Name: _______________________________________________

C1 - Differentiating and Co-ordinate Geometry

AS Mathematics

Date:

Time: 1 hour 22 minutes Total marks available: 82 Total marks achieved: ______

Q1.

(a) Find giving each term in its simplest form. (4)

(b) Find

(2) (Total 6 marks)

Q2.

Given that can be written in the form 2xp − xq,

(a) write down the value of p and the value of q.

(2)

Given that y = 5x4 − 3 + ,

(b) find , simplifying the coefficient of each term.

(4)

(Total 6 marks) Q3.

(a) Show that where A and B are constants to be found.

(3)

(b) Find f '(x).

(3)

(c) Evaluate f '(9).

(2) (Total 8 marks)

Q4. The curve C has equation

y = x3 − 9x + + 30, x > 0

(a) Find . (4)

(b) Show that the point P(4,−8) lies on C. (2)

(c) Find an equation of the normal to C at the point P, giving your answer in the form ax + by + c = 0 , where a, b and c are integers.

(6)

Q5. The curve C has equation

y = 2x − 8√ x + 5 , x ≥ 0

(a) Find , giving each term in its simplest form.

(3)

The point P on C has x-coordinate equal to

(b) Find the equation of the tangent to C at the point P, giving your answer in the form y = ax + b, where a and b are constants.

(4)

The tangent to C at the point Q is parallel to the line with equation 2x − 3y + 18 = 0

(c) Find the coordinates of Q.

(5)

Q6. The curve C has equation

y = 9 − 4x − , x > 0

The point P on C has x-coordinate equal to 2.

(a) Show that the equation of the tangent to C at the point P is y = 1 − 2x.

(6)

(b) Find an equation of the normal to C at the point P.

(3)

The tangent at P meets the x-axis at A and the normal at P meets the x-axis at B.

(c) Find the area of triangle APB.

(4)

Q7.

Figure 2

Figure 2 shows a sketch of the curve C with equation

y = 2 − , x ≠ 0

The curve crosses the x-axis at the point A.

(a) Find the coordinates of A. (1)

(b) Show that the equation of the normal to C at A can be written as

2x + 8y − 1 = 0

(6)

The normal to C at A meets C again at the point B, as shown in Figure 2.

(c) Find the coordinates of B. (4)

(Total 11 marks)

Q8. The curve C has equation

y = (x + 1)(x + 3)2

(a) Sketch C, showing the coordinates of the points at which C meets the axes.

(4)

(b) Show that

(3)

The point A, with x-coordinate −5, lies on C.

(c) Find the equation of the tangent to C at A, giving your answer in the form y = mx + c, where m and c are constants.

(4)

Another point B also lies on C. The tangents to C at A and B are parallel.

(d) Find the x-coordinate of B.

(3) (Total 14 marks)

Q1. This question was well answered with about two-thirds of the candidature achieving all 6 marks.

In part (a), a small minority of candidates struggled to deal with the fractional power when

differentiating . Some candidates incorrectly reduced the power by 1 to give a term in ; whilst other candidates struggled to multiply − 6 by 4⁄3 or incorrectly multiplied − 6 by 1⁄3. Few

candidates did not simplify to give or integrated throughout, or added a constant to their differentiated expression.

In part (b), the most common error was to for candidates to differentiate to give . Few candidates did not understand the notation for the second derivative with some integrating their differentiated result in part (a) to achieve an answer similar to the expression given in the question.

Slips occurred with the omission of x from fractional power terms with some candidates

writing in part (a) or in part (b).

Q2.

Good candidates generally had no difficulty with the division in part (a) of this question, but others were often unable to cope with the required algebra and produced some very confused solutions.

A common mistake was to "multiply instead of divide", giving 2x2 ÷ √x = 2x , and sometimes √x was interpreted as x−1 . Examiners saw a wide variety of wrong answers for p and q. Most candidates were able to pick up at least two marks in part (b), where follow-through credit was available in many cases. While the vast majority used the answers from part (a), a few differentiated the numerator and denominator of the fraction term separately, then divided.

