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AS and A Level: Core & Pure Mathematics

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Get help from 80+ teachers and hundreds of thousands of student written documents Differentiation and intergration

1. 1 It is easy to get differentiation and integration the wrong way round. Remember that the power gets smaller when differentiating.
2. 2 Differentiation allows you to find the gradient of a tangent at any point on a curve. The first derivative describes the rate of change.
3. 3 If a function is increasing then the first derivative is positive, if a function is decreasing, then the first derivative is negative.
4. 4 When asked to find the area under a curve, it is asking you to integrate that curve between two points. Even if you don’t know the points, pick two numbers. You’ll get marks for methods.
5. 5 When referring to a min/max/stationary point, the gradient equals 0. Differentiate the curve and set this to equal 0. The second derivative tells you whether it is a maximum or minimum. If the second derivative is positive, the point is a minimum, if the second derivative is negative, then the point is a maximum.

1. 1 When solving a quadratic inequality, always draw a picture. The inequality is less than 0, where the curve is below the x-axis and bigger than 0 when the curve is above the x-axis.
2. 2 Sometimes in part (a) of a question you are asked to find something, for example a radius. In part (b) you might be then asked to use the radius that you found. If you couldn’t do part (a), don’t give up, choose a random radius.

Straight lines

1. 1 To find the distance of a straight line, draw the straight line with the co-ordinates. Then make a right angle triangle, find the lengths of the horizontal and vertical lines, then use Pythagoras.
2. 2 When a question asks you for a straight line. The first thing to do is to write down the equation of a straight line. Then find out what information you know, and what information you need. Even if you don’t understand the whole question, it is important to start.

• Marked by Teachers essays 3
1. Mathematical equations can be solved in many ways; however some equations cannot be solved algebraically. I am going to show the three methods of solving these types of equations numerically

The error bounds are 0.3095 and 0.310. Change of Sign Method Failing Consider the equation f(x) =0, where f(x) = I shall try to locate the roots by making a table of values that correspond to this function of x. Rearranging f(x) =0 into x=g(x) Rearranging f(x) =0 into x=g(x) failing I am going to solve the equation f(x) =0 where f(x) =. The graph of y=f(x) is shown below. ==> ==> ==> ==> ==> ==> The sequence formed is converging; however it converges to the incorrect root meaning that it is considered as a failure.

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2. This coursework is about finding the roots of equations by numerical methods.

Now use decimal search and Excel within the interval [2.560, 2.570] x f(x) 2.561 -0.00356 2.562 0.00288 2.563 0.00933 There is a root in the range (2.561, 2.562) Now use decimal search and Excel within the interval [2.5610, 2.5620] 2.561 -0.00356 2.5611 -0.00291 2.5612 -0.00227 2.5613 -0.00163 2.5614 -0.00098 2.5615 -0.00034 2.5616 0.000304 There is a root in the range (2.5615, 2.5616) Now use decimal search and Excel within the interval [2.5615, 2.5616] x f(x) 2.56150 -0.00034 2.56151 -0.00028 2.56152 -0.00021 2.56153 -0.00015 2.56154 -8.2E-05 2.56155 -1.8E-05 2.56156 4.63E-05 There is a root in the range (2.56155, 2.56156)

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3. In my coursework I will be using three equations to investigate their solutions using three numerical methods which are: change of sign using Newton-Raphson by finding the fixed point iteration fixed point iteration after rearranging the equa

The next value I'll try is the midpoint of the new interval, -0.75 f(-0.75) = 0.34375 f(-0.875) = -0.66797 So I can see that the root lie between -0.75 and -0.875 So I now try and find the root by finding f(x) of the mid point between [-0.75, -0.875] f(-0.8125) = -0.12354 f(-0.78125) = 0.119568 to do this method I have used a calculator f(-0.79688) = 0.000365 and recorded my values in a table on excel Decimal Search A follow on from the bisection I can see that the root is between [-0.8125, -0.79688] so for this method I will try the different decimals until I become 7 d.p from the root.

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4. C3 Numerical Solutions to Equations

For example if the equation has a repeated root as shown below for the equation x�-0.96x�-5.6471x+7.48225=0. As the root between 1 and 2 is repeated, the line of the graph never crosses the x axis so the function does not ever take a negative value between 1 and 2. As a result of this no change of sign will be found using the method above and so the root will not be found. Newton-Raphson Method The Newton-Raphson method takes a value of x close to the root which is to be found and then uses and iterative formula to generate a value which is closer to the root.

