| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2010 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discriminant and conditions for roots |
| Type | Show discriminant inequality, then solve |
| Difficulty | Moderate -0.3 This is a straightforward C1 question testing completing the square and discriminant conditions. Part (a) is routine manipulation, part (b)(i) requires standard discriminant ≥ 0 setup with basic algebra, and part (b)(ii) involves solving a quadratic inequality. All techniques are standard textbook exercises with clear signposting and no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.02e Complete the square: quadratic polynomials and turning points1.02g Inequalities: linear and quadratic in single variable |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(2(x-5)^2 + 3\) | M1A1 | M1 for \(2(x-5)^2\), A1 for \(q=3\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(2(x-5)^2 \geq 0\) so \(2(x-5)^2 + 3 \geq 3 > 0\) | B1 | Minimum value is 3 |
| Therefore expression is always positive, cannot equal zero | B1 | Clear conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| For real roots: \(b^2 - 4ac \geq 0\) | M1 | Using discriminant condition |
| \((k+1)^2 - 4(2k-1)(k) \geq 0\) | M1 | Correct substitution |
| \(k^2 + 2k + 1 - 4k(2k-1) \geq 0\) | M1 | Expanding correctly |
| \(k^2 + 2k + 1 - 8k^2 + 4k \geq 0 \Rightarrow -7k^2 + 6k + 1 \geq 0\) | A1 | Hence \(7k^2 - 6k - 1 \leq 0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \((7k+1)(k-1) \leq 0\) | M1 | Factorising |
| Critical values \(k = -\frac{1}{7}\) and \(k = 1\) | A1 | Both values correct |
| \(-\frac{1}{7} \leq k \leq 1\) | M1A1 | Correct inequality, condone \(2k-1 \neq 0\) check |
## Question 7:
**Part (a)(i):** Express $2x^2 - 20x + 53$ in form $2(x-p)^2 + q$
| Working | Mark | Guidance |
|---------|------|----------|
| $2(x-5)^2 + 3$ | M1A1 | M1 for $2(x-5)^2$, A1 for $q=3$ |
**Part (a)(ii):** Explain why $2x^2 - 20x + 53 = 0$ has no real roots
| Working | Mark | Guidance |
|---------|------|----------|
| $2(x-5)^2 \geq 0$ so $2(x-5)^2 + 3 \geq 3 > 0$ | B1 | Minimum value is 3 |
| Therefore expression is always positive, cannot equal zero | B1 | Clear conclusion |
**Part (b)(i):** Show $7k^2 - 6k - 1 \leq 0$
| Working | Mark | Guidance |
|---------|------|----------|
| For real roots: $b^2 - 4ac \geq 0$ | M1 | Using discriminant condition |
| $(k+1)^2 - 4(2k-1)(k) \geq 0$ | M1 | Correct substitution |
| $k^2 + 2k + 1 - 4k(2k-1) \geq 0$ | M1 | Expanding correctly |
| $k^2 + 2k + 1 - 8k^2 + 4k \geq 0 \Rightarrow -7k^2 + 6k + 1 \geq 0$ | A1 | Hence $7k^2 - 6k - 1 \leq 0$ |
**Part (b)(ii):** Find possible values of $k$
| Working | Mark | Guidance |
|---------|------|----------|
| $(7k+1)(k-1) \leq 0$ | M1 | Factorising |
| Critical values $k = -\frac{1}{7}$ and $k = 1$ | A1 | Both values correct |
| $-\frac{1}{7} \leq k \leq 1$ | M1A1 | Correct inequality, condone $2k-1 \neq 0$ check |7
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Express $2 x ^ { 2 } - 20 x + 53$ in the form $2 ( x - p ) ^ { 2 } + q$, where $p$ and $q$ are integers.
\item Use your result from part (a)(i) to explain why the equation $2 x ^ { 2 } - 20 x + 53 = 0$ has no real roots.
\end{enumerate}\item The quadratic equation $( 2 k - 1 ) x ^ { 2 } + ( k + 1 ) x + k = 0$ has real roots.
\begin{enumerate}[label=(\roman*)]
\item Show that $7 k ^ { 2 } - 6 k - 1 \leqslant 0$.
\item Hence find the possible values of $k$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C1 2010 Q7 [12]}}