Moderate -0.8 This is a straightforward C1 question requiring basic algebraic expansion followed by term-by-term integration of powers of x. Part (a) is routine algebra, and part (b) applies standard integration rules to polynomial terms with no problem-solving insight needed. Easier than average A-level questions.
6. (a) Show that \(( 4 + 3 \sqrt { } x ) ^ { 2 }\) can be written as \(16 + k \sqrt { } x + 9 x\), where \(k\) is a constant to be found.
(b) Find \(\int ( 4 + 3 \sqrt { } x ) ^ { 2 } \mathrm {~d} x\).
- (a) e.g. \((4 + 3\sqrt{x})(4 + 3\sqrt{x})\) alone scores M1 A0 (but not \((4 + 3\sqrt{x})^2\) alone).
- e.g. \(16 + 12\sqrt{x} + 9x\) scores M1 A0.
- \(k = 24\) or \(16 + 24\sqrt{x} + 9x\) with no further evidence, scores full marks M1 A1.
- Correct solution only (cso): any wrong working seen loses the A mark.
- (b) A1: \(16x + \frac{9x^2}{2} + C\). Allow \(4.5\) or \(4\frac{1}{2}\) as equivalent to \(\frac{9}{2}\).
- A1ft: \(\frac{2k}{3} x^{\frac{3}{2}}\) (candidate's value of k, or general k).
- For this final mark, allow for example \(\frac{48}{3}\) as equivalent to 16, but do not allow unsimplified "double fractions" such as \(\frac{24}{\frac{3}{2}}\), and do not allow unsimplified "products" such as \(\frac{2}{3} \times 24\).
- A single term is required, e.g. \(8x^{\frac{3}{2}} + 8x^{\frac{3}{2}}\) is not enough.
- An otherwise correct solution with, say, C missing, followed by an incorrect solution including + C can be awarded full marks (isw, but allowing the C to appear at any stage).
**(a)** $(4 + 3\sqrt{x})(4 + 3\sqrt{x})$ seen, or a numerical value of k seen. ($k \neq 0$) | M1 |
$16 + 24\sqrt{x} + 9x$, or $k = 24$ | A1cso | (2 marks)
**(b)** $16 \to cx$ or $kx^{\frac{1}{2}} \to cx^{\frac{3}{2}}$ or $9x \to cx^2$ | M1 |
$\int [16 + 24\sqrt{x} + 9x] dx = 16x + \frac{9x^2}{2} + C$, or $16x + 16x^{\frac{3}{2}}$ | A1, A1ft | (3 marks)
**Total: 5 marks**
**Guidance:**
- (a) e.g. $(4 + 3\sqrt{x})(4 + 3\sqrt{x})$ alone scores M1 A0 (but not $(4 + 3\sqrt{x})^2$ alone).
- e.g. $16 + 12\sqrt{x} + 9x$ scores M1 A0.
- $k = 24$ or $16 + 24\sqrt{x} + 9x$ with no further evidence, scores full marks M1 A1.
- Correct solution only (cso): any wrong working seen loses the A mark.
- (b) A1: $16x + \frac{9x^2}{2} + C$. Allow $4.5$ or $4\frac{1}{2}$ as equivalent to $\frac{9}{2}$.
- A1ft: $\frac{2k}{3} x^{\frac{3}{2}}$ (candidate's value of k, or general k).
- For this final mark, allow for example $\frac{48}{3}$ as equivalent to 16, but do not allow unsimplified "double fractions" such as $\frac{24}{\frac{3}{2}}$, and do not allow unsimplified "products" such as $\frac{2}{3} \times 24$.
- A single term is required, e.g. $8x^{\frac{3}{2}} + 8x^{\frac{3}{2}}$ is not enough.
- An otherwise correct solution with, say, C missing, followed by an incorrect solution including + C can be awarded full marks (isw, but allowing the C to appear at any stage).
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6. (a) Show that $( 4 + 3 \sqrt { } x ) ^ { 2 }$ can be written as $16 + k \sqrt { } x + 9 x$, where $k$ is a constant to be found.\\
(b) Find $\int ( 4 + 3 \sqrt { } x ) ^ { 2 } \mathrm {~d} x$.\\
\hfill \mbox{\textit{Edexcel C1 2007 Q6 [5]}}