Functions, modulus and partial fractions – Year 13 A level maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   Mark scheme: a) 1 mark for each composite function; b) 1 mark for the quadratic, 1 mark for both solutions.

   1. For \(fg\), substitute \(g(x)\) into \(f\); for \(gf\), substitute \(f(x)\) into \(g\) and expand the bracket. \(fg(x)=2(x^2-3)+1=2x^2-5\); \(gf(x)=(2x+1)^2-3=4x^2+4x-2\). The two composite functions are different.
   2. Use \(gf(x)\) from part a), rearrange to zero and factorise. \(4x^2+4x-2=6\) gives \(4x^2+4x-8=0\), \(x^2+x-2=0\), \((x+2)(x-1)=0\), so \(x=-2\) or \(x=1\). Check: \(gf(1)=3^2-3=6\) and \(gf(-2)=(-3)^2-3=6\).

2. Answer to Question 2 (4 marks)

   Method: for \(|u|=v\), solve \(u=v\) and \(u=-v\), then reject any solution that makes \(v\) negative. Mark scheme: 1 mark per case, 1 mark for checking; 1 mark for part b).

   1. Case 1: \(3x-2=x+4\) gives \(x=3\); RHS \(=7\gt 0\), valid. Case 2: \(-(3x-2)=x+4\) gives \(-3x+2=x+4\), so \(x=-\tfrac12\); RHS \(=3.5\gt 0\), valid. Answer: \(x=3\) or \(x=-\tfrac12\).
   2. \(-5\le2x+1\le5\) gives \(-3\le x\le2\).

   Check (a): \(|3\times3-2|=7=3+4\) ✓ and \(|-\tfrac32-2|=3.5=-\tfrac12+4\) ✓.

3. Answer to Question 3 (4 marks)

   Method: write the fraction as a sum of simpler fractions, multiply by the denominator and substitute the roots of each linear factor. An improper fraction must be divided first. Mark scheme: 2 marks for each part.

   1. \(5x+3=A(x+3)+B(x-1)\). \(x=1\): \(8=4A\), \(A=2\). \(x=-3\): \(-12=-4B\), \(B=3\). So \(\dfrac2{x-1}+\dfrac3{x+3}\). Check at \(x=0\): \(-2+1=-1\) and \(\tfrac{3}{-3}=-1\).
   2. The degrees of numerator and denominator are equal, so the fraction is improper: write \(x^2=(x^2-1)+1\). Improper: \(\dfrac{x^2}{x^2-1}=1+\dfrac1{x^2-1}\). Then \(\dfrac1{(x-1)(x+1)}=\dfrac{1/2}{x-1}-\dfrac{1/2}{x+1}\). So the answer is \(1+\dfrac1{2(x-1)}-\dfrac1{2(x+1)}\).

4. Answer to Question 4 (4 marks)

   Mark scheme: 1 mark for each of parts a) and b), 2 marks for part c).

   1. The minimum value of \(|x-2|\) is 0, so the range is \(h(x)\ge1\). The minimum value 1 occurs at \(x=2\).
   2. Write \(|x-2|=3\), then solve \(x-2=3\) and \(x-2=-3\). \(|x-2|=3\) gives \(x=5\) or \(x=-1\). Both values give \(h(x)=3+1=4\).
   3. For \(x\ge2\) the modulus is not needed, so \(k\) is a one-to-one linear function. For \(x\ge2\), \(k(x)=x-2+1=x-1\). Inverse: \(k^{-1}(x)=x+1\), domain \(x\ge1\). The domain of \(k^{-1}\) is the range of \(k\).

5. Answer to Question 5 (4 marks)

   Mark scheme: a) 1 mark for both temperatures; b) 1 mark for isolating the exponential, 1 mark for the time; c) 1 mark.

   1. Substitute \(t=0\), then let \(t\) become very large so that the exponential tends to 0. At \(t=0\): \(T=200\) °C. As \(t\to\infty\): \(T\to25\) °C.
   2. Subtract 25, divide by 175 and take natural logarithms. \(100=25+175e^{-0.05t}\) gives \(e^{-0.05t}=\dfrac{75}{175}=\dfrac37\), so \(t=-20\ln\dfrac37=20\ln\dfrac73=\) 16.9 minutes. So the oven reaches 100 °C after about 17 minutes.
   3. The domain of the inverse is the range of \(T\) for \(t\ge0\). The inverse is defined for \(25\lt x\le200\).

Total : 20 marks