Indices and surds - Edexcel A-Level Maths (Paper 1) Exam Questions

Practise 16 exam-style questions on Indices and surds for Edexcel A-Level Maths (Paper 1). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with the Study Helper on hand if you get stuck.

16 questions
2–8 marks each
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16 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

3 marks

Solutions relying entirely on calculator technology are not acceptable.

Find the exact value of \(x\) for which
\[\dfrac{9^x}{3^{1 - x}} = 27\sqrt{3}\]

Model answer

\(9^x = \left(3^2\right)^x = 3^{2x}\)
\(\dfrac{9^x}{3^{1 - x}} = 3^{2x - (1 - x)} = 3^{3x - 1}\)
\(27\sqrt{3} = 3^3 \times 3^{\frac{1}{2}} = 3^{\frac{7}{2}}\)
So \(3x - 1 = \dfrac{7}{2}\)
\(3x = \dfrac{9}{2}\)
\(x = \dfrac{3}{2}\)

Explanation

Write everything as a power of 3: \(9^x = 3^{2x}\) and \(27\sqrt{3} = 3^{\frac{7}{2}}\), then subtract the indices when dividing and equate the powers. The B mark is for \(3^{\frac{7}{2}}\) and the M mark for forming a linear equation in \(x\). Leaving out the bracket round \(1 - x\) when subtracting is the usual error.

Mark scheme

1 mark
B1 Writes \(27\sqrt{3}\) as a power of 3: \(3^{\frac{7}{2}}\) (or \(3^3 \times 3^{\frac{1}{2}}\)). An answer with no working scores 0 marks.
An answer of \(x = \dfrac{3}{2}\) with no working scores 0 marks: solutions relying entirely on calculator technology are not acceptable.
1 mark
M1 Writes the left side as a single power of 3 and equates indices: \(3^{2x - (1 - x)} = 3^{\frac{7}{2}}\), so \(2x - (1 - x) = \dfrac{7}{2}\) oe
Allow one sign slip in the index for M1, e.g. \(2x - 1 - x\). Taking logarithms of both sides correctly and reaching a linear equation in \(x\) is also M1.
1 mark
A1 \(x = \dfrac{3}{2}\)
cao. Accept 1.5. A common error is \(2x - 1 - x = \dfrac{7}{2}\) (a bracket error), giving \(x = \dfrac{9}{2}\): M1 A0.

Question 2

4 marks

Given that \(x > 0\), write
\[\dfrac{\left(3x - 2\sqrt{x}\right)^2}{x\sqrt{x}}\]
in the form \(Ax^p + B + Cx^q\), where \(A\), \(B\), \(C\), \(p\) and \(q\) are constants.

Model answer

\(\left(3x - 2\sqrt{x}\right)^2 = 9x^2 - 12x\sqrt{x} + 4x = 9x^2 - 12x^{\frac{3}{2}} + 4x\)
\(x\sqrt{x} = x^{\frac{3}{2}}\)
\(\dfrac{9x^2 - 12x^{\frac{3}{2}} + 4x}{x^{\frac{3}{2}}} = 9x^{2 - \frac{3}{2}} - 12 + 4x^{1 - \frac{3}{2}}\)
\(= 9x^{\frac{1}{2}} - 12 + 4x^{-\frac{1}{2}}\)

Explanation

Expand the bracket fully, remembering the middle term \(-12x\sqrt{x}\), then write \(x\sqrt{x} = x^{\frac{3}{2}}\) and divide every term by it, subtracting the indices. The method marks are for the expansion and for dividing term by term. A common error is to square each term separately and lose the middle term.

