EdExcel AS-Level Maths October/November 2021 Paper 1 (Pure Mathematics)

this is the lxl as a

level

one

mathematics

paper

which covers the

pure

mod section and is from question number one of the

2021

series in this question you must show all the stages of your work Solutions that depend on calculated technology are not acceptable using algebra, we have to solve the inequality with x squared minus X minus 20. is greater than zero, we can find the critical values ​​by replacing the inequality symbol with an equal symbol if we do x squared minus to give negative 20 add to give negative one, of course, we have negative five and four, so when we factor as the product of two brackets x squared minus x minus 20 we have x minus five multiplied by X plus four and this, of course , is equal to zero and if we solve to find our critical values ​​if x minus 5 is equal to zero graph y equals x squared minus x minus 20. but ignoring the y axis we have a sketch of y equals x squared minus We are looking at where the graph is greater than zero and it is greater than zero here or here so when we come to indicate the solutions to this inequality we have X which is less than negative four or we have X which is greater than 5 and if we express this in notation of sets we have X belonging to a set of real numbers X is less than negative four or we have X belonging to a set of real numbers must show all stages of your work Solutions that depend on calculated technology are not acceptable given nine to the power of x minus one over three to the power of y plus 2 equals 81 we have to express Y in terms of X by writing your Answer in the simplest form , what we can do is rewrite both the left side and the right side so that we have a base of three that will make it easier for us to compare powers and, of course, in this question we can make use of the laws. of indices nine is the same as 3 squared, so we have 3 squared to the power of x minus 1. divided by 3 to the power of y plus 2. this is equal to 81 and 81 is equal to 3 to the power of 4 We have a base two of power to another power, so we multiply the powers we have 3 to the power of two lots of x minus one divided by 3 to the power of y plus 2.

This is equal to 3 to the power of four. If we expand the parentheses at the top we have 3 to the power of 2x minus two at the bottom 3 to the power of y Plus 2. all of this is equal to 3 to the power of 4. when we have the same base and we are dividing, we subtract the powers, so that on the left side we have 3 to the power of 2 of x minus two minus y plus 2. and this is equal to 3 to the power of four on both the left side and the right side we have the same bases, so now we can compare poses.

If we compare the powers on the left side, we have 2x minus two minus y plus 2. On the right side we have four, let's rearrange to have y in In terms of x, we have 2X minus y and then we have -2 minus another two, so is negative four and this is equal to 4. So we add four to both sides and we have 2X minus y is equal to eight, so Y is equal to 2X. minus eight we have expressed Y in terms of x question number three we have to find the integral of three x to the fourth power minus four over two x to the power of three with respect to to the power of four minus four divided by 2x cubed and we're integrating this with respect to X.

If we divide this fraction, we can divide it as 3x to the power of 4 divided by 2x cubed minus 4 over 2x cubed. then integrate this with respect to which is 2 And then, of course, here we have 2 over X cubed, which is the same as 2x to the power of negative 3. We'll integrate this result with respect to . Power, so what we end up here is three over four x squared and then if I add one to the power I get negative 2 and then if I do negative 2 divided by negative two we end up with plus one, so we have x to the power of negative two and then we add a constant of integration C question number four in this question the unit vectors I and J are towards the east and towards the north respectively a stone slides horizontally on the ice initially the stone is at the point at minus 24 I minus 10 J meters with respect to a fixed point OR after four seconds the stone is at point B 20 I plus 5j meters with respect to the fixed point OR the movement of the stone is modeled as the other particle moving in a straight line at constant speed Using the model from part A we have to show that the storm passes either we know so far or a is equal to negative 24.

I minus 10. Che and we know that or B is equal to 12i plus 5j We want to show that the stone passes through the origin, so what we want to show is we want to show air o B is a straight line, so if we know that o a is minus 24 I minus 10 J, we can conclude that air o is equal to the negative of this 24i plus 10 J and for To prove that A or B is a straight line we need to prove that a or is a multiple of a b well, what you can see from here is equal to two lots of o B of course, 24i plus 10j is equal to many 12i plus 5j, so that we can conclude that the particle passes through point O and now, for part B, we have to calculate the speed of the stone.

