Indices, surds and quadratic functions – Year 12 A level maths test answers to download as PDF

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1. Answer to Question 1 (5 marks)

   1. Use \((ab)^n=a^nb^n\), so \((3x^2)^3=27x^6\), then divide the numbers and subtract the powers. \(\dfrac{27x^6}{9x^{-1}}=3x^{6-(-1)}=\) \(3x^7\) (1 for \(27x^6\), 1 for dividing, 1 for the answer) So the expression equals \(3x^7\).
   2. Rationalise the denominator using the conjugate \(\sqrt5+1\); the denominator becomes \((\sqrt5)^2-1^2=4\). Multiply by \(\dfrac{\sqrt5+1}{\sqrt5+1}\): \(\dfrac{10(\sqrt5+1)}{5-1}\) (1) \(=\dfrac{5(\sqrt5+1)}2\) (1) Simplifying \(\tfrac{10}4=\tfrac52\) gives the required form.

   In (a), a common error is \((3x^2)^3=3x^6\), forgetting to cube the 3.

2. Answer to Question 2 (4 marks)

   1. Halve the coefficient of \(x\) to get \((x+5)^2\), then subtract \(5^2=25\). \((x+5)^2-25+7=(x+5)^2-18\) (1 for \(a=5\), 1 for \(b=-18\)) So \(f(x)=(x+5)^2-18\).
   2. The minimum occurs when the square is zero; then solve \((x+5)^2=18\) using \(\sqrt{18}=3\sqrt2\). Minimum \((-5,-18)\) (1). \((x+5)^2=18\), \(x=-5\pm3\sqrt2\) (1) The roots are \(x=-5+3\sqrt2\) and \(x=-5-3\sqrt2\).

   Check (b): \(x\approx-0.757\) or \(x\approx-9.243\), and both make \(x^2+10x+7\) zero. The exact surd form is required for the mark, so decimals alone score nothing.

3. Answer to Question 3 (4 marks)

   1. Equal roots means \(b^2-4ac=0\) with \(a=1\), \(b=k-1\), \(c=4\). \((k-1)^2-16=0\) (1), \(k-1=\pm4\) (1), so \(k=5\) or \(k=-3\) (1) Both values give equal roots.
   2. With \(k=5\) the equation is a perfect square. \(k=5\): \(x^2+4x+4=0\), \(x=-2\) (1) (With \(k=-3\) the equation is \((x-2)^2=0\), so \(x=2\).)

   Check \(k=-3\): \(x^2-4x+4\) has discriminant \(16-16=0\) ✓. Writing only \(k-1=4\) loses the negative square root and one value of \(k\); remember \(\pm\) when square-rooting.

4. Answer to Question 4 (3 marks)

   \(x^4=(x^2)^2\), so substitute \(y=x^2\) to obtain a quadratic in \(y\).

   Let \(y=x^2\): \(y^2-10y+9=0\) (1), \((y-1)(y-9)=0\) so \(y=1\) or \(9\) (1), so \(x=\pm1\) or \(\pm3\) (1).

   All four solutions are valid because 1 and 9 are positive, so each gives two square roots: \(x=-3,-1,1,3\).

   Check \(x=3\): \(81-90+9=0\) ✓. Stopping at \(y=1\) or 9 without returning to \(x\) loses the final mark, and so does giving only the positive roots.

5. Answer to Question 5 (4 marks)

   1. Half the perimeter is 50 cm, so the length is \(50-w\). Length \(=50-w\) so \(A=w(50-w)=50w-w^2\) (1) This is the required expression.
   2. Take out the factor \(-1\) and complete the square. \(A=-(w-25)^2+625\) (1), maximum 625 cm\(^2\) when \(w=25\) (1) So the poster should be a 25 cm by 25 cm square.
   3. Use the completed-square form and solve the inequality. \(-(w-25)^2+625\ge600\) gives \((w-25)^2\le25\), so \(20\le w\le30\) (1) So the width must be between 20 cm and 30 cm.

Total : 20 marks