Original post by esha06

what the heck is k

Where is the function not defined (when x>0)?

Original post by mqb2766

Where is the function not defined (when x>0)?

how would I work that out

Original post by esha06

what the heck is k

Looks like a misprint - where does the question come from?

Original post by esha06

how would I work that out

The funciton is a fraction, so when will it have a "problem"

Original post by Muttley79

Looks like a misprint - where does the question come from?

my teacher gave us a Christmas break pack so I'm not sure where its from

Original post by mqb2766

The funciton is a fraction, so when will it have a "problem"

Id say when x is 0 but it has to be greater than 0 idkk honestly

Original post by esha06

Id say when x is 0 but it has to be greater than 0 idkk honestly

x > 0 for the ln(x) functions to be defined. So thats that part sorted.

So if you have

num/denom

for what value of denom is it a problem?

Original post by mqb2766

x > 0 for the ln(x) functions to be defined. So thats that part sorted.

So if you have

num/denom

for what value of denom is it a problem?

So if you have

num/denom

for what value of denom is it a problem?

Isn't it x has to be greater than 0 as well bc the denominator has ln in it aswell

Original post by esha06

Isn't it x has to be greater than 0 as well bc the denominator has ln in it aswell

I dont really understand your argument, but I suspect the answer is no. If youre unsure (and working at home), you could sub a few values for x (smallish say > 0 and <20) into the denominator expression and see what it gives.

Obviously there will be a problem when

denom(x) = ...

So ...

(edited 6 months ago)

Original post by mqb2766

I dont really understand your argument, but I suspect the answer is no. If youre unsure (and working at home), you could sub a few values for x into the denominator expression and see what it gives.

Obviously there will be a problem when

denom(x) = ...

So ...

Obviously there will be a problem when

denom(x) = ...

So ...

i think im confusing myself too

i think that ur tryna tell me how to work out k but I don't know what it is, like it cant be the value of x bc they said that k is not equal to x so where did k come from. do you know what I mean

Original post by esha06

i think im confusing myself too

i think that ur tryna tell me how to work out k but I don't know what it is, like it cant be the value of x bc they said that k is not equal to x so where did k come from. do you know what I mean

i think that ur tryna tell me how to work out k but I don't know what it is, like it cant be the value of x bc they said that k is not equal to x so where did k come from. do you know what I mean

I really dont know what you mean. Your denominator (function, x>0) is

d(x) = ln(x) - 2

which value of d will cause a problem when you divide by it, and what is the corresponding value of x? That is k which you want to exclude from the domain of f (or g (typo)).

(edited 6 months ago)

Original post by mqb2766

I really dont know what you mean. Your denominator (function) is

d(x) = ln(x) - 2

which value of d will cause a probelm when you divide by it, and what is the corresponding value of x? That is the value of k which you want to exclude from the domain of f (or g (typo)).

d(x) = ln(x) - 2

which value of d will cause a probelm when you divide by it, and what is the corresponding value of x? That is the value of k which you want to exclude from the domain of f (or g (typo)).

OH so because its a fraction the denominator can't be 0 so u make ln(x)-2=0 and work out x which would be e^2???

Original post by esha06

OH so because its a fraction the denominator can't be 0 so u make ln(x)-2=0 and work out x which would be e^2???

Yes. Tbh, it should almost be second nature to check for division by zero, extraneous solutions, ....

(edited 6 months ago)

Original post by esha06

what the heck is k

The 2 restrictions for X would be that you can't take the natural log of a negative number, and you can't divide by zero. x>0 is the natural log restraint, the second restraint would be where ln(x)-2 = 0, so to find k you solve ln(x) - 2 = 0 for x.

Original post by mqb2766

Where is the function not defined (when x>0)?

This is a very unusual way of asking this - looks more like a misprint to me.

Original post by Muttley79

This is a very unusual way of asking this - looks more like a misprint to me.

no but i get what they're asking bc normally when they give a function like that they will have x is not equal to 0 next to it anyway

Original post by esha06

no but i get what they're asking bc normally when they give a function like that they will have x is not equal to 0 next to it anyway

Yes I know about that bit - it's the 'k' bit that looks as if the questions has a missing bit or has been miscopied. A teacher should always give a question reference ...

Original post by Muttley79

Yes I know about that bit - it's the 'k' bit that looks as if the questions has a missing bit or has been miscopied. A teacher should always give a question reference ...

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