OCR Mathematics Advanced Subsidiary GCE Core 1 (4721) January 2012

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Clyde Lewis
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1 Link to past paper on OCR website: The above link takes you to the OCR website. From there you click QUALIFICATIONS, QUALIFICATIONS BY TYPE, AS/A LEVEL GCE, MATHEMATICS, VIEW ALL DOCUMENTS, PAST PAPERS JANUARY SERIES 2012, QUESTION PAPER UNIT 4721/01 CORE MATHEMATICS 1 These solutions are for your personal use only. DO NOT photocopy or pass on to third parties. If you are a school or an organisation and would like to purchase these solutions please contact Chatterton Tuition for further details. Question 1 we need to rationalise the denominator (get rid of any from the bottom) we do this by multiplying the top and the bottom by 3+ 3 (note the change of sign) x = ( ) = = =8+3 3 ( )( ) Page 1

2 Question 2 i) this will be a reflection in the line =0 ii) whole graph is picked up and moved 2 places up translation Page 2

3 Question 3 this is completing the square but first we need to factorise out the 5 work on the LHS only Now we can complete the square on the bracket Now multiply by the Compare with the RHS 5 + 8= ( 1) + =5 = 1 Multiply by 10 = 10 8= We know = 10 8= 5 8= 13= = 10, =5, = Page 3

4 Question 4 Evaluate means actually get an answer that is a number as opposed to just simplifying i) anything to a negative power will send it to the bottom (for example 5-2 = ) 3 = = ii) =3 x = x 3 16 = 16 anything to the power of ¼ means the fourth root of it. 16 = 16 =2 =8 iii) we can cancel down the top and the bottom (as they are both ) = = = = 25= Page 4

5 Question 5 We need to solve the equation First recognise that it is a hidden quadratic If we set = then we have =0 See if it will factorise See if we can find two numbers that multiply to make -24(3 x 8) and add to make -10 Two numbers are +2 and 12 Rewrite the equation splitting the middle term into +2 and =0 Now factorise in pairs (3 +2) 4(3 +2)=0 We should have the same in both brackets we do Factorise again (3 +2)( 4)=0 3 +2=0 or 4=0 = or =4 Remember that = This means that = = = We need to square root both sides The first leads to no solutions (as we can t square root a negative) =± Page 5

6 Question 6 i) ( )= first change ( ) so that the terms are all on the top ( )= To differentiate multiply by the power and then reduce the power by 1 ( )= 4 3 ii) ( )= differentiate again ( )=8 Now substitute in = =8 =8(2) =8(8)= Page 6

7 Question 7 i) The gradient at a turning point is zero. This means we can find minimum and maximum points by differentiating and setting the differential equal to 0. first expand the brackets = Group terms = +10 to differentiate you multiply by the power of x and then reduce the power by 1 =3 2 1 Set this equal to =0 See if it will factorise See if we can find two numbers that multiply to make -3 (3 x 1) and add to make -2 Two numbers are 3 and +1 Rewrite the equation splitting the middle term into 3 and =0 Now factorise in pairs 3 ( 1)+1( 1)=0 We should have the same in both brackets we do Factorise again ( 1)(3 +1)=0 3 +1=0 or 1=0 = or =1 To decide which of these is the minimum differentiate again and substitute the values in. If this gives a positive value then it is a minimum but if it gives a negative value then it is a maximum Page 7

8 =6 2 Substitute = =6 2= 2 2= 4 Negative so the point is a maximum (and not what we want here) Substitute =1 =6(1) 2=6 2=4 positive so the point is a minimum Substitute =1 back into original equation to get the coordinate = (1)+2 ((1) 3(1)+5)=3(1 3+5)=3(3)=9 minimum point (1,9) ii) the discriminant is 4 for any quadratic + + =0 In this case =1, = 3, =5 Discriminant is ( 3) 4(1)(5)=9 20= 11 iii) because the discriminant of the quadratic factor is negative this means that the quadratic factor has no roots. The only time that the curve will meet the axis is when ( +2)=0 ie when = 2 from part i) we found that the cubic curve had a maximum when = and a minimum when =1, we know the curve crosses the axis only once when = 2 so for all other values (when > 2) the curve will be above the axis and so positive Page 8

