| Exam Board | OCR |
|---|---|
| Module | PURE |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Topic | Inequalities |
| Type | Solve quadratic inequality |
| Difficulty | Moderate -0.3 This is a multi-part question with routine techniques: (a) factorising a simple quadratic inequality, (b) substitution u=x^(3/2) to solve a disguised quadratic, (c) taking logarithms of an exponential equation. All are standard A-level methods with no novel insight required, though part (b) requires recognising the substitution pattern. Slightly easier than average due to straightforward application of well-practiced techniques. |
| Spec | 1.02g Inequalities: linear and quadratic in single variable1.02h Express solutions: using 'and', 'or', set and interval notation1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((x-2)(x+3) > 0\) | M1 | Attempt factorise |
| \(x < -3\), \(x > 2\) | A1 | Correct factors and \(> 0\) or \(y > 0\) |
| A1 | Any notation. Allow "and", "or", comma etc. But NOT \(-3 < x > 2\). \(x < -3\), \(x > 2\) seen, with no working or muddled working; SC B1 | |
| \(\{x: x < -3\} \cup \{x: x > 2\}\) | B1ft | Allow \(()\) instead of \(\{\}\). Allow \(x \in (-\infty,-3)\cup(2,\infty)\). ft their factors, dep two separate ranges |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\left(x^{\frac{3}{2}}+1\right)\left(x^{\frac{3}{2}}-8\right) = 0\) | M1 | Attempt factors of form \((x^{\frac{3}{2}} \pm k)\) or \((y \pm k)\). Or \((y+1)(y-8)\) or \(y=-1\) or \(y=8\) AND \(y=x^{\frac{3}{2}}\) soi. Allow \((x+1)(x-8)\) AND \(x = x^{\frac{3}{2}}\) seen |
| \(x^{\frac{3}{2}} = -1\) gives no solution | B1 | Condone inadequate reason |
| \(x^{\frac{3}{2}} = 8\) or \(x^3 = 64\) | A1 | \(y=8\) not enough for this mark |
| \(x = 4\) | A1 | Indep of previous A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\ln[(3^x)^2] = \ln[3 \times 2^x]\) | M1 | Attempt take logs. Allow errors, e.g. RHS \(= \ln 3 \times \ln(2^x)\) |
| \(2\ln(3^x)\) or \(\ln(3^{2x})\) or \(x\ln(3^2)\) or \(\ln(9^x)\) or \(2x\ln 3\) | A1 | LHS correct after one further step |
| \(= \ln 3 + \ln(2^x)\) | A1 | RHS correct after one further step |
| \(2x\ln 3\) or \(x\ln 9 = \ln 3 + x\ln 2\) | A1 | Both sides correct with \(x\) removed from index |
| \(x = \frac{\ln 3}{\ln\frac{9}{2}}\) | A1 | or any equivalent correct form e.g. \(\frac{1}{2-\log_3 2}\). Must be exact. ISW |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\ln(3^{2x-1}) = \ln(2^x)\) or \(\log_3(2^x) = 2x-1\) | M1 | Attempt take logs |
| \((2x-1)\ln 3 = x\ln 2\) or \(x\log_3 2 = 2x-1\) | A1 | LHS correct after one further step |
| A1 | RHS correct after one further step | |
| A1 | Both sides correct with \(x\) removed from index | |
| \(x = \frac{\ln 3}{2\ln 3 - \ln 2}\) or \(x = \frac{1}{2-\log_3 2}\) | A1 | ISW |
| Answer | Marks | Guidance |
| \(9^x = 3 \times 2^x\) | M1 | |
| \(\left(\frac{9}{2}\right)^x = 3\) | M1 | Divide by \(2^x\) and arrange into \(a^x = b\) form |
| \(4.5^x = 3\) | A1A1 | A1 for each side correct |
| \(x = \log_{4.5}(3)\) | A1 | ISW |
## Question 6:
### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(x-2)(x+3) > 0$ | M1 | Attempt factorise |
| $x < -3$, $x > 2$ | A1 | Correct factors **and** $> 0$ or $y > 0$ |
| | A1 | Any notation. Allow "and", "or", comma etc. But NOT $-3 < x > 2$. $x < -3$, $x > 2$ seen, with no working or muddled working; SC B1 |
| $\{x: x < -3\} \cup \{x: x > 2\}$ | B1ft | Allow $()$ instead of $\{\}$. Allow $x \in (-\infty,-3)\cup(2,\infty)$. ft their factors, dep two separate ranges |
### Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left(x^{\frac{3}{2}}+1\right)\left(x^{\frac{3}{2}}-8\right) = 0$ | M1 | Attempt factors of form $(x^{\frac{3}{2}} \pm k)$ or $(y \pm k)$. Or $(y+1)(y-8)$ or $y=-1$ or $y=8$ AND $y=x^{\frac{3}{2}}$ soi. Allow $(x+1)(x-8)$ AND $x = x^{\frac{3}{2}}$ seen |
| $x^{\frac{3}{2}} = -1$ gives no solution | B1 | Condone inadequate reason |
| $x^{\frac{3}{2}} = 8$ or $x^3 = 64$ | A1 | $y=8$ not enough for this mark |
| $x = 4$ | A1 | Indep of previous A1 |
### Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\ln[(3^x)^2] = \ln[3 \times 2^x]$ | M1 | Attempt take logs. Allow errors, e.g. RHS $= \ln 3 \times \ln(2^x)$ |
| $2\ln(3^x)$ or $\ln(3^{2x})$ or $x\ln(3^2)$ or $\ln(9^x)$ or $2x\ln 3$ | A1 | LHS correct after one further step |
| $= \ln 3 + \ln(2^x)$ | A1 | RHS correct after one further step |
| $2x\ln 3$ or $x\ln 9 = \ln 3 + x\ln 2$ | A1 | Both sides correct with $x$ removed from index |
| $x = \frac{\ln 3}{\ln\frac{9}{2}}$ | A1 | or any equivalent correct form e.g. $\frac{1}{2-\log_3 2}$. Must be exact. ISW |
**Alternative methods:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\ln(3^{2x-1}) = \ln(2^x)$ or $\log_3(2^x) = 2x-1$ | M1 | Attempt take logs |
| $(2x-1)\ln 3 = x\ln 2$ or $x\log_3 2 = 2x-1$ | A1 | LHS correct after one further step |
| | A1 | RHS correct after one further step |
| | A1 | Both sides correct with $x$ removed from index |
| $x = \frac{\ln 3}{2\ln 3 - \ln 2}$ or $x = \frac{1}{2-\log_3 2}$ | A1 | ISW |
| Answer | Marks | Guidance |
|--------|-------|----------|
| $9^x = 3 \times 2^x$ | M1 | |
| $\left(\frac{9}{2}\right)^x = 3$ | M1 | Divide by $2^x$ and arrange into $a^x = b$ form |
| $4.5^x = 3$ | A1A1 | A1 for each side correct |
| $x = \log_{4.5}(3)$ | A1 | ISW |6 In this question you must show detailed reasoning.
\begin{enumerate}[label=(\alph*)]
\item Solve the inequality $x ^ { 2 } + x - 6 > 0$, giving your answer in set notation.
\item Solve the equation $x ^ { 3 } - 7 x ^ { \frac { 3 } { 2 } } - 8 = 0$.
\item Find the exact solution of the equation $\left( 3 ^ { x } \right) ^ { 2 } = 3 \times 2 ^ { x }$.
\end{enumerate}
\hfill \mbox{\textit{OCR PURE Q6 [13]}}