An engineer checks a set of index and surd calculations before signing off a report. Simplify \(\dfrac{x^7 \times x^{-2}}{x^3}\) as a single power of \(x\)....

Assessment: Mathematics Specification A 4MA1 | Paper 1 Mock 01 | Written Paper 1 (1F/1H) Subject: Mathematics Specification A - 4MA1

Question 1 Report

An engineer checks a set of index and surd calculations before signing off a report.

  1. Simplify \(\dfrac{x^7 \times x^{-2}}{x^3}\) as a single power of \(x\). (2)
  2. Solve \(2^{3x-1} = 32\). (3)
  3. Using your value of \(x\) from part (b), find \(x^{-3}\) as a fraction. (2)
  4. Simplify \(\sqrt{75} - \sqrt{12}\) in the form \(a\sqrt{b}\), where \(b\) is prime. (2)
  5. Explain why \(\sqrt{75} - \sqrt{12}\) cannot be found by working out \(\sqrt{75 - 12}\). (2)

Answer Details

This question chains index laws, an exponential equation, a negative index, and surd subtraction, finishing by testing why roots cannot be subtracted "inside" a single root.

  1. (a) [2]. \(\dfrac{x^7 \times x^{-2}}{x^3} = x^{7 + (-2) - 3} = x^{2}\). One mark for combining the indices in the numerator (\(7-2=5\)), one for the final subtraction giving \(x^2\).
  2. (b) [3]. \(32 = 2^5\), so \(2^{3x-1} = 2^5\) gives \(3x - 1 = 5\), so \(3x = 6\) and \(x = 2\). One mark for \(32=2^5\), one for the equation \(3x-1=5\), one for \(x=2\).
  3. (c) [2]. Using \(x=2\) from part (b), \(x^{-3} = 2^{-3} = \dfrac{1}{2^3} = \dfrac{1}{8}\). One mark for rewriting as a reciprocal, one for the fraction \(\dfrac{1}{8}\).
  4. (d) [2]. \(\sqrt{75} = \sqrt{25\times3} = 5\sqrt3\) and \(\sqrt{12} = \sqrt{4\times3} = 2\sqrt3\), so \(\sqrt{75} - \sqrt{12} = 5\sqrt3 - 2\sqrt3 = 3\sqrt3\). One mark for correctly simplifying both surds, one for the final difference \(3\sqrt3\).
  5. (e) [2]. \(\sqrt{75-12} = \sqrt{63} = \sqrt{9\times7} = 3\sqrt7 \approx 7.94\), which is not equal to \(3\sqrt3 \approx 5.20\) from part (d). One mark for evaluating \(\sqrt{63}\) and comparing it numerically with \(3\sqrt3\), one for the correct explanation: subtracting the numbers under separate roots is not the same operation as subtracting the roots themselves, so \(\sqrt{a}-\sqrt{b} \ne \sqrt{a-b}\) in general.

Square roots do not distribute over subtraction (or addition); always simplify each surd separately and combine like terms in \(\sqrt{k}\), rather than combining the numbers underneath a single root sign.

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