I am in my first year of my mathematics degree at the University of Exeter and one of my favourite things about maths is how it can appear in unexpected situations. For example, in **cryptography **(the writing and solving of codes of which the strongest use more maths than you may realise) are used in the decipherment of ancient languages, such as the ones present on the Rosetta stone (a rock found in Eqypt where the Greek inscriptions were used to decipher the hieroglyphs). Cryptography also played a major role in the World Wars; a famous example is the enigma machine.

One of my favourite books on this topic which is suitable for whatever your level of maths may be is **Simon Singh's 'The Code Book'**. I read this book in my first year of A-levels and it is the book which developed my interest further in the applications of maths. I also recommend any of Simon Singh's books.

These constant links that maths has with everything around us is one of the things I love so much about it. Maths makes me question things around me and sometimes an answer to one question just leads to so many more, but the more questions there are,** the more beautiful maths becomes**.

My strongest area of maths is **algebra** and I especially enjoy **pure maths **( I studied **FP1 **& **FP2 **at A-level). My weakest areas are statistics and decision maths.

In my A-levels I got **100% in C2, C3 and C4. **The exam board I was taught was **OCR.**

I would be happy to tutor you in **C1, ****C2, C3, C4 **and **FP1****.**

Through my tutorials with you I hope to develop your **understanding, interest and confidence** in maths.

I am in my first year of my mathematics degree at the University of Exeter and one of my favourite things about maths is how it can appear in unexpected situations. For example, in **cryptography **(the writing and solving of codes of which the strongest use more maths than you may realise) are used in the decipherment of ancient languages, such as the ones present on the Rosetta stone (a rock found in Eqypt where the Greek inscriptions were used to decipher the hieroglyphs). Cryptography also played a major role in the World Wars; a famous example is the enigma machine.

One of my favourite books on this topic which is suitable for whatever your level of maths may be is **Simon Singh's 'The Code Book'**. I read this book in my first year of A-levels and it is the book which developed my interest further in the applications of maths. I also recommend any of Simon Singh's books.

These constant links that maths has with everything around us is one of the things I love so much about it. Maths makes me question things around me and sometimes an answer to one question just leads to so many more, but the more questions there are,** the more beautiful maths becomes**.

My strongest area of maths is **algebra** and I especially enjoy **pure maths **( I studied **FP1 **& **FP2 **at A-level). My weakest areas are statistics and decision maths.

In my A-levels I got **100% in C2, C3 and C4. **The exam board I was taught was **OCR.**

I would be happy to tutor you in **C1, ****C2, C3, C4 **and **FP1****.**

Through my tutorials with you I hope to develop your **understanding, interest and confidence** in maths.

No DBS Check

Firstly set 2sinhx+3coshx=5

Now using the exponential definitions of sinhx and coshx rewrite the equation to give:

2(1/2(e^x-e^-x))+3(1/2(e^x+e^-x))=5

Simplify the equation by expanding out the brackets, multipling by 2 to eliminate fractions and collecting like terms together, as so:

e^x-e^-x+3/2e^x+3/2e^-x=5

2e^x-2e^-x+3e^x+3e^-x=10

5e^x+e^-x=10

5e^x+e^-x-10=0

e^-x is equivalent to 1/e^x therefore multiply through by e^x to get a quadratic equation in e^x

5e^2x-10e^x+1=0

Now using the quadratic equation (where a=5, b=-10 and c=1) solve for e^x

I will indicate 'plus or minus' by +/- (not to be confused with plus, divide, minus)

e^x=(-(-10)+/-√(-10)^2-4(5)(1))/2(5)

e^x=(10+/-√80)/10

e^x=1+/-(2√5)/5

To solve for x you must take the natural logarithm of both sides as (ln^e=1) so

x=ln(1+(2√5)/5

or

x=ln(1-(2√5)/5)

Firstly set 2sinhx+3coshx=5

Now using the exponential definitions of sinhx and coshx rewrite the equation to give:

2(1/2(e^x-e^-x))+3(1/2(e^x+e^-x))=5

Simplify the equation by expanding out the brackets, multipling by 2 to eliminate fractions and collecting like terms together, as so:

e^x-e^-x+3/2e^x+3/2e^-x=5

2e^x-2e^-x+3e^x+3e^-x=10

5e^x+e^-x=10

5e^x+e^-x-10=0

e^-x is equivalent to 1/e^x therefore multiply through by e^x to get a quadratic equation in e^x

5e^2x-10e^x+1=0

Now using the quadratic equation (where a=5, b=-10 and c=1) solve for e^x

I will indicate 'plus or minus' by +/- (not to be confused with plus, divide, minus)

e^x=(-(-10)+/-√(-10)^2-4(5)(1))/2(5)

e^x=(10+/-√80)/10

e^x=1+/-(2√5)/5

To solve for x you must take the natural logarithm of both sides as (ln^e=1) so

x=ln(1+(2√5)/5

or

x=ln(1-(2√5)/5)

You must be careful with these sorts of questions as although 1/(5-2x) is equivalent to (5-2x)^-1, when you integrate you would add one to the power and divide by the new power. But if you were to add one to (5-2x)^-1 you would get zero. Therefore, when you are integrating a fraction with a linear expression as the denominator (meaning a denominator where the greatest power of x is 1), it integrates to the natural logarithm (ln) of the denominator, multiplied by the differential of the denominator.

So in this example, 1/(5-2x) would integrate to

[ln(5-2x)/(-2)] (as 5-2x differentiates to -2) for 3≤x≤4

Then you would sub in the limits of x and subtract as usual:

ln(5-2(4))/(-2) - ln(5-2(3))/(-2)

= -1/2ln(5-8) - -1/2ln(5-6)

Remeber that you cannot take the ln of a negative number, so it is best to write it as:

= -1/2ln|-3| - -1/2ln|-1|

= -1/2ln(3) - -1/2ln(1)

ln(1)=0 so our answer is

-1/2ln3

You must be careful with these sorts of questions as although 1/(5-2x) is equivalent to (5-2x)^-1, when you integrate you would add one to the power and divide by the new power. But if you were to add one to (5-2x)^-1 you would get zero. Therefore, when you are integrating a fraction with a linear expression as the denominator (meaning a denominator where the greatest power of x is 1), it integrates to the natural logarithm (ln) of the denominator, multiplied by the differential of the denominator.

So in this example, 1/(5-2x) would integrate to

[ln(5-2x)/(-2)] (as 5-2x differentiates to -2) for 3≤x≤4

Then you would sub in the limits of x and subtract as usual:

ln(5-2(4))/(-2) - ln(5-2(3))/(-2)

= -1/2ln(5-8) - -1/2ln(5-6)

Remeber that you cannot take the ln of a negative number, so it is best to write it as:

= -1/2ln|-3| - -1/2ln|-1|

= -1/2ln(3) - -1/2ln(1)

ln(1)=0 so our answer is

-1/2ln3