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

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1. Answer to Question 1

   1. Complete the square: \(f(x)=(x-3)^2+1\). A square is at least 0, so the range is \(f(x)\ge1\).
   2. We need \(2x-4\ge0\), so the domain is \(x\ge2\). At \(x=2\), \(g=0\) and \(g\) increases without limit. Range: \(g(x)\ge0\).
   3. The fraction \(\dfrac3{x-2}\) takes every value except 0 (a numerator of 3 can never produce 0). Range: \(h(x)\in\mathbb{R}\), \(h(x)\ne0\).

2. Answer to Question 2

   Method: \(fg(x)\) means apply \(g\) first, then \(f\). Replace every \(x\) in the outer function by the whole inner expression.

   1. For \(fg\), substitute \(x^2+2\) into \(f\); for \(gf\), substitute \(3x-1\) into \(g\). \(fg(x)=3(x^2+2)-1=3x^2+5\). \(gf(x)=(3x-1)^2+2=9x^2-6x+3\). The two results are different: composition is not commutative.
   2. Set the two expressions equal and collect everything on one side. The quadratic does not factorise, so use the formula. \(3x^2+5=9x^2-6x+3\) gives \(6x^2-6x-2=0\), i.e. \(3x^2-3x-1=0\). By the formula \(x=\dfrac{3\pm\sqrt{9+12}}{6}=\dfrac{3\pm\sqrt{21}}6\). So \(x=\dfrac{3+\sqrt{21}}6\) or \(x=\dfrac{3-\sqrt{21}}6\).
   3. Work from the inside: first find \(f(2)\), then apply \(f\) to the result. \(f(2)=5\) and \(f(5)=14\), so \(ff(2)=14\). So \(ff(2)=14\).

3. Answer to Question 3

   1. Let \(y=\dfrac{2x+1}{x-3}\). Then \(y(x-3)=2x+1\), \(xy-3y=2x+1\), \(x(y-2)=3y+1\), so \(x=\dfrac{3y+1}{y-2}\). Hence \(f^{-1}(x)=\dfrac{3x+1}{x-2}\), \(x\ne2\) (the range of \(f\) excludes 2).
   2. \(y=e^{2x}-1\) gives \(e^{2x}=y+1\), \(2x=\ln(y+1)\). So \(g^{-1}(x)=\tfrac12\ln(x+1)\), domain \(x>-1\) (the range of \(g\)).
   3. \(y=x^2+4\) with \(x\ge0\) gives \(x=\sqrt{y-4}\). So \(h^{-1}(x)=\sqrt{x-4}\), domain \(x\ge4\).

4. Answer to Question 4

   1. \(3x-1=8\) gives \(x=3\). \(3x-1=-8\) gives \(x=-\tfrac73\).
   2. Case 1: \(x+2=3x-2\) gives \(x=2\). Check: \(3(2)-2=4>0\) and \(|2+2|=4\). Case 2: \(-(x+2)=3x-2\) gives \(-x-2=3x-2\), so \(x=0\). Check: \(3(0)-2=-2<0\), impossible since a modulus cannot be negative, so reject. Only \(x=2\).
   3. Square both sides: \((2x-1)^2=(x+4)^2\), so \(4x^2-4x+1=x^2+8x+16\), \(3x^2-12x-15=0\), \(x^2-4x-5=0\), \((x-5)(x+1)=0\). So \(x=5\) or \(x=-1\). Check: \(|9|=|9|\) and \(|-3|=|3|\).

5. Answer to Question 5

   1. \(-7<2x-5<7\) gives \(-2<2x<12\), so \(-1\lt x<6\).
   2. \(x+3\ge4\) or \(x+3\le-4\), so \(x\ge1\) or \(x\le-7\).
   3. Both sides must be considered: the right-hand side \(2x\) must be positive, so \(x>0\). Squaring: \((x-1)^2<4x^2\), so \(3x^2+2x-1>0\), \((3x-1)(x+1)>0\), giving \(x<-1\) or \(x>\tfrac13\). Combined with \(x>0\): \(x>\tfrac13\). Check \(x=1\): \(|0|<2\) true; \(x=0.2\): \(0.8<0.4\) false.