To do this we need to find the vector a b and that will tell us the distance and then we can find the velocity using velocity equals distance divided by time so we have in this diagram points A and B and we have this origin here what we can do is find a vector a b, it starts here and ends at B so we can figure out what the vector a B is, it is the vector a b, well, from -24 to 12 we go 36. in the direction I and from -10 to 5 we go 15 in the J direction, so what we have found here is the vector a b.

What we now need to do, since we found the vector a b, we can conclude that the distance the particle travels will be given by the magnitude of a b. equals 36 squared plus 15 square root, so the distance we have is 36 squared plus 15 square root, which is 39. So we have 39 meters for the distance and therefore the speed is equal to the distance of 39 meters over time 4 seconds the speed is equal to 9.75 meters per second question number five we have figure one that shows part of the curve for the equation y is equal to 3x squared minus 2 and we have the point p with coordinates 210 and this is located on the curve for the part here we have to find the gradient of the tangent to the curve at point p for part a if we were to find the gradient of the tangent to the curve up here, we would first have to figure out what d y over DX is what the gradient function of the tangent d y over DX tells us which is equal to 6x y at the point p with coordinates 10. d y over DX is equal to 6 lots of two, which is 12.

So this 12 from here represents the gradient of the tangent at the point p Another Part B the point Q with quadruple x coordinate two plus h is also found in the Curve we have to find the slope of the line PQ giving your answer in terms of H in its form simplest well, we have We have this point p here and then we have point Q with coordinates 2 plus h for the x coordinate and for the y coordinate we replace the X with 2 plus h, so for the y coordinate we would have three lots of 2 more h all squared minus two, so what we can do now is find the gradient of paq, so if we were to find the gradient of peq, we can do it by considering the change in y from the change in x the gradient of PQ, so we have the change in y well, we have three lots of two plus h squared minus two, minus 10. so we have three lots of two plus h squared minus two, then we subtract ten, so that's the change in y and then the change in x we have two plus h minus two and if we were to solve this we have three lots of four plus two h plus h squared minus 12. divided by h 12 plus 6h plus three h squared instead of here we have four so here we have 12. so we have 12 plus 12h plus 3H squared minus 12 at the top and at the bottom we have h so we can conclude what we end up as the gradient function we have got 12h plus 3H squared divided by the hitch which is just 12 plus 3H, so we have 12 plus 3H.

Another part C of the question briefly explains the relationship between Part B and the answer to the party, as H becomes smaller and smaller, in other words, as H. tends towards zero, then 12 plus 3H would tend towards 12, of course, 3 H would tend towards zero, so the gradient of the chord tends to the gradient of the tangent to the curve question number six in this question, you should show all the stages of your working solutions. relying on calculated technology is not acceptable for the party, we have to use algebra to find all the solutions to the equation 3x cubed minus 17 x squared minus 6x equals zero the first thing we will do is factor an X many 3x squared minus 17x minus six is ​​equal to zero and if we try to factor the quadratic we have 3x plus 1 multiplied by X is equal to zero if 3x plus 1 is 0 then for Therefore, part B finds all the real solutions of three lots of Y minus two, all to the power of six minus 17. lots of Y minus 2 to the power of four minus six lots of Y minus two, all squared equals to zero, maybe if we rewrite the equation well, let's start by copying the equation, we have three lots of y minus 2 to the power of six minus 17 knots of Y minus two to the power of four minus six lots of y minus two squared, all of this is the same to zero and if we use the laws of indexes, you can see that we have three lots of y minus 2. squared to the power of 3 minus 17 lots of y minus 2 squared minus six lots of y minus 2 squared and all of this is equal to zero, what we've done here is rewritten everything in terms of y minus 2, everything squared.

