Edexcel Core 3 - 21st June 2016 AM

Domain - consider the function and look for x values that will give you 'undefined' values. You can often compare it to normal functions and work from there.
What I mean by that is, for example,

If you had ln(2x+5) then you know that for something simple like lnx, x has to be >0 so you know that in ln(2x+5), 2x+5 > 0 and so x> -5/2.

Also for something like tanf(x) where f(x) is any function of x, f(x) must not be equal to any multiple of pi/2 as it is undefined (easy way to remember - tanx = sinx/cosx, and cosx = 0 when x = pi/2, hence tanx is .../0 = undefined.

And lastly for fractions, you can't divide by 0 - so if you had $\frac{x}{x-3}$ for example, then x can take any value but 3

For the range, it just takes practice. Once you know the domain, it may help to put in numbers close to the domain to see if you can spot the range. If, for example,
$\frac{x}{x-3}$ had the domain $4 \leq x \leq 7$ (for whatever reason, rather than just x = anything but 3) then you could find f(4) = 4 and f(7) = (7/4) and deduce that it is a decreasing function. You could also write the function as $\frac{x-3+3}{x-3} = 1 - \frac{1}{x-3}$ to see it more easily.

Also, test out large values of x. If you had $\frac{3x + 1}{x-2}$ then as x gets really large (eg 10000, but ideally infinity) y becomes $\frac{3\times 10000 - 1}{10000 - 2}$ which is pretty much equal to 3*10000/10000 = 3, and you can tell that it's always going to be greater than 3 as x gets larger as the numerator is slightly greater than 3x, and the numerator is slightly less than 1x so it will always be 3.0000'..12321566 or whatever and you can deduce that y>3.

Domain - consider the function and look for x values that will give you 'undefined' values. You can often compare it to normal functions and work from there.
What I mean by that is, for example,

If you had ln(2x+5) then you know that for something simple like lnx, x has to be >0 so you know that in ln(2x+5), 2x+5 > 0 and so x> -5/2.

Also for something like tanf(x) where f(x) is any function of x, f(x) must not be equal to any multiple of pi/2 as it is undefined (easy way to remember - tanx = sinx/cosx, and cosx = 0 when x = pi/2, hence tanx is .../0 = undefined.

And lastly for fractions, you can't divide by 0 - so if you had $\frac{x}{x-3}$ for example, then x can take any value but 3.

For the range, it just takes practice. Once you know the domain, it may help to put in numbers close to the domain to see if you can spot the range. If, for example,
$\frac{x}{x-3}$ had the domain $4 \leq x \leq 7$ (for whatever reason, rather than just x = anything but 3) then you could find f(4) = 4 and f(7) = (7/4) and deduce that it is a decreasing function. You could also write the function as $\frac{x-3+3}{x-3} = 1 + \frac{3}{x-3}$ to see it more easily.

Also, test out large values of x. If you had $\frac{3x + 1}{x-2}$ then as x gets really large (eg 10000, but ideally infinity) y becomes $\frac{3\times 10000 - 1}{10000 - 2}$ which is pretty much equal to 3*10000/10000 = 3, and you can tell that it's always going to be greater than 3 as x gets larger as the numerator is slightly greater than 3x, and the numerator is slightly less than 1x so it will always be 3.0000'..12321566 or whatever and you can deduce that y>3.

Are there any proofs we need to remember for C3? Tyyy in advance

If cot2x=0 then why are the x values 45 and 135


Posted from TSR Mobile

I'm assuming the range is 0 < x < 360.

cot2x = 1/tan2x

so 1/tan2x = 0

When you flip it over to get tan2x on top, you are doing 0/0 ---> 0/0 = undefined. So you're basically looking at wherever there are asymptotes on the y = tanx graph. This is where x = 90 and 270 for tanx. So, for tan2x, it's 45 and 135 degrees because you divide 90 and 270 by two

Wouod Tanx=0/0 not just be 0,180 and 360 though


Posted from TSR Mobile

No because 0/0 does not equal 0 Try it on your calculator; you get "MATH ERROR".

Thanks that's just me being retarded


Posted from TSR Mobile

If u get below an E (40ums) does that mean u get automatically 0 UMS (U) and it counts as if u've never done the exam?

No, you still get UMS, provided it is at least 1 UMS.

Hi! Would someone help out in 9(b) and (c)?
http://www.madasmaths.com/archive/iygb_practice_papers/c3_practice_papers/c3_d.pdf

b this is when denominator = 0
c just use the Rcos form you derived

b this is when denominator = 0
c just use the Rcos form you derived

Could you explain b a bit further? Is it when the denominator = 0 because anything / 0 = undefined (i.e. asymptote?)

So for the equation y = (3x + 5)/(6x + 2) you would find the asymptote by setting 6x + 2 = 0 and solve for x?