# Edexcel A2 C3 Mathematics 12th June 2015Watch

This discussion is closed.
3 years ago
#501
(Original post by Elliemay97)
Hi guys I'm really struggling with domain and range questions can anyone help me with the first one I have worked out the inverse function. But not sure how to do the domain and range. Thanks Attachment 405145
(a) Domain of f^-1 is the same as the Range of f.... so you need to figure out what the graph of f(x) looks like. Since f(x) is restricted to x<2 you should have an asymptote, work out some y-values to see the shape of the graph.

(b) Range of f^-1 is the same as the Domain of f... which is stated as x<2.
So the Range of f^-1 is f^-1(x) <2 or could write y<2.
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3 years ago
#502
(Original post by Elliemay97)
Hi guys I'm really struggling with domain and range questions can anyone help me with the first one I have worked out the inverse function. But not sure how to do the domain and range. Thanks Attachment 405145

best suggestion! if you are finding the range DRAW THE GRAPH!! but for this make sure you know what basic curves like x^2, or e^x look like so you can have a sketch of what they ask you in the question.

Secondly.. domain of the inverse is ALWAYS THE RANGE OF THE ACTUAL FUNCTION (NON-INVERSE), and as you may have already guessed, the range of the inverse is ALWAYS THE DOMAIN OF THE ACTUAL FUNCTION (NON INVERSE)

also!!

when writing out the domains and ranges of the inverse dont forget to change the x and y's (DOMAINS ALWAYS HAVE THE X IN THEM WHILE RANGE ALWAYS HAS Y IN THEM)
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3 years ago
#503
(Original post by arsenalfc97)
it seems silly but the thing i find hardest in c3 is sketching inverse functions even though I know that it's a reflection in y=x and that when you do this the x and y co-ordinates of any point switch around. i just find it hard to visualise the reflection. Any tips for this?
If you don't have a funny invigilator, take in tracing paper into your exam
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3 years ago
#504
Hello! I'm doing some worksheets ready for the exam, (solomon Worksheet B - Differentiation) and I'm struggling to simplify the equations when they involve 'e' and 'ln' - has any one got any tips/ suggestions on this? For example,

y= 5e^x-3lnx
dy/dx= 5e^x-3/x

But then I'm trying to work out what Y is if X is 0, which apparently is 3... but I'm just trying to make that leap to understand how Y=3
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3 years ago
#505
(Original post by Viggi)
Hello! I'm doing some worksheets ready for the exam, (solomon Worksheet B - Differentiation) and I'm struggling to simplify the equations when they involve 'e' and 'ln' - has any one got any tips/ suggestions on this? For example,

y= 5e^x-3lnx
dy/dx= 5e^x-3/x

But then I'm trying to work out what Y is if X is 0, which apparently is 3... but I'm just trying to make that leap to understand how Y=3
Hi, you can't have x=0 in that because there is an asymptote for the ln part. You can also see the differential doesn't exist there because 3/x
1
3 years ago
#506
(Original post by Viggi)
Hello! I'm doing some worksheets ready for the exam, (solomon Worksheet B - Differentiation) and I'm struggling to simplify the equations when they involve 'e' and 'ln' - has any one got any tips/ suggestions on this? For example,

y= 5e^x-3lnx
dy/dx= 5e^x-3/x

But then I'm trying to work out what Y is if X is 0, which apparently is 3... but I'm just trying to make that leap to understand how Y=3

are you sure abt x being 0 because ln0 is a math error/assymptote
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3 years ago
#507
https://4738aa85dc81f6cb672cd1f6a998...0Questions.pdf

I think I'm doing it wrong then - is there any chance you can give me some help? It's question 2b.
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3 years ago
#508
(Original post by Viggi)
https://4738aa85dc81f6cb672cd1f6a998...0Questions.pdf

I think I'm doing it wrong then - is there any chance you can give me some help? It's question 2b.

yeah first put Y as 0 to get the x coordinate of R. This will be in terms of e. After this find length OQ AND PR Then use 0.5(b-a)x height(OR)
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3 years ago
#509

I haven't checked, but this should be right.
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3 years ago
#510
I know I'm going to sound dumb - so with the equation y= (5e-3)x + 3, the '+3' is the y intercept...? is that the same as being the y coordinate??
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3 years ago
#511
how does 6lnx-4x^1/2=0 become x=20-24ln2??
0
3 years ago
#512
(Original post by Viggi)
I know I'm going to sound dumb - so with the equation y= (5e-3)x + 3, the '+3' is the y intercept...? is that the same as being the y coordinate??
Yep. It's in the form y=mx+c but the m (gradient) is skanky looking
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3 years ago
#513
(Original post by rayquaza17)
You're not allowed a calculator like that in your exam!!
(Original post by samb1234)
Yes you are. You can't have a calculator that does symbolic calculus but numerical is fine

Posted from TSR Mobile
(Original post by Krollo)
Fx 991es?

Posted from TSR Mobile
Sorry just to confirm, is CASIO 991ES allowed or not?
0
3 years ago
#514
(Original post by JaySP)
Sorry just to confirm, is CASIO 991ES allowed or not?
Yep it is
1
3 years ago
#515
(Original post by JaySP)
Sorry just to confirm, is CASIO 991ES allowed or not?
100%, definitely yes
But, examiners are wise to it, and if you use it for integration, you won't get the marks without showing the method. Only use it to check your final answer.
1
3 years ago
#516
(Original post by Viggi)
how does 6lnx-4x^1/2=0 become x=20-24ln2??
Oh that's a really weird one
Wait a minute that's not possible. If you sub back in x= 20 - 24ln2 you would get a non zero result. So it can't be that. I think what it's referring to is the y intercept of the tangent rather than the curve because the curve doesn't have a y intercept (otherwise it would include a constant). So yeah the equation of the tangent to the curve is 2y= x + 24ln2 -20. When y=0, the equation can be rearranged to give x=20 - 24ln2. Hope that helps (This smiley looks quite creepy XD)
1
3 years ago
#517
can someone explain how you would go about doing part b and c, thanks
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3 years ago
#518
(Original post by studentwiz)
can someone explain how you would go about doing part b and c, thanks
Just a hint - use part a, square that, then consider what the max and min values for the sine squared are.
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3 years ago
#519
(Original post by Mr T Pities You)
Just a hint - use part a, square that, then consider what the max and min values for the sine squared are.
how do you find the max of 2root5sin(3thetha-1.107)^2
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3 years ago
#520
(Original post by studentwiz)
can someone explain how you would go about doing part b and c, thanks

my advice- draw the graph or at least know the shapes of the graph and know where they intersect the x axis becaue those are the guiding points
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