Simultaneous equations, inequalities and polynomials – Year 12 A level maths worksheet answers to download as PDF

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1. Answer to Question 1

   1. Both equations are linear, so use elimination: make the coefficients of \(x\) equal. Multiply the second equation by 2: \(4x+10y=2\). Subtract from the first: \(-13y=13\), so \(y=-1\). Then \(2x-5=1\) gives \(x=3\). Solution: \(x=3\), \(y=-1\). Check: \(12+3=15\) ✓ and \(6-5=1\) ✓.
   2. One equation is linear and one is quadratic, so substitute the linear one into the quadratic. Substitute: \(x^2+(2x-1)^2=10\), so \(5x^2-4x-9=0\), \((5x-9)(x+1)=0\), giving \(x=\tfrac95\) or \(x=-1\). Then \(y=2x-1\): \((\tfrac95,\tfrac{13}5)\) and \((-1,-3)\). Check: \(1+9=10\) ✓.
   3. Make \(y\) the subject of the linear equation and substitute into \(xy=12\). From the second equation \(y=7-x\). Then \(x(7-x)=12\), so \(x^2-7x+12=0\), \((x-3)(x-4)=0\). \((3,4)\) and \((4,3)\).

2. Answer to Question 2

   1. Collect the terms, then divide by the coefficient of \(x\). \(-3x\ge15\), and dividing by \(-3\) reverses the sign: \(x\le-5\).
   2. Expand the bracket and collect the \(x\) terms on one side. \(2x+8<3x-1\), so \(9\lt x\), i.e. \(x>9\).
   3. Factorise to find the critical values, then use a sketch of the parabola. \(x^2-x-12=(x-4)(x+3)>0\). The graph is U-shaped and positive outside the roots: \(x<-3\) or \(x>4\).
   4. Do not divide by \(x\), because \(x\) could be negative or zero; bring every term to one side and factorise instead. \(3x^2-12x\le0\), \(3x(x-4)\le0\). U-shape, negative between the roots: \(0\le x\le4\).

3. Answer to Question 3

   1. By the factor theorem, \((x-a)\) is a factor if \(f(a)=0\); try small divisors of the constant term 6. \(f(1)=1-7+6=0\), so \((x-1)\) is a factor. Divide: \(x^3-7x+6=(x-1)(x^2+x-6)\), and \(x^2+x-6=(x+3)(x-2)\). So \(f(x)=\) \((x-1)(x-2)(x+3)\). Check: \(f(2)=8-14+6=0\) ✓.
   2. Try divisors of 6 again: \(g(1)=-4\ne0\), but \(g(2)=0\). \(g(2)=16+4-26+6=0\), so \((x-2)\) is a factor. Divide: \(g(x)=(x-2)(2x^2+5x-3)\) and \(2x^2+5x-3=(2x-1)(x+3)\). So \(g(x)=\) \((x-2)(2x-1)(x+3)\). Check: \(g(\tfrac12)=\tfrac14+\tfrac14-\tfrac{13}2+6=0\) ✓.

4. Answer to Question 4

   1. Use long division, dividing the leading term each time. \(x^3+2x^2-5x+1=(x-2)(x^2+4x+3)+7\). Quotient \(x^2+4x+3\), remainder 7. (Working: \(x^3\div x=x^2\), subtract \(x^3-2x^2\) to leave \(4x^2-5x\); \(4x^2\div x=4x\), subtract \(4x^2-8x\) to leave \(3x+1\); \(3x\div x=3\), subtract \(3x-6\) to leave 7.)
   2. By the remainder theorem, the remainder on dividing by \((x-a)\) is \(f(a)\). \(f(2)=8+8-10+1=7\). ✓ This agrees with the long division.
   3. Dividing by \((x+1)\) means \(a=-1\), so evaluate \(f(-1)\). \(f(-1)=-1+2+5+1=7\). The remainder is 7 (a coincidence in this case).

5. Answer to Question 5

   1. Use the factor theorem for \((x-2)\) and the remainder theorem for \((x+1)\). \(p(2)=0\): \(8+4a+2b-6=0\), so \(2a+b=-1\). \(p(-1)=-12\): \(-1+a-b-6=-12\), so \(a-b=-5\).
   2. Add the two equations to eliminate \(b\). Add: \(3a=-6\), so \(a=-2\) and \(b=3\). Check: \(2(-2)+3=-1\) ✓ and \(-2-3=-5\) ✓.
   3. Factorise \(p(x)\) by grouping. \(p(x)=x^3-2x^2+3x-6=x^2(x-2)+3(x-2)=(x-2)(x^2+3)\). The quadratic \(x^2+3\) has discriminant \(0-12<0\), so no real roots. The only real root is \(x=2\). So the graph of \(y=p(x)\) crosses the \(x\)-axis only once, at \(x=2\).