Mark scheme

1 mark
M1 Expands \(\left(3x - 2\sqrt{x}\right)^2\) to three terms, e.g. \(9x^2 - 12x\sqrt{x} + 4x\); or a correct final answer of \(9x^{\frac{1}{2}} - 12 + 4x^{-\frac{1}{2}}\) with no working
An answer of \(9x^{\frac{1}{2}} - 12 + 4x^{-\frac{1}{2}}\) with no working scores 4 marks. \(\left(3x - 2\sqrt{x}\right)^2 = 9x^2 - 4x\) (two terms) is M0.
1 mark
A1 \(9x^2 - 12x^{\frac{3}{2}} + 4x\)
Accept \(9x^2 - 12x\sqrt{x} + 4x\).
1 mark
M1 Writes \(x\sqrt{x} = x^{\frac{3}{2}}\) and divides each term by it, subtracting indices correctly for at least one term
\(x\sqrt{x} = x^{\frac{1}{2}}\) is a common error: M0 unless the division is otherwise correct.
1 mark
A1 \(9x^{\frac{1}{2}} - 12 + 4x^{-\frac{1}{2}}\)
cao. Accept \(A = 9\), \(p = \tfrac{1}{2}\), \(B = -12\), \(C = 4\), \(q = -\tfrac{1}{2}\) (or the terms in another order, with \(p\) and \(q\) swapped accordingly). Do not accept \(9\sqrt{x}\) or \(\dfrac{4}{\sqrt{x}}\) in the final answer unless the values of \(p\) and \(q\) are also stated, as the form asks for powers of \(x\).

Question 3

5 marks

Using algebra, find the values of the constants \(p\) and \(q\) for which
\[2^p \times 8^q = 8 \qquad \text{and} \qquad \dfrac{25^p}{5^q} = 25\sqrt{5}\]

Model answer

\(2^p \times 8^q = 2^p \times 2^{3q} = 2^{p + 3q}\) and \(8 = 2^3\), so \(p + 3q = 3\)
\(\dfrac{25^p}{5^q} = \dfrac{5^{2p}}{5^q} = 5^{2p - q}\) and \(25\sqrt{5} = 5^2 \times 5^{\frac{1}{2}} = 5^{\frac{5}{2}}\), so \(2p - q = \dfrac{5}{2}\)
From the second equation, \(q = 2p - \dfrac{5}{2}\).
\(p + 3\left(2p - \dfrac{5}{2}\right) = 3\)
\(7p = \dfrac{21}{2}\), so \(p = \dfrac{3}{2}\)
\(q = 3 - \dfrac{5}{2} = \dfrac{1}{2}\)

Explanation

Write each equation using one base, 2 for the first and 5 for the second, and equate the indices to get two linear equations. Solving them simultaneously gives \(p = \tfrac{3}{2}\) and \(q = \tfrac{1}{2}\). Common errors are writing \(8^q\) as \(2^{q + 3}\) instead of \(2^{3q}\), and writing \(25\sqrt{5}\) as \(5^{\frac{3}{2}}\).

Mark scheme

1 mark
M1 Uses a common base in one equation to form a linear equation in \(p\) and \(q\), e.g. \(2^p \times 2^{3q} = 2^3\). An answer with no working scores 0 marks.
An answer of \(p = \dfrac{3}{2}\), \(q = \dfrac{1}{2}\) with no working scores 0 marks: the question says "using algebra". Taking logarithms correctly is an acceptable method.
1 mark
A1 \(p + 3q = 3\)
1 mark
A1 \(2p - q = \dfrac{5}{2}\) (or \(4p - 2q = 5\))
From \(5^{2p - q} = 5^{\frac{5}{2}}\). This mark needs the first M mark (a correct method for one of the equations). Common errors: \(25\sqrt{5} = 5^{\frac{3}{2}}\) or \(5^{2.5}\) written as \(5^{2} \times 5^{0.5}\) and then mishandled.
1 mark
M1 Solves their two linear equations simultaneously to find a value for \(p\) or \(q\)
This may be done on a calculator. Their equations must both be linear in \(p\) and \(q\).
1 mark
A1 \(p = \dfrac{3}{2}\) and \(q = \dfrac{1}{2}\)
Both needed. Accept 1.5 and 0.5. cso.

Question 4

8 marks

A football is modelled as a sphere. A sphere of radius \(r\) cm has volume \(V\) cm³ and surface area \(S\) cm², where
\[V = \dfrac{4}{3}\pi r^3 \qquad \text{and} \qquad S = 4\pi r^2\]
Show that \(S^3 = 36\pi V^2\).

A football has volume 5400 cm³. Use the model to find its surface area, correct to 3 significant figures.

A second ball, also modelled as a sphere, has a surface area that is twice the surface area of the football. Find the exact value of
\[\dfrac{\text{volume of the second ball}}{\text{volume of the football}}\]
giving your answer in the form \(k\sqrt{2}\).