6. Answer to Question 6

   1. Write \(\dfrac A{x+1}+\dfrac B{x-2}\), so \(7x+1=A(x-2)+B(x+1)\). \(x=-1\): \(-6=-3A\), \(A=2\). \(x=2\): \(15=3B\), \(B=5\). Answer: \(\dfrac2{x+1}+\dfrac5{x-2}\).
   2. Write \(\dfrac A{x-1}+\dfrac B{(x-1)^2}\), so \(4x-1=A(x-1)+B\). \(x=1\): \(B=3\). Coefficient of \(x\): \(A=4\). Answer: \(\dfrac4{x-1}+\dfrac3{(x-1)^2}\).
   3. Numerator and denominator both have degree 2, so divide first: \(x^2+1=1\cdot(x^2+x)+(1-x)\). So the fraction is \(1+\dfrac{1-x}{x(x+1)}\). Then \(\dfrac{1-x}{x(x+1)}=\dfrac Ax+\dfrac B{x+1}\) with \(1-x=A(x+1)+Bx\): \(x=0\): \(A=1\); \(x=-1\): \(2=-B\), \(B=-2\). Answer: \(1+\dfrac1x-\dfrac2{x+1}\).

7. Answer to Question 7

   Method: to find an inverse, write \(y=f(x)\), make \(x\) the subject, then swap the letters. The domain of \(f^{-1}\) is the range of \(f\).

   1. Look at what happens to the exponential term as \(t\) becomes very large. As \(t\to\infty\), \(e^{-0.08t}\to0\) so \(T\to22\) °C (room temperature). At \(t=0\), \(T=90\). Range: \(22\lt T\le90\). So the coffee never quite reaches room temperature.
   2. Subtract 22, divide by 68 and take natural logarithms. \(y=22+68e^{-0.08t}\) gives \(e^{-0.08t}=\dfrac{y-22}{68}\), so \(t=-\dfrac1{0.08}\ln\dfrac{y-22}{68}=-12.5\ln\dfrac{y-22}{68}\). So \(T^{-1}(x)=-12.5\ln\left(\dfrac{x-22}{68}\right)\) with domain \(22\lt x\le90\). The domain of \(T^{-1}\) is the range of \(T\) found in part a).
   3. Substitute 60 into the inverse function. \(t=-12.5\ln\dfrac{38}{68}=12.5\times0.5817=\) 7.27 minutes. So the coffee reaches 60 °C after about 7.27 minutes.
   4. Substitute 50 into the inverse function. \(t=-12.5\ln\dfrac{28}{68}=12.5\times0.8873=\) 11.1 minutes. So the customer should wait about 11.1 minutes.

8. Answer to Question 8

   1. \(gf(x)=1.17\times0.98x=1.1466x\). It converts £\(x\) to euros after the fee.
   2. \(1.1466\times250=\) €286.65.
   3. Let \(y=1.1466x\), so \((gf)^{-1}(y)=\dfrac{y}{1.1466}\). For €500: \(\dfrac{500}{1.1466}=\) £436.07.
   4. With a percentage fee the order does not change the final answer, but a flat fee of, say, £3 must be taken from the pounds before conversion, since £3 taken from euros would be worth less. The order of composition is a modelling decision.

9. Answer to Question 9

   1. \(-8\le m-500\le8\), so \(492\le m\le508\).
   2. A (491.5) is below 492 and C (508.5) is above 508, so both are rejected. B (496.0), D (505.0), E (499.2) and F (507.8) all lie inside the range. Accepted: B, D, E, F.
   3. Reading \(r=m+2.5\) satisfies \(|r-500|\le8\), so \(492\le m+2.5\le508\) and \(489.5\le m\le505.5\).
   4. A V-shape with vertex \((500,0)\). The accepted region is the part of the graph below the line \(y=8\), between \(m=492\) and \(m=508\).