The reason for this is that if we compare with what we have in our original equation, you can see that X is equal to Y minus 2. Everything squared, so compare with part A we can see that squared, so we have y minus two all squared equals zero minus one third or six, but of course here and minus 2 all squared cannot equal negative one third, we win. We don't get any real solution, so if we get rid of this and minus 2, everything squared can be equal to zero or six, let's take note of why that is a and minus 2, everything squared is equal to negative one third, it has more real solutions, of course, we cannot get square root of a. negative number and then we have y minus 2 all squared is equal to zero, from which we can see that Y is equal to 2. or we have and minus 2 all squared is equal to six and if you apply the square root and then add two , we can conclude that Y is equal to plus minus square root 6. and therefore, if we state the solutions we have y is equal to and Y is equal to 2 plus minus square root 6. these are our only real solutions question number seven the parallelogram p q r s has an area of ​​50 centimeters squared since PQ has a length of 14 centimeters and QR has borrowed seven centimeters and the angle spq is obtuse what we have to do for part A is find the angle spq in degrees to two decimal places here There is a diagram of the parallelogram that we are given that p q measures 14 centimeters we are also given the QR has a length of seven centimeters we are given s p q is obtuse and we have to find this angle here well, how would we go about finding the area of ​​a parallelogram ?

Well, the area of ​​a parallelogram is double. the area of ​​a triangle now, to start, we can find this angle here the angle p q r we will call it a theater funnel how can we do it right to find the area of ​​one of these triangles that is half times seven? times 14 times the sine of theta and then we multiply this by two because we have twice as many triangles and this area is equal to 50. so the sine of theta is equal to 50 over 7 times 14, which is 25 over 49. which we can conclude that Theta is equal to the inverse sine of this which is 30.67 Etc and therefore the uncle spq because it is obtuse and here we have cointerior angles we have 180 minus this 30.67 Etc so we have 180 minus this which is 149 0.32 degrees given with two decimals Another part B we have to find the length of the diagonal SQ give your answers with one decimal well we know that this length here is also seven we know that the angle here is one hundred and let's underline it, we have 149.32 for this angle here, so that this angle is 149.32 degrees given to onegiven to two decimal places and what we need to find is this length SQ and we can do it using the cosine rule, so for part B of the question, if we use the cosine rule, SQ squared is equal to 14 squared plus 7 al squared minus two times times 14 times times 7 times times the cosine of 149.3 Etc.

So if we calculate what SQ is, we have 14 squared plus 7 squared. minus 2 times 14 times 7. times the cosine of this, which is 413.5 ETC and therefore SQ, which is equal to the square root of this, which is 20.3 centimeters to one decimal, so if we add this to our diagram, we have 20. 3 question number eight G of five we have to find the value of a well it would make sense for us to find the value of a by calculating the coefficient of x to the power of five, if we go to the formula book, we can find the coefficient of x to the power of five using the results we see here, for so the results here will allow us to find the coefficient of x to the power of five. the five then the coefficient of x to the power of five and this here the coefficient of a to the power of 5 is equal to 3 4 or 2 divided by four four eight, which is 243 over 25 or rather, over 32 e is the fifth root of this, which is 3 over 2. so a is equal to 3 over 2 . and now, for Part B, using this value of a, what we need to do is find the constant term in the expansion of 1 plus 1 over x up to 4 multiplied by two plus x to the power of 8. so, for Part B we have foreign x to the fourth power, so we have one plus one plus one over What to do is somehow find the constant term.

Now what we will do is write a couple of terms of the expansion of two plus three x over 2 to the power of eight. in a certain order we have one plus one over . raised to the power of zero and then for the next term eight, choose one two to the power of seven 3x over 2. raised to one now the question is when exactly can we stop. write the full expansion, but what we have to do is find the constant term now here, the one multiplied by the result here will give a constant term, but then we have this over x to the power of four and if I multiply it by something involving a raised x to the fourth in the numerator, they will cancel to give a constant term, so we could continue with this, but as soon as we have eight, choose four Times times 2 to the power of four multiplied by 3x over 2. power of four those terms here will give one term next to the four, let's just move all of this around so that we have several different terms, so what is multiplied to give a constant term?

