Easy -1.3 This is a straightforward C1 indices question requiring basic recall of index laws. Part (a) is direct evaluation of a fractional power, and part (b) applies standard rules for negative and fractional indices to a simple expression. Both parts are routine textbook exercises with no problem-solving or insight required, making this easier than average.
(b) For \(2^{-2}\) or \(\frac{1}{4}\) or \(\left(\frac{1}{2}\right)^2\) or \(0.25\) as coefficient of \(x^k\), for any value of \(k\) including \(k = 0\)
M1
Correct index for \(x\) so \(Ax^{-2}\) or \(\frac{A}{x^2}\) o.e. for any value of \(A\)
B1
\(= \frac{1}{4x^2}\) or \(0.25 x^{-2}\)
A1 cao
(3 marks)
Notes:
(a)
B1: Answer 2 must be in part (a) for this mark
(b)
Look at their final answer
M1
For \(2^{-2}\) or \(\frac{1}{4}\) or \(\left(\frac{1}{2}\right)^2\) or \(0.25\) in their answer as coefficient of \(x^k\) for numerical value of \(k\) (including \(k = 0\)) so final answer \(\frac{1}{4}\) is M1 B0 A0
B1
\(Ax^{-2}\) or \(\frac{A}{x^2}\) or equivalent e.g. \(Ax^{\frac{10}{-5}}\) or \(Ax^{\frac{50}{-25}}\) i.e. correct power of \(x\) seen in final answer. May have a bracket provided it is \((Ax)^{-2}\) or \(\left(\frac{A}{x}\right)^2\)
A1
\(\frac{1}{4x^2}\) or \(\frac{1}{4}x^{-2}\) or \(0.25 x^{-2}\) oe but must be correct power and coefficient combined correctly and must not be followed by a different wrong answer.
Poor bracketing:
\(2x^{-2}\) earns M0 B1 A0 as correct power of \(x\) is seen in this solution (They can recover if they follow this with \(\frac{1}{4x^2}\) etc)
Special case:
\((2x)^{-2}\) as a final answer and \(\left(\frac{1}{2x}\right)^2\) can have M0 B1 A0 if the correct expanded answer is not seen. The correct answer \(\frac{1}{4x^2}\) etc. followed by \(\left(\frac{1}{2x}\right)^2\) or \((2x)^{-2}\), treat \(\frac{1}{4x^2}\) as final answer so M1 B1 A1 isw. But the correct answer \(\frac{1}{4x^2}\) etc clearly followed by the wrong \(2x^{-2}\) or \(4x^{-2}\), gets M1 B1 A0 do not ignore subsequent wrong work here
**(a)** $32^{\frac{1}{5}} = 2$ | B1 |
**(b)** For $2^{-2}$ or $\frac{1}{4}$ or $\left(\frac{1}{2}\right)^2$ or $0.25$ as coefficient of $x^k$, for any value of $k$ including $k = 0$ | M1 |
Correct index for $x$ so $Ax^{-2}$ or $\frac{A}{x^2}$ o.e. for any value of $A$ | B1 |
$= \frac{1}{4x^2}$ or $0.25 x^{-2}$ | A1 cao |
| | **(3 marks)** |
| **Notes:** | |
|---|---|
| **(a)** | B1: Answer 2 must be in part (a) for this mark |
| **(b)** | Look at their final answer |
| M1 | For $2^{-2}$ or $\frac{1}{4}$ or $\left(\frac{1}{2}\right)^2$ or $0.25$ in their answer as coefficient of $x^k$ for numerical value of $k$ (including $k = 0$) so final answer $\frac{1}{4}$ is M1 B0 A0 |
| B1 | $Ax^{-2}$ or $\frac{A}{x^2}$ or equivalent e.g. $Ax^{\frac{10}{-5}}$ or $Ax^{\frac{50}{-25}}$ i.e. correct power of $x$ seen in final answer. May have a bracket provided it is $(Ax)^{-2}$ or $\left(\frac{A}{x}\right)^2$ |
| A1 | $\frac{1}{4x^2}$ or $\frac{1}{4}x^{-2}$ or $0.25 x^{-2}$ oe but must be correct power and coefficient combined correctly and must not be followed by a different wrong answer. |
| **Poor bracketing:** | $2x^{-2}$ earns M0 B1 A0 as correct power of $x$ is seen in this solution (They can recover if they follow this with $\frac{1}{4x^2}$ etc) |
| **Special case:** | $(2x)^{-2}$ as a final answer and $\left(\frac{1}{2x}\right)^2$ can have M0 B1 A0 if the correct expanded answer is not seen. The correct answer $\frac{1}{4x^2}$ etc. followed by $\left(\frac{1}{2x}\right)^2$ or $(2x)^{-2}$, treat $\frac{1}{4x^2}$ as final answer so M1 B1 A1 isw. But the correct answer $\frac{1}{4x^2}$ etc clearly followed by the wrong $2x^{-2}$ or $4x^{-2}$, gets M1 B1 A0 do not ignore subsequent wrong work here |
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