Standard +0.3 This is a straightforward binomial expansion problem requiring students to find coefficients of x² in two binomial expressions, set up an equation equal to 70, and solve a quadratic. It involves standard application of the binomial theorem with minimal algebraic manipulation, making it slightly easier than average but still requiring careful execution of multiple steps.
4 The coefficient of \(x ^ { 2 }\) in the expansion of \(\left( 1 + \frac { 2 } { p } x \right) ^ { 5 } + ( 1 + p x ) ^ { 6 }\) is 70 .
Find the possible values of the constant \(p\).
Coefficient of \(x^2\) in \((1+\frac{2}{p}x)^5\) is \(10(\frac{2}{p})^2 = \frac{10\times2^2}{p^2}[=\frac{40}{p^2}]\)
B1
Accept with \(x^2\) present. Must evaluate \(^5C_2\)
Coefficient of \(x^2\) in \((1+px)^6\) is \(15(p)^2[=15p^2]\)
B1
Accept with \(x^2\) present. Must evaluate \(^6C_2\)
\(\frac{40}{p^2}+15p^2=70\)
\*M1
Forming an equation in \(p\) with *their* coefficients, the given 70, no \(x\) terms and no extra terms.
\(15p^4-70p^2+40[=0]\) or \(3p^4-14p^2+8[=0]\)
DM1
Forming a 3-term equation in \(p\) (or another variable) with all terms on one side and *their* coefficients.
\([5](p^2-4)(3p^2-2)[=0]\) or \(\frac{70\pm\sqrt{70^2-4(15)(40)}}{30}\) or \(\frac{14\pm\sqrt{14^2-4(3)(8)}}{6}\)
DM1
Attempt to solve 3-term quartic (or quadratic in another variable) by factorisation, formula or completing the square.
\(p=\pm2,\ \pm\sqrt{\frac{2}{3}}\)
A1
OE e.g. \(\pm\frac{\sqrt{6}}{3}\) or AWRT \(\pm0.816\). If \*M1 DM1 DM0, allow SC B1 for 4 correct values.
6
## Question 4:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Coefficient of $x^2$ in $(1+\frac{2}{p}x)^5$ is $10(\frac{2}{p})^2 = \frac{10\times2^2}{p^2}[=\frac{40}{p^2}]$ | **B1** | Accept with $x^2$ present. Must evaluate $^5C_2$ |
| Coefficient of $x^2$ in $(1+px)^6$ is $15(p)^2[=15p^2]$ | **B1** | Accept with $x^2$ present. Must evaluate $^6C_2$ |
| $\frac{40}{p^2}+15p^2=70$ | **\*M1** | Forming an equation in $p$ with *their* coefficients, the given 70, no $x$ terms and no extra terms. |
| $15p^4-70p^2+40[=0]$ or $3p^4-14p^2+8[=0]$ | **DM1** | Forming a 3-term equation in $p$ (or another variable) with all terms on one side and *their* coefficients. |
| $[5](p^2-4)(3p^2-2)[=0]$ or $\frac{70\pm\sqrt{70^2-4(15)(40)}}{30}$ or $\frac{14\pm\sqrt{14^2-4(3)(8)}}{6}$ | **DM1** | Attempt to solve 3-term quartic (or quadratic in another variable) by factorisation, formula or completing the square. |
| $p=\pm2,\ \pm\sqrt{\frac{2}{3}}$ | **A1** | OE e.g. $\pm\frac{\sqrt{6}}{3}$ or AWRT $\pm0.816$. If \*M1 DM1 DM0, allow **SC B1** for 4 correct values. |
| | **6** | |
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4 The coefficient of $x ^ { 2 }$ in the expansion of $\left( 1 + \frac { 2 } { p } x \right) ^ { 5 } + ( 1 + p x ) ^ { 6 }$ is 70 .\\
Find the possible values of the constant $p$.\\
\hfill \mbox{\textit{CAIE P1 2022 Q4 [6]}}