10. Answer to Question 10

    Method: a distance between two positions is always the modulus of their difference. An equation \(|u|=k\) gives \(u=k\) or \(u=-k\); an inequality \(|u|\lt k\) gives \(-k\lt u\lt k\).

    1. Subtract the two positions and take the modulus, since a distance cannot be negative. \(d=|(30-3t)-2t|=|30-5t|\). This is the required result.
    2. Solve \(|30-5t|=10\) by considering both signs. \(30-5t=10\) gives \(t=4\); \(30-5t=-10\) gives \(t=8\). So at 4 s and 8 s. At 4 s they have not yet met; at 8 s they have already passed each other.
    3. Solve the double inequality. \(|30-5t|<5\) gives \(-5<30-5t<5\), so \(-35<-5t<-25\), i.e. \(5\lt t<7\). They are within 5 m for 2 s. So they are within 5 m of each other between 5 s and 7 s.
    4. They pass when the distance between them is zero. When \(d=0\): \(t=6\) s. So they pass each other after 6 seconds, 12 m from the west end.

11. Answer to Question 11

    Method: to find an inverse, write \(y=f(x)\), make \(x\) the subject, then swap the letters. The domain of \(f^{-1}\) is the range of \(f\).

    1. Rearrange the lens equation to make \(\dfrac1v\) the subject, then take the reciprocal. \(\dfrac1v=\dfrac1{10}-\dfrac1u=\dfrac{u-10}{10u}\), so \(v=\dfrac{10u}{u-10}\). This is the required formula.
    2. Substitute each value of \(u\). \(g(15)=\dfrac{150}5=30\); \(g(20)=\dfrac{200}{10}=20\); \(g(30)=\dfrac{300}{20}=15\). The image distances are 30 cm, 20 cm and 15 cm.
    3. Substitute \(g(u)\) into \(g\) and multiply the top and bottom by \(u-10\). \(gg(u)=\dfrac{10\cdot\frac{10u}{u-10}}{\frac{10u}{u-10}-10}=\dfrac{100u}{10u-10(u-10)}=\dfrac{100u}{100}=u\). So \(g\) is self-inverse: \(g^{-1}=g\). Swapping object and image distances gives the same lens equation. Part b) illustrates this: \(g(15)=30\) and \(g(30)=15\).
    4. Write \(g(u)\) as a constant plus a fraction. \(g(u)=10+\dfrac{100}{u-10}\) with \(u>10\), so \(g(u)>10\). Range: \(g(u)>10\). So the image is always more than 10 cm from the lens.

12. Answer to Question 12

    Method: to find an inverse, write \(y=f(x)\), make \(x\) the subject, then swap the letters. The domain of \(f^{-1}\) is the range of \(f\).

    1. Apply \(f\) to the expression \(1.03x+200\). \(ff(x)=1.03(1.03x+200)+200=1.0609x+406\). It gives the balance after two years. So \(ff(x)=1.0609x+406\).
    2. Subtract 200 and divide by 1.03. \(y=1.03x+200\) gives \(x=\dfrac{y-200}{1.03}\). So \(f^{-1}(y)=\dfrac{y-200}{1.03}\). The inverse function undoes one year of saving.
    3. Apply the inverse function twice, working backwards one year at a time. Start of year 2: \(f^{-1}(5000)=\dfrac{4800}{1.03}=\) £4660.19. Start of year 1: \(f^{-1}(4660.19)=\dfrac{4460.19}{1.03}=\) £4330.29. Check: \(\dfrac{5000-406}{1.0609}=4330.29\). So Maya started with £4330.29.

13. Answer to Question 13

    Method: to find an inverse, write \(y=f(x)\), make \(x\) the subject, then swap the letters. The domain of \(f^{-1}\) is the range of \(f\).