6. Answer to Question 6

   1. Solve each inequality separately, then find the values that satisfy both, for example on a number line. \(2x>-6\) gives \(x>-3\). \(x^2\le16\) gives \(-4\le x\le4\). Both: \(-3\lt x\le4\). So \(x\) must be greater than \(-3\) and at most 4.
   2. The quadratic \((x-1)(x-4)\) is negative between its roots. \((x-1)(x-4)<0\) gives \(1\lt x<4\). \(3x>3\) gives \(x>1\). Both: \(1\lt x<4\). Here the linear condition is already satisfied by every \(x\) in \(1\lt x<4\), so the solution is \(1\lt x<4\).

7. Answer to Question 7

   1. Set the costs of A and B equal, with \(m\) the number of months. \(25+18m=23m\), so \(5m=25\) and \(m=\) 5 months (both cost £115).
   2. Substitute \(m=12\) into each cost formula. A: \(25+18\times12=£241\). B: \(23\times12=£276\). C: \(60+14\times12=£228\). For a full year gym C is the cheapest.
   3. Form an inequality and round up to the next whole month. \(60+14m<25+18m\) gives \(35<4m\), \(m>8.75\). So from month 9 gym C is cheaper than A (at 9 months: C £186, A £187).
   4. Substitute \(m=6\) into each formula. 6 months: A \(=25+108=£133\), B \(=£138\), C \(=60+84=£144\). Gym A is cheapest.

8. Answer to Question 8

   1. The perimeter is twice the sum of length and width, and the area is length times width. \(2l+2w=34\), i.e. \(l+w=17\), and \(lw=60\).
   2. Substitute \(l=17-w\) into \(lw=60\). \(l=17-w\), so \(w(17-w)=60\), \(w^2-17w+60=0\), \((w-5)(w-12)=0\). The garden is 12 m by 5 m. Check: perimeter \(2(12+5)=34\) ✓, area \(12\times5=60\) ✓.
   3. Use Pythagoras. Diagonal \(=\sqrt{12^2+5^2}=\sqrt{169}=\) 13 m. The path is 13 m long.
   4. Multiply the perimeter by the cost per metre. Perimeter 34 m, so cost \(=34\times22=\) £748. Fencing costs £748.

9. Answer to Question 9

   1. Profit means \(P>0\): \(-0.5x^2+12x-50>0\). Multiply by \(-2\) (reversing the sign): \(x^2-24x+100<0\). This is the required inequality.
   2. Use the quadratic formula to find the critical values, then remember that a U-shaped graph is negative between its roots. Roots: \(x=\dfrac{24\pm\sqrt{576-400}}2=12\pm\sqrt{44}=12\pm6.633\). So \(5.37\lt x<18.63\).
   3. Complete the square for \(P\). \(P=-0.5(x-12)^2+22\). Greatest profit at \(x=12\) (1 200 loaves), \(P=22\), i.e. £2 200.
   4. Combine the profit interval with the production limits. Allowed: \(6\le x\le20\). Profit when \(5.37\lt x<18.63\), so within the allowed range between 600 loaves and 1 863 loaves (about 1 860).

10. Answer to Question 10

    1. Total cost \(=420+8n\le1000\), so \(8n\le580\) and \(n\le72.5\), so \(n\le72\) for a whole number of students.
    2. Since \(n>0\), multiplying by \(n\) keeps the direction of the inequality. Cost per student \(=\dfrac{420+8n}{n}\le14\). Multiply by \(n\): \(420+8n\le14n\), so \(420\le6n\) and \(n\ge70\).
    3. Combine the two conditions \(70\le n\le72\). Also \(n\ge40\) is automatically satisfied. So \(n=\) 70, 71 or 72.
    4. Divide the total cost by the number of students. \(n=70\): \(\dfrac{980}{70}=£14.00\). \(n=71\): \(\dfrac{988}{71}=£13.92\). \(n=72\): \(\dfrac{996}{72}=£13.83\). All three values are within the £14 limit.

