OCR Mathematics Advanced Subsidiary GCE Core 3 (4723) June 2012

Transcription

1 Link to past paper on OCR website: The above link takes you to OCR s website. From there you click QUALIFICATIONS, QUALIFICATIONS BY TYPE, AS/A LEVEL GCE, MATHEMATICS, VIEW ALL DOCUMENTS, PAST PAPERS JUNE SERIES 2012, QUESTION PAPER UNIT 4723/01 CORE MATHEMATICS 3 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 Solve these two separate inequalities and then put them back together again +1<2 5 inequality 1 +1<2 5 inequality 2 Inequality 1 Subtract from both sides 1< 5 Add 5 to both sides 6< Inequality 2 Add to both sides 1<3 5 Add 5 to both sides 4<3 Divide both sides by 3 < unlikely to be the correct way around as contradicts the first inequality Page 1

2 We have ended up with two inequalities we need to check that the inequality sign is correct. Likely to be incorrect as contradicting each other. Using this method often muddles up the signs. Check both by putting back into the original and seeing if it works Try 6< We could use = > > 11 which works Try < I have switched the sign We could use = > > 1 5>1 which works >6 And < Page 2

3 Question 2 i) split the logarithm ln =ln ln=ln+ln ln Bring the power to the front We have ln+2ln ln * We are given that = and = Take natural logarithms of both sides ln=ln =280ln=280 And ln=ln =300ln=300 Substitute these back into * ln =ln+2ln ln= = =261 As required ii) take logs of both sides ln5 >ln ln5>ln+ln ln5> ln5>580 Divide both sides by 5 > > The smallest integer is Page 3

4 Question 3 i) secɵ= Ɵ and cotɵ= Ɵ so we have Ɵ xsinɵ= Ɵ tanɵ= Ɵ Multiply both sides by tanɵ tan Ɵ=36 Square root both sides tanɵ= ±6 Now we know that Ɵ is acute so tanɵ must be positive tanɵ=6 as required ii) a) from formulae sheet tanɵ 45= Ɵ Ɵ Substitute in tanɵ=6 and tan45=1 tanɵ 45= = = b) tan 2Ɵ=tanƟ+Ɵ= ƟƟ = = ƟƟ = Page 4

5 Question 4 a) solve by substitution let =6+1 =6 Multiply both sides by =6 Divide both sides by 6 = Limits =0 =60+1=1 =4 =64+1=25 Substitute into the integral = 3 = =6 25 1=625 ) 61 =30 6=24 as required Page 5

6 b) solve by substitution let = = Multiply both sides by = Divide both sides by = = Limits =0 = =1 =1 = = Substitute into the integral +2 = = = +4+ = +4+4lne 1= +4+4ln +41+4ln1 = = Page 6

7 Question 5 i) the blue curve is =14 it is the usual shape for a negative quadratic the red curve is =ln We can see that the two curves will only ever meet once and so there is exactly one root to the equation 14 =ln ii) first we need to set the equation to ln=0 We need to find two consecutive integers where one integer gives a negative value and the other gives a positive value. Then if we have our sign change we will know that the root lies between these two integers Try = ln2=7.92 Try = ln3=1.70 Try = ln4= 6.16 Sign change from =3 and =4 3< < Page 7

8 b) start by putting 3 into your calculator. Press = and then AC. Now type in 14 3 ln and press = This gives = Press = again to get = Press = again to get = Press = again to get = Press = again to get = Press = again to get = Press = again to get = The values aren t changing much any more so =3.24 to 2 decimal places Page 8

9 Question 6 i) = 3h +4 x 6h=9h3h +4 when h=0.6 = = = = ii) we are given that =0.015 we want from the chain rule = x = =0.18 h Page 9

10 Question 7 We are given that 12=9 We work from the inside out 12= 12 12= 12 = =9 Subtract 5 from both sides 2 12 =4 Divide both sides by 2 12 =2 Cube both sides 12 =2 =8 Add a to both sides 12=8+ Subtract 8 from both sides =12 8=4 =4 We have the inverse function of g, we need to find the inverse inverse function (the original function) If we start with = 4 Swap the and rearrange to make the new the subject = 4 Cube both sides = 4 Add 4 to both sides == Page 10

11 To find we work from the inside out =68 2+5= =68 Subtract 4 from both sides 2+5 =64 Cube root both sides 2+5=4 Subtract 5 from both sides 2= 1 Divide both sides by 2 = Page 11

12 Question 8 i) From formulae sheet Sin (θ + α) = sin θ cos α + cos θ sin α So Rsin (θ + α) = Rsin θ cos α + Rcos θ sin α Comparing coefficients Rcos α = 3 and Rsin α = 4 Square both R 2 cos 2 α = 9 and R 2 sin 2 α = 16 Add the two together R 2 cos 2 α + R 2 sin 2 α = 9+16=25 Factorise out the R 2 R 2 (cos 2 α + sin 2 α) = 25 cos 2 α + sin 2 α = 1 R 2 = 25 Square root both sides R = 5 Divide the two equations = tan α = Inverse tan (arctan) both sides α = arctan ( ) = 53.1⁰ 5sin (θ ⁰) R = 5, α = 53.1⁰ Page 12

13 ii) a) ⁰+ 1 = 0 subtract 1 from both sides = 1 divide both sides by = take inverse sine (arcsin) of both sides ⁰ = arcsin ( ) = 11.54⁰,180⁰ 11.54⁰=191.54ᵒ ⁰ = ⁰, ⁰ subtract 53.1 from each side = 64.6⁰,138.4⁰ Page 13

14 b) 37 5sinƟ+53.1ᵒ+ 43 subtract c from all three sides 37 5sinƟ+53.1ᵒ 43 Divide all three sides by 5 sinɵ+53.1ᵒ We know that sine of anything can only be between -1 and +1 = 1 =1 Multiply both sides of both equations by 5 37 = 5 43 =5 Add the two equations together to make the 5 disappear = 5+5=0 6 2=0 Add 2c to both sides 6=2 Divide both sides by 2 =3 Put back into either equation to solve 37 3= 5 40= 5 Divide both sides by -5 =8 =3, = Page 14

15 Question 9 i) we need to use the product rule on the first part = + Let = = 2 Differentiate both =1 = = Substitute into the formulae ln2=ln2+ =ln2+1 Now we can differentiate the whole expression and get ln2 =ln 2+1 1=ln 2 as required Page 15

16 ii) to find a volume of revolution then we use the following Note that as we are rotating about the y axis we use Limits Lower limit: when =0 = = = Upper limit: when = = Multiply both sides by 2 2= Take the natural log of both sides ln2= Volume = ln2 From part i) we know that this is the same as ln2 = ln2 ln2 = ln ln1 = 4 0 =2 + = + = Page 16

17 iii) this volume will be the volume of a cylinder minus the volume calculated in part ii) the cyclinder has radius of 2 and height of volume of a cylinder is h volume = x 2 x 3 +1= 4 3 1= Page 17

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