So let's write this in the constant term, we have one multiplied. for the result here we have one multiplied by H, choose zero two to the power of eight multiplied by 3x over 2. to the power of 0. and what gives the most a constant term will be the one over x to the power of four multiplied by this here we have one over x to the power of four multiplied by 8 choose four two to the power of 4. multiplied by 3x over 2. to the power of four why is that the kiss? Well, let's continue solving all this. we have one multiplied by one multiplied by 2 to the power of eight, which is 256. multiplied by one, that's the result here foreign is 1 over x to the power of four multiplied by well, let's write this down, we have one over x to the power of four multiplied by choosing four , which is 70. multiplied by 2 to the power of four, so we have 2 to the power of four, which is 16. and then we have multiplied this by 3x over 2, all to the power of force here we have 1.5 power of 4, which is 81 over 16.

So we multiply this by 81 over 16. or 81 x to the power of four over 16. So these terms are multiplied together and happily for us this and this cancel out, so the constant term is equal to 256 plus 70. multiplied by 16 times 81 over 16. it turns out that we obtain five thousand nine hundred and twenty-six as a constant term question number nine, we have defined the value of the constant k where K is between 0 and 9, so that the integral of 6 over the root square of x with respect to I'm integrating this with this bet, so we add 1 to the power we get x to the power of half and if we divide it by the new power, six divided by half is 12.

So what we need to calculate is 12x to the power half evaluated between K and 9. So we have 12 lots of 9 halved minus 12K halved and if we order this even further 3 times 12 is 36 we have 36 minus 12. K halved and if we equal this result to 20. we have 36 minus 12K halved, this is equal to 20. so if we rearrange we have 12. K halved is equal to 16. K halved is equal to 16. divided by 12. which is 4 over 3. and if we square both sides k then it is equal to 16. over 9. then we have k is equal to 16 over 9. foreigner a student is investigating the following statement about the natural numbers n cubed minus N is a multiple of 4.

We have to prove using algebra that the statement is always true for odd numbers, so the first thing we will do is indicate the shape of an odd number. We know that odd numbers have the form 2x plus 1. So if we let y equal X plus one, then n cubed minus n will equal n cubed, so we have 2x plus 1 cubed minus 2x plus 1 .And of course, here X is an integer, so if we expand this right here. we need to use the binomial, so expanding we have one lot of 2x to the power of 3 multiplied by 1 to the power of zero Plus three lots of 2x, so squared many of one to the power of one Plus three lots of 2 of x to the power of one times 1 squared Plus one lots of 2x to zero one cubed and then B we subtract 2X and then we subtract the one as well, the term here would cancel with the term here, so we have a nicer cubed text, we We have 12 x al squared plus 6x and then we have rather here is just the minus one that console, so here we have 2x minus one and rather this cancels out, then we subtract 2X, so we have 8X cubed plus 12x squared. plus 4X and if I take a 4 as a common factor we have four lots of 2x cubed plus 3x squared plus one, so what we have here is four lots of an integer because X is an integer 2x cubed plus three x squared plus one is also an integer and this result here is a multiple of four.

Another part B, we have to use a counterexample to show that the statement is not always true. We can use any integer, for example, what would happen when n equals? 2. Well, we would have 2 cubed minus two, which is equal to 6. and 6. is not a multiple of four, so this whole statement is sometimes true. Question number eleven: The owners of the nature reserve decided to increase the area covered in the reserve by trees. The planting of trees began on January 1, 2005. The area of ​​the nature reserve covered by trees is one cubic kilometer squared and is modeled by the equation a is equal to 80 minus 45 e to the power of c t where C is a constant and T is the number of years after January 1, 2005.

Using the model we have to find the area of ​​the nature reserve that was covered by trees just before the tree planting for the party began, just before the tree planting began, that of course is when t equals zero, so when T equals zero a equals 80. minus 45 e to the power of 0 e to the power of 0 is 1. 80 minus 45 is 35. so we have 35 foreign square meters on January 1, 2019 an area of ​​60 kilometers of the nature reserve was covered by trees use this information to find a complete equation for the model it will give you a value of c23 significant figures, well, this time it is equal to 60. and T is equal to 14. the ratio T is equal to 14. from 2005 to 2019 this over here 14. yes and knowing that a is equal to 60 we have 60 is equal to 80. minus 45 foreign to the power of 14 C and what we have to do is figure out what C is if we rearrange 45 e to the power of 14c is equal to 20. so 14c, instead of dividing it times 45 to start, e to the power of 14c is equal to 4 over 9, so 14 C is equal to the natural logarithm of for over nine from which we can conclude that C is equal to 1 over 14. comes from the natural logarithm of 4 over 9 and if we calculate the decimal equivalent of this 1 over 14 multiplied by the natural logarithm of 4 over 9. to find this to equal minus zero point no five seven nine Etc. and giving the value of C to three significant figures the equation of model is given by a is equal to 80 minus 45 e to the power of negative zero point no five seven nine t 2 3 significant figures and now, for part C of the question, on January 1, 2020, the owners of the nature reserve announced a long-term plan to have 100 square kilometers of the nature reserve covered with trees.