    1. Divide each term of the numerator by \(n\). Splitting the fraction: \(\dfrac{500}n+\dfrac{20n}n=\dfrac{500}n+20\). \(C(100)=20+5=\) £25. So 100 T-shirts cost £25 each on average.
    2. Set \(C(n)=22\) and solve for \(n\). \(20+\dfrac{500}n=22\) gives \(\dfrac{500}n=2\), \(n=\) 250. So the club needs 250 T-shirts.
    3. Make \(n\) the subject of \(y=20+\dfrac{500}n\). \(y=20+\dfrac{500}n\) gives \(n=\dfrac{500}{y-20}\). So \(C^{-1}(x)=\dfrac{500}{x-20}\), domain \(20\lt x\le520\) (since \(C(1)=520\)). The domain of the inverse is the range of \(C\).
    4. Look at the sign of the fraction. \(\dfrac{500}n>0\) for every \(n\), so \(C(n)>20\); the average approaches the print cost £20 as \(n\) grows. So the set-up charge is spread over more shirts but never disappears.

14. Answer to Question 14

    1. In one hour the fast pump fills \(\dfrac1a\) of the pool and the slow pump \(\dfrac1{a+6}\), and together \(\dfrac14\). Multiply by \(4a(a+6)\): \(4(a+6)+4a=a(a+6)\), so \(8a+24=a^2+6a\), \(a^2-2a-24=0\).
    2. \((a-6)(a+4)=0\), so \(a=6\) (a time cannot be \(-4\)). The faster pump takes 6 h and the slower 12 h.
    3. Put over a common denominator: \(\dfrac{(a+6)-a}{a(a+6)}=\dfrac6{a(a+6)}\).
    4. \(\dfrac16+\dfrac1{12}=\dfrac{2+1}{12}=\dfrac14\). They fill \(\dfrac14\) of the pool per hour, so 4 hours in total.

15. Answer to Question 15

    1. The graph is a V-shape with vertex \((3,-2)\). y-intercept: \(f(0)=3-2=1\), so \((0,1)\). x-intercepts: \(|x-3|=2\), so \(x=1\) and \(x=5\): \((1,0)\) and \((5,0)\).
    2. The minimum value is \(-2\): range \(f(x)\ge-2\).
    3. \(|x-3|=\dfrac x2+2\). Case 1: \(x-3=\dfrac x2+2\) gives \(\dfrac x2=5\), \(x=10\); check RHS \(=7>0\). Case 2: \(3-x=\dfrac x2+2\) gives \(1=\dfrac{3x}2\), \(x=\dfrac23\); check RHS \(=\dfrac73>0\). So \(x=10\) or \(x=\dfrac23\).
    4. For \(x\ge3\), \(g(x)=x-3-2=x-5\). Let \(y=x-5\), so \(x=y+5\): \(g^{-1}(x)=x+5\), domain \(x\ge-2\) (the range of \(g\)).
    5. \(|x-3|-2<1\) gives \(|x-3|<3\), so \(0\lt x<6\).

16. Answer to Question 16

    1. \(fg(x)=2(x^2+1)-3=2x^2-1\) with \(x\ge0\). The minimum is \(-1\) at \(x=0\), so the range is \(fg(x)\ge-1\).
    2. The domain of \(g\) is \(x\ge0\), so the input \(f(x)=2x-3\) must satisfy \(2x-3\ge0\), that is \(x\ge\tfrac32\). Then \(gf(x)=(2x-3)^2+1=4x^2-12x+10\).
    3. \(2x^2-1=4x^2-12x+10\) gives \(2x^2-12x+11=0\), so \(x=\dfrac{12\pm\sqrt{144-88}}4=\dfrac{12\pm\sqrt{56}}4=3\pm\dfrac{\sqrt{14}}2\), i.e. \(x=4.871\) or \(1.129\). Only \(x\ge\tfrac32\) is allowed, so \(x=4.871\).
    4. \(y=x^2+1\) with \(x\ge0\) gives \(x=\sqrt{y-1}\). \(g^{-1}(x)=\sqrt{x-1}\) with domain \(x\ge1\) and range \(g^{-1}(x)\ge0\).