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5. Best shape for gutter and further alegbra - using Excel to solve some mathematical problems.

where h = (w - a) cos (?) (height) a = a (first parallel side) b = a + 2(w - a) sin (?) (second parallel side) To calculate angle ?, I have taken the vertical perpendicular to the base as the starting point at ? = 0, Firstly, due to the 3 variables I fixed a and w-a to be equal length and varied ? by 1 degree. A snapshot of the values obtained using excel is below. ?� with horizontal a b Area 28 3.333 3.333 14.41629 29 3.333 3.333 14.42938 30 3.333 3.333 14.43376 31 3.333 3.333 14.42935 32 3.333 3.333 14.41606 Highlighted in red using conditional formatting is the greatest area using these values and varying ?

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6. OCR MEI C3 Coursework - Numerical Methods

This is because they are so close together that there is no sign change between f(0) and f(0.1). The graph on the next page shows the graph of y=f(x), which shows that the roots cannot be found using a decimal search. Newton-Raphson Finding a root The graph of y=f(x): Roots of f(x)=0 exist in the intervals [-2,-1] and [0,1]. f(x)=(x2+0.9x-2.52)-1+1 f'(x)=-(2x+0.9)(x2+0.9x-2.52)-2 We shall find a root by taking x1=1 xn f(xn) f'(xn) x1 1 -0.61290 -7.54422 x2 0.91876 -0.17786 -3.79788 x3 0.87193 -0.02563 -2.78114 x4 0.86271 -0.00071 -2.62915 x5 0.86244 0.00000 -2.62488 x6 0.86244 0.00000 -2.62488 f(0.862435)=0.000014 f(0.862445)=-0.000012 The change of sign indicates that the root is 0.86244�0.000005.

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7. Functions. Mappings transform one set of numbers into another set of numbers. We could display a mapping in three ways

Inverse Functions If a function is one-to-one then an inverse function exists, called f-1(x) For Simple Functions eg f(x) = 4x + 1, x?R x Alternatively, swap y and x throughout and rearrange to make y the subject. Many-to-one functions can be made into one-to-one functions by restricting the domain. f-1(x) can then be found Range of function = Domain of Inverse Function Domain of Function = Range of Inverse Function Sketching the graph can be helpful to determine the range and the domain. y = f(x) is a reflection of y = f-1(x) in the line y = x An asymptote is formed where there is a "forbidden" number.

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8. Numerical Methods coursework

The approximate methods of definite integrals may be determined by numerical integration using: 1. The Trapezium Rule: The Trapezium Rule divides the area underneath the curve into trapeziums. We can then use the formula (where a, b are the bases and h is the height of the trapezium) to estimate the area. Dependent on the amount of trapezia (n) used the general formula is: 2. The Midpoint Rule: The Midpoint Rule divides the area underneath the curve into rectangles. We can then use the formula (where a, b are the 2 different sides of the rectangle)

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9. C3 Mei - Numerical Methods to solve equations

0.7] [0.66, 0.67] [0.662, 0.663] Root intervals [0.6621, 0.6622] I know that the root is 0.662 to 3 decimal places In order to test the boundaries, I put them back into the equation: (0.6621)4 - 5(0.6621)2 +2 = 0.00292 (0.6622)4 - 5(0.6622)2 + 2 = -0.0025 As the values have different signs, it further indicates that the root is close to 0.662 to 3 decimal places. Failure The decimal search will generally fail when two roots are close together or when there is a double root present.

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10. numerical solutions-Comparison of the three methods and Newton Raphson

I will now illustrate one of these roots graphically, showing closely the changes in the tangent to the curve. I will use the root in interval [2, 3] Error bounds for interval [2, 3] 2.3301 is the root in this interval to 5 significant figures. Therefore the error is 2.3301 � 0.00005 I will now perform the change of sign test to confirm it is within these limits. Lower limit is 2.33005 then f (2.33005) = -0.000098648 Upper limit is 2.33015 then f (2.33015) = 0.0010302 There is a change of sign which confirms root in interval is 2.3301 � 0.00005 When does this method fail?