We have to remain a reason why the model is not appropriate. for this plan Well, what we're going to do to start part C is draw a sketch of air versus here's a sketch of the graph of a versus t. I don't have an asymptote, of course, a horizontal asymptote, we have the initial one. area 35 we calculated that before and what we can conclude then is that the maximum area covered by trees alone is 80 square kilometers question number 12 in this question you must show all the stages of your work Solutions that completely depend on calculated technology are not acceptable for The first part we have to solve between 0 and 450.

The equation 5 cos theta squared is equal to 6 sinusoidal Theta, giving answers with a decimal, we start with five cos theta squared is equal to 6. Sinusoidal theta if we replace cos al square with 1 minus sine. squared we would have five lots of one minus sine theta squared equals six sine Theta and if we expand the brackets on the left side we have five minus 5 sine theta squared which equals six sine Theta, so we rearrange so that one side be equal to zero we have five sine theta squared plus six sine theta minus five equals zero. Now we need to solve this, so if you figure out what sine Theta is we can use our equation solver on the calculator.

We have five multiplied squared plus six sine minus 5. is equal to zero, we get a value of zero point five six six Etc and the other value that is equal to minus 1.76 Etc, so we have sine Theta equal to minus one point seven six six Etc. Now, can this work sin Theta equals negative 1.766 Etc, although it can't because sine is between negative one and one, so sine Theta is not equal to negative 1.766. Etc., our only value here is this, let's store this in our calculator and to figure out what theater is with a decimal, then we have Inverse sine of this, so we have the inverse sine of that value, which is 34 point five degrees , we add 360 because if we remember for our domain, the value of theta is between 0 and 450.

So here we get a value. of 300 and 94.5 degrees and for the stick with 360. we can do 180 minus this, which is one hundred forty-five point five degrees. We have given all of our solutions to one decimal. Now for the next part of the question, students try to solve. the equation to solve x is between minus 90 and 90 the equation three times x minus 5 sine look at the first line, we have 3 Tan x minus 5 sin of cos x leaving us with 3 sine online so that cancel by sine to three over five, so if I do the inverse cos of three over five, they get 53.1 which is what they got, but if we take a look at the domain minus 19 to 90, we also have a solution of minus 53.1 and that is due to the symmetry of the graph, so let's write the second suit, they don't find everything. the solutions of cos x are three or five in the domain minus 90 to 90, so you missed the solution Alpha plus 40 degrees equals three - the fifths are alpha one alpha two alpha three and alpha four we have to find the highest degreeclose the value of alpha four so for part B we have cos of 5 alpha plus 40. is equal to three over five alpha plus 40. we would expect to get multiple solutions here is the first solution 53.13 Etc. the next solution, well, let's store this in to the next solution will be 360 ​​minus this is 306 point eight six Etc. let's store this in B the next solution will be this value 53.1 ETC plus 360 So we have a plus 360, which is 413.13 ETC, store this in C and for the solution final we have the value of B plus 360, which is 666 point eight six, etc., we will start this today and then if we divide. each of them, if we subtract 40 and then divide each of them by five, you will find that Alpha is equal, first of all we have negative 40 over 5. the first angle is 2.62, etc., the next one we have 53 point three . seven Etc, the next one we have 74 point six two Etc and the final one we get 125.37 Etc, we have here everything for an Alfa two Alfa three and Alfa 4.