Model answer

\(S^3 = \left(4\pi r^2\right)^3 = 64\pi^3 r^6\)
\(V^2 = \left(\dfrac{4}{3}\pi r^3\right)^2 = \dfrac{16}{9}\pi^2 r^6\)
\(36\pi V^2 = 36\pi \times \dfrac{16}{9}\pi^2 r^6 = 64\pi^3 r^6\)
So \(S^3 = 36\pi V^2\).
For the football, \(S^3 = 36\pi \times 5400^2\), so \(S = \sqrt[3]{36\pi \times 5400^2} = 1488.49\ldots\)
Surface area \(= 1490\) cm² (3 s.f.)
For the second ball, \(\left(2S\right)^3 = 36\pi V_2^2\), so \(8S^3 = 36\pi V_2^2\). With \(S^3 = 36\pi V_1^2\):
\(8 \times 36\pi V_1^2 = 36\pi V_2^2\), so \(\dfrac{V_2^2}{V_1^2} = 8\)
\(\dfrac{V_2}{V_1} = \sqrt{8} = 2\sqrt{2}\)

Explanation

Cube \(S\) and square \(V\) so that both contain \(r^6\); then \(36\pi \times \dfrac{16}{9}\pi^2 = 64\pi^3\) shows the result. The result gives \(S\) from \(V\) directly, and doubling \(S\) multiplies \(S^3\) by 8, so \(V^2\) is multiplied by 8 and \(V\) by \(\sqrt{8} = 2\sqrt{2}\). Common errors are \(\left(4\pi r^2\right)^3 = 4\pi^3 r^6\), rounding the radius too early, and giving the ratio as a decimal.

Mark scheme

1 mark
M1 Eliminates \(r\) correctly in principle, e.g. finds \(S^3 = 64\pi^3 r^6\) and \(V^2 = \dfrac{16}{9}\pi^2 r^6\) (or makes \(r\) the subject of one formula and substitutes into the other). An answer with no working scores 0 marks for the show-that; a surface area of awrt 1490 cm² with no working scores M1 A1 (2 marks), and \(2\sqrt{2}\) with no working scores M1 dM1 A1 (3 marks).
An answer with no working scores 0 marks for the show-that part. For the surface area, awrt 1490 (cm²) stated without working scores M1 A1 (2 marks), and \(2\sqrt{2}\) stated without working scores M1 dM1 A1 (3 marks), so correct answers to the last two parts alone, without any working, score 5 of the 8 marks. Substituting a numerical value of \(r\) to check the result is not a proof: M0.
1 mark
A1 \(36\pi V^2 = 36\pi \times \dfrac{16}{9}\pi^2 r^6 = 64\pi^3 r^6\) (or an equivalent correct intermediate line)
1 mark
A1* \(S^3 = 64\pi^3 r^6 = 36\pi V^2\), with no errors
cso. The final line must be \(S^3 = 36\pi V^2\).
1 mark
M1 Uses \(S^3 = 36\pi \times 5400^2\) to find \(S\), or finds \(r = \sqrt[3]{\dfrac{3 \times 5400}{4\pi}}\) and uses \(S = 4\pi r^2\)
Independent of the show-that marks.
1 mark
A1 \(S =\) awrt 1490 (cm²), i.e. in [1485, 1495)
The value is 1488.49... Premature rounding of \(r\) (\(r = 10.9\) gives 1493; \(r = 10.88\) gives 1487.6) is condoned if the answer is awrt 1490. Units are not required, but wrong units lose this mark.
1 mark
M1 Uses the relationship between \(S\) and \(V\) for both balls, e.g. \(\left(2S\right)^3 = 36\pi V_2^2\) and \(S^3 = 36\pi V_1^2\), so \(\dfrac{V_2^2}{V_1^2} = 2^3\); or \(V \propto S^{\frac{3}{2}}\), so the ratio is \(2^{\frac{3}{2}}\)
Alternative: the radius is multiplied by \(\sqrt{2}\) (since \(S \propto r^2\)), so the volume is multiplied by \(\left(\sqrt{2}\right)^3\). Numerical work from 1490 cm² (e.g. finding both volumes on a calculator) scores M1 only if it leads to an exact form.
1 mark
M1 (dM1) Takes the positive square root, or evaluates \(2^{\frac{3}{2}} = \sqrt{8}\) oe
Depends on the previous M mark.
1 mark
A1 \(2\sqrt{2}\)
cao (\(k = 2\)). Do not accept 2.83 or \(\sqrt{8}\) as the final answer.

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