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11. Finding the root of an equation

X Y -0.3 0.179 -0.29 0.133233 -0.28 0.087744 -0.27 0.042551 -0.26 -0.00233 -0.25 -0.04688 -0.24 -0.09107 -0.23 -0.1349 -0.22 -0.17834 -0.21 -0.22138 -0.2 -0.264 There is a change of sign and therefore a root in the region between -0.27 and -0.26. The next part of this investigation is finding where, between -0.27 and -0.26, a change of sign occurs. X Y -0.27 0.042551 -0.269 0.038049 -0.268 0.03355 -0.267 0.029054 -0.266 0.024561 -0.265 0.020071 -0.264 0.015585 -0.263 0.011102 -0.262 0.006622 -0.261 0.002145 -0.26 -0.00233 There is a change of sign and therefore a root in the region between -0.261 and -0.260.

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12. decimal search

curve crosses the x-axis changing its sign from positive ( + ) to negative ( - ). An initial interval of where a root lies can be obtained from a sketch. By taking the values of the initial interval we can increase the value of x by increments of 0.1 within the initial interval. Then if each of the increments of x is substituted into function of x a value can be obtained, and where there is change in sign from the values of f(x); it will state a closer interval (second interval)

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13. Investigation into combined transoformations of 6 trigonometric functions

I will also be studying the symmetry, maximum/minimum points and asymptotes of each graph. The first transformation combination I will be looking at is ab using the trigonometric function of sine. Sine is shown by a blue line in Figure 1.0 A sine graph has no asymptotes but has a rotational symmetry or 180o about the origin(0,0), therefore the only remaining thing to look at is the maximum and minimum values. y=asin(bx) I am going to change a and b to the following values Graph 1 Graph 2 Graph 3 Graph 4 Graph 5 a 1 2 2 -2 -2 b 1 2 -2 2 -2 Predictions Graph 1 - Has a maximum of 1 by point (0.5?,1)

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14. Numerical Methods Coursework

I intend to use the trapezium and Mid-point rule as these are on of the basis in which the approximation can be found. Note: X is in radians. 1. Mid - point rule: The mid point rule was adopted, because it is used to approximate the area underneath the graph using rectangles. Below is the formula which is involved in the calculation. 2. Trapezium rule; This is also similar to the mid - point rule. It was adopted, because it would help me to approximate the region under the graph using strips of trapezium and calculating their area.

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15. Sequence & Series

is a. b. 5. Find the sums of the following series a. 5+9+13+...+101 b. -17-12-7-...+33 c. 1+ 1 1/4 + 1 1/2 + ... + 9 3/4 The sum of an A.P. with n terms, where (this is a rearrangement of the formula for the last term) a. It can be seen that a=5, l=101 and d=4, therefore and so. b. Here a=-17, l=33 and d=5. So and . c. a=1, l=9 1/4 and d= 1/4 and so . The sum is then given by . 6. Evaluate a. b. a. b. 7. Find the sum of the A.P.

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16. Investigation of the Phi Function

Part 1 a) i The value of ?(3) is 2. This is because 3 is a prime number, so its prime factor is naturally 3 and it cannot share a factor with any of the numbers smaller than it. In this case, 1 and 2 are the only numbers smaller than 3 and they are both co-prime with 3, so the value of ?(3) is 2, because there are two numbers (1 and 2) that are co-prime with 3. ii The value of ?(8) is 4, because the prime factor of 8 is 2 and1, 3, 5 and 7 are the only four positive integers less than 8 which do not have 2 as a factor.

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17. Numerical Solutions of Equations

I tabulated this one more time. x y -1.9238 0.002248669 -1.92379 0.002003871 -1.92378 0.001759078 -1.92377 0.001514289 -1.92376 0.001269505 -1.92375 0.001024725 -1.92374 0.00077995 -1.92373 0.000535179 -1.92372 0.000290412 -1.92371 4.56502E-05 -1.9237 -0.000199108 The other root of the equation can be found using this method. It is, to 5 significant figures, 1.1144. Failure of the Change of Sign Method The Change of Sign method only works for certain equations, e.g. where there is a change of sign or where a change of sign is shown in the first table of values.