Therefore, we can conclude that Alfa 4 is equal to 125 degrees to the nearest . grade question number 13. the resting heart rate of a mammal measured in beats per minute is modeled by the equation H is equal to pm to the power of Q, repeat and give a constant and M is the mass of the mammal measured in kilograms figure 2 which is what we see here illustrates the linear relationship between log base 10 of H and log base 10 of M. The line meets the vertical log base 10 of the H axis is 2.25 and has a gradient of minus 0.235 for the part, we have than find three significant ones. calculate the values ​​of p and Q, think of this as being in the form y equals MX plus C instead of us having an X, instead we have log base 10 of M and instead of having a y, we have log in base 10 of H now we can normally write y equals minus 0.235 which is the Times gradient times X plus 2.25, but instead of having those X and Y variables, we have log in base 10 of M and log in base 10 of H, so what we can do for part A is replace Y with log base 10 of H and we can replace the X with log base 10 of M, so if we do that we will have log base 10 of H is equals 2.25 minus zero point two three five times log in base 10 of M, then if anti-log on both sides we have 10.

Rather we have h is equal to 10 to the power of 2.25 minus zero point two three five times log in base 10. of M and if we use the power of many logs we can rewrite here as 10 to the power of 2.25 More log base 10 of M to the power of minus zero point two three five and so on, as we have 10 to the power of two results that add up, we can use many indices to the power. of m plus n is equal to a to the power of M times a to the power of n when we have the same base when we multiply we add the powers so that H is equal to 10 to the power of 2.25 multiplied by 10 to the log power in base 10 of M raised to the power of minus zero point two three five now 10 raised to the power of log base 10 that's just going to be something it's going to be um so let's write this Mr. yes We have x raised to the power of the log base , this is equal to y, so if we use that idea we can see that H is equal to 10 to the power of 2.25 multiplied by m to the power of negative zero point two. three five, so p is equal to 10 to the power of 2.25, which if we calculate this we get 178 to 3 significant figures, so this is the value of P and then the value of Q which is equal to minus, not to point two, three, five. this is given to three significant figures, so we have our value for p, we have the value of Q, so now, with that particular mammal that has a mass of five kilograms and a resting heart rate of 119 beats per minute, we have to comment on suitability.

From the model for this mammal, this time we are given that the mass is five kilograms, so with our model we can replace M with five, so H is equal to 10. to the power of 2.25 multiplied by 5 to the power of negative zero point two three five well, of course, this is the value of P and this is the value of Q, so if we figure out what H is, we have 10 to the power of 2.25 multiplied by 5 to the power of negative 0.235, we get 121.8 etc. and if we look at the units we have beats per minute, so we have 121.8 etc. beats per minute and if we take a look at what it actually is, it's actually 119 beats per minute, so let's write this down, it's actually 119. beats per minute, these values ​​are close, so what can we say?

So we can say that it is reasonably accurate to two significant figures if around the H value here and 119 to two significant figures we get 120. is per minute, if of course two significant figures, so the model is suitable now for part C of the question with reference to the model, we have to interpret the value of the constant p very simply p is the resting heart rate in beats per minute of a mammal with a mass of one. kilogram question number 14 a curves here you have equation y is equal to f of x but f of x is equal to minus three x squared plus 12x plus 8. we have to write f of x in the batch form of X plus B all squared plus c a b and c are constants that must be found, so for the group we start with negative 3x squared plus 12x plus 8 and we must write this in the form given to us here, the form given to us here suggests that we must complete the square, so if we factor the coefficient of x of the first two terms look at the coefficient of x squared of the first term and the second term we have negative three on the outside and inside this big bracket you would have x squared minus 4X Clause big bracket and then We have the plus a, so if we complete the square of this big parenthesis, we have negative 3 on the outside and then here we would have x minus 2, half of the four is two and then we square this and subtract the square. of two which is four and then we have plus eight, so we have minus three lots of x minus two all squared minus three times minus 4 is plus 12. and then we have plus eight, so if we order With all this we have minus three lots of x minus two all squared plus twenty, from which we can conclude that f of x is equal to negative three lots of x minus 2 all squared plus 20.