11. Answer to Question 11

    1. The water hits the hillside where the two heights are equal. \(4x-x^2=0.5x\) gives \(3.5x-x^2=0\), \(x(3.5-x)=0\). The jet meets the hill at \(x=0\) (the nozzle) and \(x=3.5\), \(y=1.75\). Point \((3.5,\,1.75)\).
    2. Complete the square for the jet. \(y=-(x-2)^2+4\), so the greatest height is 4 m at \(x=2\).
    3. Complete the square for the gap. \(g=-(x-1.75)^2+3.0625\). Greatest gap 3.06 m at \(x=1.75\).
    4. Set the gap equal to 3 and solve the quadratic. \(3.5x-x^2=3\) gives \(x^2-3.5x+3=0\), \((x-1.5)(x-2)=0\), so \(x=\) 1.5 m or 2 m. So the jet is 3 m above the hillside at two places.

12. Answer to Question 12

    1. After cutting the squares, the base is \((20-2x)\) by \((14-2x)\) and the height is \(x\). Length \(=20-2x\), width \(=14-2x\), height \(=x\). \(V=x(20-2x)(14-2x)=x(280-68x+4x^2)=4x^3-68x^2+280x\).
    2. Substitute \(x=2\) into the factorised form. \(V(2)=2\times16\times10=\) 320 cm\(^3\).
    3. Set \(V=288\) and divide by 4; use the factor theorem with \(x=4\). \(4x^3-68x^2+280x-288=0\Rightarrow x^3-17x^2+70x-72=0\). At \(x=4\): \(64-272+280-72=0\) ✓. Divide by \((x-4)\): \(x^3-17x^2+70x-72=(x-4)(x^2-13x+18)\). The quadratic gives \(x=\dfrac{13\pm\sqrt{97}}2=11.42\) or \(1.58\). \(11.42>7\) is impossible (the card is only 14 cm wide), so the other solution is \(x=1.58\).
    4. The width needs \(14-2x>0\), so \(0\lt x<7\).

13. Answer to Question 13

    1. Let an adult ticket cost £\(A\) and a child ticket £\(C\). \(2A+3C=47.5\) and \(3A+2C=52.5\).
    2. Use elimination by making the coefficients of \(A\) equal. Multiply the first by 3 and the second by 2: \(6A+9C=142.5\) and \(6A+4C=105\). Subtract: \(5C=37.5\), \(C=7.5\). Then \(2A=47.5-22.5=25\), \(A=12.5\). Adult £12.50, child £7.50. Check: \(3\times12.5+2\times7.5=52.5\) ✓.
    3. Substitute the prices. \(2\times12.50+4\times7.50=25+30=\) £55.
    4. Compare the family ticket with the normal price. Yes, the family ticket saves £55 \(-\) £50 \(=\) £5.

14. Answer to Question 14

    1. Substitute the speeds into the formula. \(v=30\): \(0.02\times900+18=36\) ✓. \(v=50\): \(0.02\times2500+30=80\) ✓. Both entries are correct.
    2. Set \(d\le60\) and clear the decimals. \(0.02v^2+0.6v\le60\). Multiply by 50: \(v^2+30v\le3000\), i.e. \(v^2+30v-3000\le0\). This is the required inequality.
    3. Use the quadratic formula; a U-shaped graph is negative between its roots. \(v=\dfrac{-30\pm\sqrt{900+12000}}2=\dfrac{-30\pm113.58}2\), so \(v=41.79\) or \(-71.79\). The inequality holds between the roots, so \(v\le\) 41.8 mph.
    4. A speed cannot be negative. So only the positive root gives the maximum speed of 41.8 mph.

15. Answer to Question 15

    1. \(150a+130c=3270\) and \(200a+120c=3880\).
    2. Use elimination by making the coefficients of \(a\) equal. Multiply the first by 4 and the second by 3: \(600a+520c=13080\) and \(600a+360c=11640\). Subtract: \(160c=1440\), \(c=9\). Then \(150a=3270-1170=2100\), \(a=14\). Adults £14, children £9. Check: \(200\times14+120\times9=2800+1080=3880\) ✓.
    3. Write the takings in terms of the number of adult tickets, then use the restriction on child tickets. Let \(x\) be the number of adult tickets; then children \(=400-x\ge150\), so \(x\le250\). Takings \(=14x+9(400-x)=5x+3600\), greatest when \(x=250\): \(5\times250+3600=\) £4 850 (250 adults and 150 children: \(3500+1350\)).
    4. Each adult ticket brings £14 but costs £2.50, so the fixed cost of £2 900 must be covered. Profit per ticket (all adult) \(=14-2.5=£11.50\). Need \(11.5n>2900\), so \(n>252.17\). At least 253 tickets.