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18. newton raphson

This process can now be continued using intervals of width 0.01 then by using intervals of width 0.001 etc. Below I will show those intervals and I will bold and highlight those where change of sign occurs in. x f(x) x f(x) x f(x) x f(x) 0.6 -0,48224 0,68 -0,00501 0,68 -0,00501 0,680700 -0,000447 0.61 -0,42714 0,681 0,00151 0,6801 -0,00436 0,680710 -0,000382 0.62 -0,37079 0,682 0,00804 0,6802 -0,0037 0,680720 -0,000316 0.63 -0,31316 0,683 0,014585 0,6803 -0,00305 0,680730 -0,000251 0.64 -0,25423 0,684 0,021144 0,6804 -0,0024 0,680740 -0,000186 0.65 -0,19397 0,685 0,027718 0,6805 -0,00175 0,680750 -0,000121 0.66 -0,13237 0,686 0,034306 0,6806 -0,0011

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19. linear correlation

Reciprocal graphs equation would be f(x) a/x In this part I will have to divide x by a, a would be the same number and x will change. Out of all these formula I am looking at graph to see what is difference if I change a number or symbol e.g. plus I change it to a minus etc. - What am I going to do - Aim: investigate the functions in the graph also talk about shapes and the positions that has coefficients in the function.

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20. Henna Night

At a Henna function it is traditional for the bride to wear the colour yellow or golden. The drole of the drum was getting louder and louder, with the beat gradually getting faster. There were two lines of girls waiting for Yasmin to pass by. They were holding little baskets with party poppers and flowers inside. Included in this line Nadia and her cousins. People looked left and right, up and down, to see where the beat of the drum was coming from. The drummer was at the front, slowly walking and behind was Yasmin. A yellow shimmering sheet with golden threads at each corner of the sheet was carried over Yasmin's head by her two brothers, Wasim and Aseem.

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21. GCSE Math Coursework: Triminoes

3. I will continue to do this until I feel I have used enough numbers. 4. On each set of data I will record the largest number used, the sum of all the cards added together, and the number of cards used. 5. I will then make a results table. 6. I will collect the data from each group and place it together then I will find out whether the equation is a linear, quadratic, cubic or quartic. 7. By replacing the correct numbers with the correct letters I will find out the formula. 8. I will then write down the formula for each set of sequence.

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22. Borders Coursework

These values will then be substituted in the equation with 'n' to check that the formulas are working. These are the 'values of n'. All formulas that I will be calculating will be tested to make sure that they are correct. Part 1: 2D Shapes: 1st term: 2nd term: 3rd term: 4th term: 5th term: 6th term: 7th term: 8th term: 9th term: Part 1: 2D Shapes: Table to show the amount of squares in total and the amount of extra squares used for the 2D drawing: Value of 'n' 1 2 3 4 5 6 7 8 9 Extras

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23. Sequences and series investigation

2(3 squares) 1(5 squares) The Patterns I Have Noticied in Carrying Out the Previous Method I have now carried out ny first investigation into the pattern and have seen a number of different patterns. Firstly I can see that the number of squares in each pattern is an odd number. Secondly I can see that the number of squares in the pattern can be found out by taking the odd numbers from 1 onwards and adding them up (according to the sequence).

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24. Investigate the relationships between the lengths of the 3 sides of the right angled triangles and the perimeters and areas of these triangles.

4 , 12 , 24 , 40 8 , 12 , 16 4 , 4 5 , 13 , 25 , 41 8 , 12 , 16 4 , 4 Length of shortest side: Term no 1 2 3 4 5 Sequence 3 5 7 9 11 Sequence 2n 1 4 6 8 10 Sequence 1 1 1 1 1 2n + 1 Length of the middle side: F (n) = an� + bn +c F= (1) = a x 1� + b x 1 + c = a + b + c = 4 - eqn1 F= (2)

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25. Numerical Methods used to solve those equations which cannot be solved analytically.

Therefore the root of the equation is 1.685 � 0.05 Third Interval x f(x) 1.68 -0.01101 1.681 -0.009628 1.682 -0.008243 1.683 -0.006854 1.684 -0.005459 1.685 -0.004059 1.686 -0.002654 1.687 -0.001243 1.688 0.0001725 1.689 0.001593 1.69 0.003019 Therefore, we can say that the one root of the equation; y=0.6938x3 - 0.9157x2 - 1.421x + 1.671 is 1.6885 � 0.0005 However, decimal search cannot be used in all circumstances. This can be demonstrated using the equation: y=0.6918x3 - 0.2159x2 - 3.019x + 2.77 GRAPH If we zoom in, the roots of the graph can be seen more clearly: TABLE OF VALUES x f(x)

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