Part B of the question, curve C has an inflection point maximum in M ​​and we have to find the coordinates of M, while the inflection point will be negative B C, so the inflection point would have negative coordinates B C, so we have negative negative two, so it's 20. Rather , are two and then for the quadrilateral Y, not the inflection point, we have 20. So M has coordinates 20. And now, for part C of the question, figure three shows a sketch of curve C, that is what we see here and the line L passes through it. M y is parallel to the x axis, the region R shown shared in figure 2 is bounded by c l and the y axis we have to use algebraic integration to find the exact area of ​​r, well L would have the equation y equals 20, is parallel to the x-axis and is tangent to the curve, so L has the equation y is equal to 20.

Now let's figure out the area of ​​the rectangle, what do we mean that the area of ​​the rectangle is the area here between zero and two? the coordinates of M are two twenty, so we have 2 over 20 above, we can find the area of ​​the rectangle, so the area of ​​the rectangle is 2 multiplied by 20, which is 40. units squared, now we can find the area under the curve between 0 and 2. so what we can do now is find the area under the curve between zero and two so that the area under the curve we can find using integration, we can integrate between 0 and 2. minus three x squared plus 12x 8.

So if we integrate this because we add 1 to the power divided by the new power, we have negative X cubed plus 6X squared plus 8x and this will evaluate between zero and two, so we have negative 2 cubed plus six. Slots of two squared plus eight lots of two for the upper limit for the lower limit would have zero another zero and then another zero so it turns out that here we get 32 ​​units squared and we saw therefore the area of ​​R is equal to 40. minus 32 it turns out that the area of ​​r is equal to eight units squared question 15.

Figure 4 shows a sketch of a Circle C with Center n with coordinates seven four and the line L with equation y is equal to one third C at point P for part A we have to find the equation of the line NP in the form y is equal to MX plus C where M and C are constants well for part we start with the line L this has an equation and is equal to a third of X and so what? we can see here that the gradient is equal to one third, so the gradient of the PN line will be the negative reciprocal of this, so the gradient of PN is the negative reciprocal of one third, which is -3, so that we use Y. minus y1 is equal to m lots of x minus X1 we have y minus well, it goes through 0.74, so we have y minus four is equal to minus three lots of equation y is equal to here We have negative 3x and we have negative 21 plus 4.

Rather, we have plus 21 plus 4, which is 25. So we'll call this equation one. We'll call this equation two now for Part B of the question. We have to find an equation for C before we do anything else. We need to find the length of NP, which is the radius. Now we know that n has coordinates 7 4 and we know that P has coordinates. We can find the coordinates. of P by solving this equation with this equation simultaneously, so if I do that, we have this line here which is the line y P and from there we can find the point P by solving NP simultaneously with l, in other words, solve equations one and two simultaneously and we can do it by equating them to each other so that we have one third of X is equal to negative 3x plus 25. so we have 10 over 3 , the corresponding y coordinate is equal to 5 over 2. so we have the coordinates of P 15 over 2. and for the y coordinate we have five over two, so we can find out the length of RP, which of course is just the radius , so this is the radius, so R is equal to if we use the length formula, we have 15 over 2 minus 7 squared plus 4 minus five over two squared with square root, so If we find out what the radius is, we have the square root of 7.5 minus 7 squared plus 4 minus 2.5 squared the integer square root which is root of 10 over 2.

Therefore, the radius is root of 10. over 2 and if squared this is for the equation of C we have x minus 7 all squared plus y minus 4 all squared is equal to 5 over 2. and now for our part C, the line with the equation y is equal to one third of K is a previous constant, it is also a tangent to C we have to find the value of the constant k now, if we go back to our diagram, let's add this tangent line, we will do it here here we have our line, we will call it L2 L2 intersexy here somewhere moment is fine and what we can do is figure out at what point what are the coordinates of Q and we can do this using vectors, so if we were to use vectors, let's draw a quick diagram to illustrate what we need to do. for part C here is our circle, we have Q n n p now what we can do is determine how we go from n to p and then we can use that information to figure out how we go from Q to n in the X direction to go from seven to 15 over 2. we add the half, so we add 0.5 in the Do the same here, we add 0.5 and here we would remove 1.5.