Q3. Most candidates made a good attempt at expanding the brackets but some struggled with

with answers such as being quite common. The

next challenge was the division by √x and some thought that Many, but not all, who had difficulties in establishing the first part made use of the given expression and there were plenty of good attempts at differentiating. Inevitably some did not interpret f '(x) correctly and a few attempted to integrate but with a follow through mark here many scored all 3 marks. In

part (c) the candidates were expected to evaluate correctly and then combine the fractions - two significant challenges but many completed both tasks very efficiently.

Q4.

As a last question this enabled good candidates to demonstrate an understanding of the techniques of gradients, applying problem solving and logical skills to achieve the final equation. 26% achieved full marks in this question. There were very few blank scripts or evidence of candidates who did not have time to complete the question. Usually if candidates did have difficulty, it was because they had made a mistake in answering the early part of the question. In part(a) most candidates were able to differentiate the equation correctly, although there were some problems with coefficients. Most mistakes occurred when differentiating with candidates being unable to rewrite it as 8x−1 prior to differentiation, or losing the term completely on differentiation. This term also caused candidates problems in the subsequent substitution of numbers which resulted in many strange results. Again, as in Q2, an inability to deal with fractions was seen.

In part (b) the usual approach was to substitute (4, −8) into the equation and show that −8 = −8. Cases where candidates substituted x = 4 mistakenly into their gradient instead of the equation of the curve C were frequent, although sometimes corrected. Substituting into fractional items proved to be too much for some candidates and consequently elementary mistakes were made. Simplification of the third term to −72 caused the most problems (many getting 54).

In part (c) there was again the occasional mistake of substitution into the wrong expression. Those candidates who correctly found the gradient of the curve, at the point P, usually went on and found the equation of the normal without any trouble. Arithmetic was often poor and it was common to see 24 − 27 − = − and other numerical slips. However even those candidates who had made anerror initially then attempted to find a perpendicular gradient and went on to use it successfully in finding the equation of their normal. Very few used the gradient of the tangent in error. Where candidates used y=mx+c the calculations for c were often numerically incorrect and followed long, complex (often messy) workings. Presentation in this question varied from some excellent easily followed solutions to some with little coherence.

Q5. On the whole (a) was very well done with the majority of candidates gaining full marks. Only a very small minority attempted integration and hardly anyone received less than two marks from the

three available. The majority of candidates reached 2 − 4x . Common errors seen were 2 − 4x

or 2 − 4x + 5x. The fractional powers were usually dealt with correctly on this part of the question.

In (b) many reached the correct answer of y = −6x + 3. Errors were made substituting x = into

4x to obtain gradient and further errors made substituting into the expression for y. Some

candidates found working with fractions challenging, e.g. = 2, so gradient equal to 2 − 4/2 = 0.

Some did not substitute x = into the function to get a y value but used (0, 5) to find the equation.

More able candidates answered (c) well, realising that they were required to set their gradient

function obtained in (a) to , the gradient of the given line. Some who got as far as = 2 − 4x

made errors in their algebra and these included = , leading to x = or even x = 3 and √x = 3 leading to x = √3. Of those who successfully reached x = 9, some attempted to find the y value by

substituting into y = x + 6 instead of substituting into the original equation. There was a significant proportion of the candidates who, after rearranging the equation of the straight line into the form y = mx + c, were unable to progress to gain any marks at all for (c). Of those who

proceeded unsuccessfully, it was common to see y = 0, so x + 6 = 0 leading to x = −9. Others found the points of intersection of 2x − 3y + 18 = 0 and y = −6x + 3 or found the co-ordinates of points of intersection of 2x − 3y + 18 = 0 with the x and y axes thus getting (0, 6) and (−9, 0). These answers did not answer the question set and gained no credit.

Q6.

Responses to this question varied considerably, ranging from completely correct, clear and concise to completely blank. Most candidates who realised the need to differentiate in part (a) were able to make good progress, although there were occasionally slips such as sign errors in the differentiation. A few lost marks by using the given equation of the tangent to find the y-coordinate of P. Those who used no differentiation at all were limited to only one mark out of six in part (a). Even candidates who were unsuccessful in establishing the equation of the tangent were sometimes able to score full marks for the normal in part (b). Finding the area of triangle APB in part (c) proved rather more challenging. Some candidates had difficulty in identifying which triangle was required, with diagrams suggesting intersections with the y-axis instead of the x-axis. The area calculation was sometimes made more difficult by using the right angle between the tangent and the normal, i.e.

