}\right)$ is negative | B1 | Award for boundary value even if only evaluated. Correct inequality |
| $x > -2$ | A1 [3] | |
**OR method:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Solve $\left(\frac{1}{2}\right)^x = 4$ by taking logs base $\frac{1}{2}$ | M1 | Using log base $\frac{1}{2}$ |
| $\log_{\frac{1}{2}}(4) = -2$ | B1 | Award for boundary value even if only seen as part of an equation or incorrect inequality |
| Test value e.g. $x=0$: $\left(\frac{1}{2}\right)^0 = 1 < 4$, so $x > -2$ | A1 [3] | Correct inequality |
**Part (c):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Using laws of logs: $\log_2(x+8)^2 - \log_2(x+6) = 3$ | M1 | At least one correct use of laws of logs |
| $\log_2 \dfrac{(x+8)^2}{(x+6)} = 3$ | | |
| $\dfrac{(x+8)^2}{(x+6)} = 2^3$ | M1 | Clearing logs to obtain $2^3$ or $8$ seen in an equation |
| $(x+8)^2 = 8(x+6)$ | A1 | Correct quadratic |
| $x^2 + 8x + 16 = 0$; Discriminant is $8^2 - 4 \times 1 \times 16 = 0$ | M1 | Attempt to find discriminant of their quadratic (allow one slip). Allow M1 for attempt to solve their quadratic |
| So there is only one solution | A1 [5] | Correct argument from zero discriminant or repeated root $x = -4$ found |11 In this question you must show detailed reasoning.
\begin{enumerate}[label=(\alph*)]
\item A student is asked to solve the inequality $x ^ { \frac { 1 } { 2 } } < 4$.
The student argues that $x ^ { \frac { 1 } { 2 } } < 4 \Leftrightarrow x < 16$, so that the solution is $\{ x : x < 16 \}$.\\
Comment on the validity of the student's argument.
\item Solve the inequality $\left( \frac { 1 } { 2 } \right) ^ { x } < 4$.
\item Show that the equation $2 \log _ { 2 } ( x + 8 ) - \log _ { 2 } ( x + 6 ) = 3$ has only one root.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI AS Paper 1 2020 Q11 [9]}}