9 Question 8 We can first work out the equation of line l We have the gradient and a point it goes through so we can use the formula = ( ) 5= 2( 3) Expand the brackets 5= 2 +6 Add 5 to both sides = Let point B be (, ) length between two points is given by ( ) +( ) length AB will be ( 3) +( 5) =6 5 square both sides ( 3) +( 5) =36 x 5=180 equation 1 We know that point B is on the line = equation 2 Substitute equation 2 into equation 1 ( 3) +(( 2 +11) 5) =180 ( 3) +( 2 +6) =180 Expand the brackets =180 Group terms =180 Subtract 180 from both sides =0 Divide both sides by = Page 9

10 Try to factorise Find two numbers that multiply to give -27 but add to give -6 The two numbers are -9 and +3 ( 9)( +3)=0 =9 = 3 Substitute back into equation 2 to get y When =9 = 2(9)+11= 18+11= 7 When = 3 = 2( 3)+11=6+11=17 B could be (9, 7) ( 3,17) Page 10

11 Question 9 i) The graph is a negative quadratic so will be the usual shape Meets axis Substitute =0 into the equation of the curve =0+0+12=12 (0,12) Meets axis Substitute =0 into the equation of the curve +12=0 Multiply all terms by =0 See if it will factorise Find two numbers that multiply to give -12 and add to give +1 The two numbers are +4 and -3 ( +4)( 3)=0 =3 = Page 11

12 ii) rearrange so that the leading term is positive add to both sides 12 > Add x to both sides 12 > + Subtract 12 from both sides 0 > <0 Factorise ( +4)( 3)<0 Find the critical values (where it equals 0) = 4 = +3 if we put these two x values in order on a number line then we need to work out whether we want the bit in between them or the bits either side We can see that we want the bits in between the two x values (this is because we want the curve to be below the line) We could also see this by trying any x value. For example when =0 then we have <0 which works. 4< < Page 12

13 iii) this is simultaneous equations We already have as the subject in the first equation so substitute this into the second equation 3 +(12 )=4 Expand the brackets =4 Group terms =4 add to both sides 2 +12= +4 Subtract 2 from both sides 12= 2 +4 Subtract 12 from both sides 2 8=0 Factorise Find two numbers that multiply to make -8 but add to make -2 The two numbers are -4 and +2 ( 4)( +2)=0 =4 = 2 Put these both back into either equation to get the corresponding y values I shall use the first equation When =4 =12 (4) (4) = = 8 When = 2 =12 ( 2) ( 2) =12+2 4=10 =4 = 8 = 2 =10 Check these by substituting into the second equation 3(4)+( 8)=12 8=4 it works 3( 2)+(12)= 6+10=4 it works Page 13

14 Question 10 i) Equation of a circle is given by ( ) +( ) = ( 2) +( 4) =5 ( +2) +( 4) =25 Expand the brackets =25 Group terms = Page 14

15 ii) the tangent will be perpendicular to the radius at this point if we first find the gradient of the radius then we can take the negative reciprocal to find the gradient of the tangent gradient radius P ( 5,8), ( 2,4) Gradient is Gradient radius = Gradient of tangent We now have the gradient and a point that it goes through (P) So we can use the formula for a straight line = ( ) 8= ( 5) Multiply both sides by =3 +15 Substract 4 from both sides 32= Add 32 to both sides 0= =0 as required iii) substitute =3 into the equation and hopefully we get =14 3(3) 4 +47= =0 Add 4y to both sides 56=4 Divide both sides by 4 =14 as required Page 15

16 iv) triangle CPT will be a right angled triangle with the right angle at P this is because a tangent will always make a right angle with the radius at the point where it touches the circle area of a triangle is h h let base be length CP and height be length PT length between two points is given by ( ) +( ) length CP ( 2 5) +(4 8) = (3) +( 4) = 9+16= 25=5 Length PT ( 5 3) +(8 14) = ( 8) +( 6) = 64+36= 100=10 Area triangle CPT x 5 x 10= Page 16

17 If you found these solutions helpful and would like to see some more then visit our website Chatterton Tuition is a tuition agency providing tuition in all subjects in the area local to North Yorkshire. However we also offer online tuition via Skype. For more information call, or visit our website. It should be noted that Chatterton Tuition is responsible for these solutions. The solutions have not been produced nor approved by OCR. In addition these solutions may not necessarily constitute the only possible solutions Page 17

Mathematics (MEI) Advanced Subsidiary GCE Core 1 (4751) June 2010

Mathematics (MEI) Advanced Subsidiary GCE Core 1 (4751) June 2010 Link to past paper on OCR website: www.ocr.org.uk The above link takes you to OCR s website. From there you click QUALIFICATIONS, QUALIFICATIONS BY TYPE, AS/A LEVEL GCE, MATHEMATICS (MEI), VIEW ALL DOCUMENTS,