Easy -1.2 This is a straightforward two-part question testing basic manipulation of indices and rationalising surds. Part (i) requires simple index law application (49 = 7², √7 = 7^(1/2)), while part (ii) is a standard 'show that' requiring rationalising the denominator by multiplying by the conjugate—both are routine textbook exercises with no problem-solving insight needed.
2. (i) Given that \(\frac { 49 } { \sqrt { 7 } } = 7 ^ { a }\), find the value of \(a\).
(ii) Show that \(\frac { 10 } { \sqrt { 18 } - 4 } = 15 \sqrt { 2 } + 20\)
You must show all stages of your working.
Way 1: \(\frac{49}{\sqrt{7}}=\frac{7^2}{7^{\frac{1}{2}}}=7^{2-\frac{1}{2}}\) Way 2: \(7\sqrt{7}=7^{1+\frac{1}{2}}\) Way 3: \(7^a=\frac{49}{\sqrt{7}}\Rightarrow a=\frac{\log\frac{49}{\sqrt{7}}}{\log 7}\) or \(a=\log_7\frac{49}{\sqrt{7}}\)
M1
Way 1: Subtracts their powers of 7. Way 2: Cancels fraction to \(7\sqrt{7}\) and adds their powers. Way 3: Correct use of logs to obtain correct expression for \(a\)
\((a=)1\frac{1}{2}\) (oe) or \(7^{1\frac{1}{2}}\) seen
A1
cao — do not allow inexact decimals e.g. \(1.4999...\Rightarrow a=1.5\) scores M1A0
Part (ii):
Answer
Marks
Guidance
Working
Mark
Guidance
Way 1: \(\frac{10(\sqrt{18}+4)}{(\sqrt{18}-4)(\sqrt{18}+4)}\) Way 2: \((15\sqrt{2}+20)(\sqrt{18}-4)\)
M1
Way 1: Multiply numerator and denominator by \(\sqrt{18}+4\) or equivalent. Way 2: Attempts to expand to at least 3 terms
\(=\frac{\cdots}{2}\) or \(=15\sqrt{36}-60\sqrt{2}+20\sqrt{18}-80\)
B1
Way 1: Correctly obtains \(\pm2\) in denominator (must follow M1). Way 2: All 4 terms correct (must follow M1)
\(\frac{10}{\sqrt{18}-4}=5(3\sqrt{2}+4)=15\sqrt{2}+20^*\) or \(=90-60\sqrt{2}+60\sqrt{2}-80=10\) so \(\frac{10}{\sqrt{18}-4}=15\sqrt{2}+20^*\)
A1cso
Way 1: Correct result with no errors and \(\sqrt{18}=3\sqrt{2}\) used before final answer. Way 2: Obtains 10 with no errors and \(\sqrt{18}=3\sqrt{2}\) seen, and conclusion stated
## Question 2:
### Part (i):
| Working | Mark | Guidance |
|---------|------|----------|
| **Way 1:** $\frac{49}{\sqrt{7}}=\frac{7^2}{7^{\frac{1}{2}}}=7^{2-\frac{1}{2}}$ **Way 2:** $7\sqrt{7}=7^{1+\frac{1}{2}}$ **Way 3:** $7^a=\frac{49}{\sqrt{7}}\Rightarrow a=\frac{\log\frac{49}{\sqrt{7}}}{\log 7}$ or $a=\log_7\frac{49}{\sqrt{7}}$ | M1 | Way 1: Subtracts their powers of 7. Way 2: Cancels fraction to $7\sqrt{7}$ and adds their powers. Way 3: Correct use of logs to obtain correct expression for $a$ |
| $(a=)1\frac{1}{2}$ (oe) or $7^{1\frac{1}{2}}$ seen | A1 | cao — do not allow inexact decimals e.g. $1.4999...\Rightarrow a=1.5$ scores M1A0 |
### Part (ii):
| Working | Mark | Guidance |
|---------|------|----------|
| **Way 1:** $\frac{10(\sqrt{18}+4)}{(\sqrt{18}-4)(\sqrt{18}+4)}$ **Way 2:** $(15\sqrt{2}+20)(\sqrt{18}-4)$ | M1 | Way 1: Multiply numerator and denominator by $\sqrt{18}+4$ or equivalent. Way 2: Attempts to expand to at least 3 terms |
| $=\frac{\cdots}{2}$ **or** $=15\sqrt{36}-60\sqrt{2}+20\sqrt{18}-80$ | B1 | Way 1: Correctly obtains $\pm2$ in denominator (must follow M1). Way 2: All 4 terms correct (must follow M1) |
| $\frac{10}{\sqrt{18}-4}=5(3\sqrt{2}+4)=15\sqrt{2}+20^*$ **or** $=90-60\sqrt{2}+60\sqrt{2}-80=10$ so $\frac{10}{\sqrt{18}-4}=15\sqrt{2}+20^*$ | A1cso | Way 1: Correct result with no errors and $\sqrt{18}=3\sqrt{2}$ used before final answer. Way 2: Obtains 10 with no errors and $\sqrt{18}=3\sqrt{2}$ seen, and conclusion stated |
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2. (i) Given that $\frac { 49 } { \sqrt { 7 } } = 7 ^ { a }$, find the value of $a$.\\
(ii) Show that $\frac { 10 } { \sqrt { 18 } - 4 } = 15 \sqrt { 2 } + 20$
You must show all stages of your working.\\
\hfill \mbox{\textit{Edexcel C12 2016 Q2 [5]}}