17. Answer to Question 17

    1. \(3x^2+9x-3=A(x+2)^2+B(x-1)(x+2)+C(x-1)\). \(x=1\): \(9=9A\), \(A=1\). \(x=-2\): \(12-18-3=-9=-3C\), \(C=3\). Coefficient of \(x^2\): \(3=A+B\), so \(B=2\). Answer: \(\dfrac1{x-1}+\dfrac2{x+2}+\dfrac3{(x+2)^2}\).
    2. At \(x=2\): original \(=\dfrac{12+18-3}{(1)(16)}=\dfrac{27}{16}\). Partial fractions: \(1+\dfrac24+\dfrac3{16}=\dfrac{16+8+3}{16}=\dfrac{27}{16}\). They agree.
    3. Each term tends to 0, and the dominant term is \(\dfrac{1}{x}+\dfrac2x=\dfrac3x\), so \(F(x)\to0\) from above (the graph approaches the x-axis).
    4. For \(x>1\): \(x-1>0\), \(x+2>0\) and \((x+2)^2>0\), so every term is positive and their sum is positive.

18. Answer to Question 18

    1. \(119.6\le L\le120.4\). Ring 1 (0.3 away) accepted, ring 2 (0.5) rejected, ring 3 (0.2) accepted, ring 4 (0.5) rejected, ring 5 (0.4) accepted, ring 6 (0.1) accepted.
    2. Cost \(=5\times\) distance: ring 1 £1.50, ring 2 £2.50, ring 3 £1.00, ring 4 £2.50, ring 5 £2.00, ring 6 £0.50.
    3. \(|L-120|<0.3\) so \(119.7\lt L<120.3\).
    4. The reading is \(r=L+0.15\), so \(|L+0.15-120|\le0.4\) gives \(-0.4\le L-119.85\le0.4\), so \(119.45\le L\le120.25\).

19. Answer to Question 19

    1. \(e^{-2}=0.13534\), so \(3e^{-2}=0.40601\) and \(P(5)=\dfrac{800}{1.40601}=569\), which matches the table. As \(t\to\infty\), \(e^{-0.4t}\to0\) and \(P\to800\).
    2. \(P(0)=200\) and \(P\) increases towards 800, so the range is \(200\le P<800\).
    3. \(N=\dfrac{800}{1+3e^{-0.4t}}\) gives \(1+3e^{-0.4t}=\dfrac{800}N\), so \(e^{-0.4t}=\dfrac{800/N-1}3\), and \(t=-2.5\ln\left(\dfrac{800-N}{3N}\right)\) with domain \(200\le N<800\).
    4. \(N=600\): \(t=-2.5\ln\dfrac{200}{1800}=-2.5\ln\frac19=2.5\ln9=\) 5.49 years.
    5. The model predicts a carrying capacity of 800, which is realistic for limited food and space, and it starts at 200 as observed. It cannot allow for sudden events such as disease, and a real population fluctuates.

20. Answer to Question 20

    1. \(ff(x)=\dfrac{3\left(\frac{3x+k}{x-3}\right)+k}{\frac{3x+k}{x-3}-3}=\dfrac{3(3x+k)+k(x-3)}{(3x+k)-3(x-3)}=\dfrac{9x+3k+kx-3k}{k+9}=\dfrac{x(9+k)}{k+9}=x\) (provided \(k\ne-9\)). So \(f\) is self-inverse: it undoes itself.
    2. \(f^{-1}(x)=f(x)=\dfrac{3x+k}{x-3}\).
    3. \(\dfrac{3x+k}{x-3}=x\) gives \(3x+k=x^2-3x\), so \(x^2-6x-k=0\).
    4. Discriminant \(36+4k>0\) gives \(k>-9\).
    5. For \(k=7\) the equation \(f(x)=f^{-1}(x)\) is the same as \(f(x)=x\) because \(f=f^{-1}\), so \(x^2-6x-7=0\), \((x-7)(x+1)=0\), so \(x=7\) or \(x=-1\).