(AP × BP) , rather than using AB as a base.

Q7.

This was a substantial question to end the paper and a number of candidates made little attempt beyond part (a). Part (c) proved quite challenging but there were some clear and succinct solutions seen. Some stumbled at the first stage obtaining x = 2 or even 1 instead of to the solution of 2 −

= 0 but most scored the mark for part (a).

The key to part (b) was to differentiate to find the gradient of the curve and most attempts did try this but a number had −x−2. Some however tried to establish the result without differentiation and this invariably involved inappropriate use of the printed answer. Those who did differentiate correctly sometimes struggled to evaluate ( )−2 correctly. A correct "show that" then required clear use of the perpendicular gradient rule and the use of their answer to part (a) to form the equation of the normal. There werea good number of fully correct solutions to this part but plenty of cases where multiple slips were made to arrive at the correct equation.

Most candidates set up a correct equation at the start of part (c) but simplifying this to a correct quadratic equation proved too challenging for many. Those who did arrive at 2x2 +15x −8 = 0 or 8y2 −17y = 0 were usually able to proceed to find the correct coordinates of B but there were

sometimes slips here in evaluating 2 − for example. There were a few candidates who used novel alternative approaches to part (c) such as substituting xy for 2x −1 from the equation of the curve into the equation of the normal to obtain the simple equation 8y + xy = 0 from whence the two intersections y = 0 and x = −8 were obtained.

Q8. Most candidates were able to make a good attempt in part (a). The most common mistake was not finding the intercept on the y-axis, or finding the wrong one; (0,10) was a common answer, with a few candidates writing 1 × 3 × 3 =10. Another common error was to draw the graph touching the x-axis at −1 and crossing it at −3 rather than the other way round, or to draw a negative cubic graph (i.e. wrong orientation). Others also showed the graph touching/crossing at x = +3 and +1. Multiple attempts produce a confused answer which usually gains no credit.

The majority of candidates were able to obtain full marks on part (b). Most understood that they needed to expand the brackets, with surprisingly few errors, although the absence of any negatives probably helped. Most candidates obtained the required cubic polynomial and differentiated correctly to get the printed answer. Those who made errors in their expansion were usually honest and differentiated their expansion, thereby gaining the method mark for differentiation. A few candidates did not know what to do, and tried to integrate backwards.

Many candidates clearly understood what to do and correctly substituted x = −5 into f(x)

and , to get the correct y value and gradient in part (c), resulting in the correct equation.

There were more simple arithmetical mistakes than expected, including a significant number who wrote "−4 × 4 = 0" for their y value and also for their gradient wrote "75 − 70 + 15 = 10". As well as poor arithmetic skills, many candidates made the arithmetic harder than necessary by choosing to calculate their value of y from their expanded cubic expression, rather than using the factorised form given in the question. Not only did this make the arithmetic more difficult, but it also introduced the possibility of an

incorrect y value being found if their expanded cubic were incorrect. y = 20x + 116 was also seen as a result of poor re-arrangement. Weaker candidates attempted to find a second point on the curve and used the gradient of a line segment rather than the gradient of the tangent. Others automatically thought they had to begin by

differentiating to find the equation of a tangent, and so they used to find their gradient.

Part (d) seemed to differentiate between the better candidates and the rest. Good ones usually scored full marks, others either did not attempt it, or did not know how to begin.

Some weak candidates put their equation from (c) equal to or the original equation of curve C and attempted to solve. Many thought they had to produce a second equation of a line, usually using a gradient of "20" and choosing a different point, or a value of "c", without any apparent reason. Others thought that this next part must be to

do with the normal and so found the equation of a line with gradient . Some others understood that they needed to use their value for the gradient and the statement

'parallel so same gradient' was common, but candidates did not relate this to .

Almost all candidates who correctly equated to 20 went on to correctly factorise

and, pleasingly, many candidates then explained why they chose rather than x = − 5 as their answer showing that they had good understanding of the context.

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