c1 exam questions

The exact steps you need to get full marks in each question- according to mark schemes.

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\displaystyle \sum_{r = 1}^4 3r = 3\times 1 + 3\times 2 + 3\times 3 + 3\times 4 = 3 + 6 + 9 + 12 = 30 (http://www.thestudentroom.co.uk/latexrender/pictures/16/16472a967b35bf6b6ddb43c6cd9cd717.png)

This is the sum of '3r' for values of 'r' from r = 1 to r = 4.


\displaystyle \sum_{r = 1}^n U_r (http://www.thestudentroom.co.uk/latexrender/pictures/cc/cc059ed24dc1d2624d50901efd9d9239.png)

This is the general case. For the sequence Ur, this means the sum of the terms obtained by substituting in 1, 2, 3, ... n in turn for r in Ur.


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Sum of the series and recurrence relationship.

The recurrence relationship in an arithmetic sequence is:

Un= a + (n - 1)d (http://www.thestudentroom.co.uk/latexrender/pictures/9f/9f5f367bd783ca116c675addaf47965b.png)

The sum of the arithmetic sequence is:

\displaystyle S_n = \frac{1}{2}n[2a + (n - 1)d] (http://www.thestudentroom.co.uk/latexrender/pictures/1d/1d3bc1b5dcf87b4b03aa926d334ab9d5.png)


When they give you the sum of the series and the recurrence relationship formula, and ask you to find the amount on day x:

1. Use simultaneous equations and solve for x.


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When they tell you to find f(x) from a differentiated expression:

1. Integrate the expression

2. Substitute the numbers of the given point to find C.

3. Add the value of C to your integrated equation.

When they say find the normal to C at a given point:

1. Differentiate the expression.

2. Substitute the numers in to find M

3. As the gradient of the normal is -1/M, do this and use the equation

        y-y1=m(x-x1) along with the coordinates given.

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Proving the sum of the series

If they ask you to prove this sum of the series:

 \displaystyle n^{th}\ \mathsf{term} = a + (n-1)d (http://www.thestudentroom.co.uk/latexrender/pictures/12/12b9d95986168b50a629da84e77cf3bf.png)

\displaystyle S = a + (a+d) + (a+2d) + (a+3d) + ... + (L-d) + L  (http://www.thestudentroom.co.uk/latexrender/pictures/cd/cdff735ba47e1c6b87482afdca4ce8d5.png)

\displaystyle S = L + (L-d) + (L-2d) + (L-3d) + ... + (a+d) + a (http://www.thestudentroom.co.uk/latexrender/pictures/b8/b8ba00cb435f5d743e2eed7add24aed4.png)

Add these two:

\displaystyle 2S = n(a + L) (http://www.thestudentroom.co.uk/latexrender/pictures/1d/1dc39b50aaf04ae746560ab0a635859d.png)

Since L is the last term, we know it equals \displaystyle (a + (n-1)d) (http://www.thestudentroom.co.uk/latexrender/pictures/28/2877f69e3b1b009aa8e41cd5f5b6db78.png), where n is the number of terms of the series in the sum.

\displaystyle 2S = n(2a + (n-1)d) (http://www.thestudentroom.co.uk/latexrender/pictures/3e/3e9829b4dc1d7af442c504774d4a1840.png) , divide this by 2 and the formula is proved.

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Translations and transformations of graphs:

Translations and transformations:

f(x-a)= move a to left

f(x+a)= move a to right

f(x)+a= move a upwards.

f(ax)= multiply x coordinates by 1/a

af(x)= multiply y coordinates by a

f(-x)= the graph reflect on the Y AXIS

-f(x)= the graph reflects on the X AXIS

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Coordinate geometry formulae:

m= y2-y1/x2-x1

y2-y1/y-y1= x2-x1/x-x1


the gradient of a perpendicular line is -1/m

the normal to a line is perpendicular to the tangent

Use pythagoras to find the distance between two points.

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Indices& other rules

When you get a fractional indice REMEMBER to first do the square/cubed/etc. root and then multiply it by the power given.

Anything to the power of 0= 1

Integrate the following expression x^n --->>>  x^n+1/n+1

Differentiate the following expression: ax^n --->>> n(ax)^n-1

Completing the square: 



Even if you have extra numbers that aren't part of B just add them to the general equation!

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When using inequality signs, you need to know which part of the graph they're looking for.

1. If 2x+3y+1<0, they're looking for the inside region- the part of the curve under the x-axis.

2. If 2x+3y+1>0, they're looking for the outside region- the part of the curve which is above the x-axis. 

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