16. Answer to Question 16

    1. Set the two expressions for \(y\) equal. \(x^2-4x+1=x+k\Rightarrow x^2-5x+1-k=0\).
    2. Two distinct points need a positive discriminant. Discriminant \(=25-4(1-k)=21+4k>0\), so \(k>-\tfrac{21}4\).
    3. Substitute \(k=-5\) and factorise. \(x^2-5x+6=0\), \((x-2)(x-3)=0\). Points \((2,-3)\) and \((3,-2)\) (using \(y=x-5\)). Distance \(=\sqrt{1^2+1^2}=\sqrt2\approx\) 1.41.
    4. A tangent meets the curve at exactly one point, so the discriminant is zero. Tangent when \(21+4k=0\), \(k=-\tfrac{21}4\). Then \(x^2-5x+\tfrac{25}4=0\), \((x-\tfrac52)^2=0\), \(x=\tfrac52\), \(y=\tfrac52-\tfrac{21}4=-\tfrac{11}4\). Point \((\tfrac52,-\tfrac{11}4)\).

17. Answer to Question 17

    1. Use the factor theorem, then divide by \((x-3)\) and factorise the quadratic quotient. \(f(3)=54-27-33+6=0\), so \((x-3)\) is a factor. Divide: \(f(x)=(x-3)(2x^2+3x-2)=\) \((x-3)(2x-1)(x+2)\).
    2. Use the roots, the \(y\)-intercept and the sign of the leading coefficient. Roots at \(x=-2,\ \tfrac12,\ 3\); \(y\)-intercept \(f(0)=6\); positive \(x^3\) coefficient so the curve rises on the right and falls on the left (as in the figure).
    3. Read from the sketch where the curve is above the \(x\)-axis. \(f(x)>0\) for \(-2\lt x\lt \tfrac12\) or \(x>3\).
    4. Use the remainder theorem with \(a=-1\). \(f(-1)=-2-3+11+6=\) 12. The remainder is 12.

18. Answer to Question 18

    1. Substitute \(x=4\): the length is 8 m and the width is 3 m. \(x=4\): \(8\times3=24\) and \(2(8+3)=22\) ✓.
    2. Expand the area and subtract 50. \((x+4)(x-1)<50\Rightarrow x^2+3x-4<50\Rightarrow x^2+3x-54<0\). \((x+9)(x-6)<0\), so \(-9\lt x<6\).
    3. The perimeter is twice the sum of length and width. Perimeter \(=2[(x+4)+(x-1)]=2(2x+3)=4x+6\ge20\), so \(x\ge3.5\).
    4. Both: \(3.5\le x<6\) (and \(x>1\) for a positive width). Whole numbers: \(x=4\) (8 m by 3 m) and \(x=5\) (9 m by 4 m).

19. Answer to Question 19

    1. Expand the factorised form. \(x(x-3)^2=x(x^2-6x+9)=x^3-6x^2+9x\). The root at \(x=3\) is repeated, so the track touches the ground level at \((3,0)\) without crossing it.
    2. Set \(x^3-6x^2+9x=4\): \(x^3-6x^2+9x-4=0\). Expand \((x-1)^2(x-4)=(x^2-2x+1)(x-4)=x^3-6x^2+9x-4\) ✓. Beam meets the track at \(x=1\) (touching) and \(x=4\).
    3. A repeated root means the beam is a tangent at \(x=1\): the track reaches its first peak at height 4 and just touches the beam there before descending.
    4. \(x(x-3)^2>0\) for \(x>0\) with \(x\ne3\). So above ground for \(0\lt x<3\) and \(3\lt x\le4\).

20. Answer to Question 20

    1. Use the factor theorem twice to obtain two equations in \(a\) and \(b\). \(p(2)=16+8a-28+2b+6=0\Rightarrow4a+b=3\). \(p(-3)=81-27a-63-3b+6=0\Rightarrow9a+b=8\). Subtract: \(5a=5\), \(a=\) 1, \(b=\) \(-1\).
    2. \(p(x)=x^4+x^3-7x^2-x+6\). Divide by \((x-2)(x+3)=x^2+x-6\): quotient \(x^2-1\). So \(p(x)=(x-2)(x+3)(x-1)(x+1)\).
    3. Roots at \(-3,-1,1,2\) and the quartic is positive for large \(|x|\). Sign pattern: positive for \(x<-3\), negative for \(-3\lt x<-1\), positive for \(-1\lt x<1\), negative for \(1\lt x<2\), positive for \(x>2\). So \(p(x)<0\) when \(-3\lt x<-1\) or \(1\lt x<2\).
    4. Use the remainder theorem with \(a=3\). \(p(3)=81+27-63-3+6=\) 48. The remainder is 48.