Currently we have the coordinates of n being seven four if we go back to the question we have. seven four here we have 7.5 2.5 so we fly backwards we take 0.5 from 7. so here we get 6.5 and then we add 1.5 to 4 to get 5.5 so we've used vectors to essentially work backwards so just move this here so far down that those coordinates will leave them as fractions for now we have 13 over 2. and then we have 11. over 2. now, if we substitute this point in y is equal to one third X plus K we can find the value of K, so the foreign y is equal to one third of third lot of x, so maybe if we put a parenthesis here ten over three and therefore foreign X plus 10 over 3. that is the equation of the line that is also tangent to C question 16 the curves here have the equation y is equal to f of x where f of cube plus 15x squared minus 39x plus b MBR constant but since the point 2 10 is at C and the gradient of the curve at 2 10 is -3 we have to show that the value of a is -2 and find the value of B we start with f of x which is equal to ax cubed plus 15 now F of 2 is given to be 10.

So if we use this information and substitute it in here, we have an x ​​cubed, so we have a lot of 2 cubed plus 15 lots of 2 squared minus 39 lots of two plus b this . is equal to 10. we have eight here plus 60. less good here we have negative 78 plus b, all of this is equal to 10. So, eight air plus b is equal to 28, we'll call thisequation 1. and We have also been given information about the gradient: the gradient is equal to -3 at the point x is equal to, so F dash of 2 is equal to -3, so we have three lots of two at square plus 30 lots of two minus 39 this equals negative three 12a plus sixty minus 39 equals negative three twelve a equals negative 24. equals negative two we'll call it three, let's move this here and now if we substitute three, let's better call it two if now we substitute two into one, we have eight here, so we have eight lots of negative two plus b, this equals 28. so we have negative 16 plus b equals 28. and therefore, be equals a 44. and knowing that a is equal to negative 2 and B is equal to 44. therefore, we can conclude that f of x is equal to negative two x cubed plus 15x squared minus 39x plus 44.

We can also conclude that F dash of The stationary points would suggest that F dash of X is equal to zero, we want to show that F dash of X is not equal to zero. Can we do this right? We can use the discriminant for Part B if we use the discriminant B squared minus 4AC. We have B squared, so we have 30 squared minus 4 lots of negative six lots of negative 39. We wonder if this is greater. that 0 equals zero or less than zero 30 squared minus 4 times -6 times minus 39 that is minus 36 Which is less than zero, we can conclude that there are no real roots for F dash of X to be equal to zero and because F dash of x not equal to zero we can conclude from this that f of x has no inflection points now for part C of the question we have to write f of x in the form x minus four lots of Q quadratic which is fine for part C we know what f of x is, that's negative two x cubed plus 15 x squared minus 39x plus 44. this is equivalent to x minus four lots of some Lambda quadratic x squared plus beta gamma now if we compare coefficients concentrating on , therefore gamma is equal to negative 11 and if we compare the set x squared we have 15 of them on the left side and on the right side we have negative four times Lambda and then we have x times Beta X or the coefficient of x squared is beta now we know what Lambda is that is -2, so we have 15 equals -4 times -2 plus beta 15 equals 8 plus beta beta equals 7.

Knowing the values ​​of Lambda beta and gamma, we can conclude that negative 2x cubed plus 15 x squared minus 39x plus 44 is equivalent to x minus 4 multiplied by Lambda x squared, so we have negative two x squared plus beta minus 11. So f of x is equal to negative two x squared plus 7x minus 11. times x minus 4. now that we've done that, derive the coordinates of the points of intersection of the curve with the equation F of the zero point two x and the coordinate axis well, let's see where the curve would intersect the x and y axes, it intersects the y axis when F of 0 when we have zero is equal to 44. and therefore one of the points is zero 44. and if I solve the equation f of x is equal to zero, we have x minus four lots of negative two x squared plus 7X minus 11 equals zero, so x minus 4 equals 0, from which we can conclude , we can conclude that x equals four x equals 4 is the only real solution, so The point where it intersects the let's write this.

F of zero point two x 0.2 is the same as a fifth F of a fifth X well, let's write this correctly if we go from f of x to F of a fifth stretch five in the nothing happens in the y direction, so zero 44 is where F of 0.2