Mathematics: Post your doubts here!

[Mushraf altaf]

ill appreciate help with this one,

 

[A*****]

PURE MATHS 1 FEB/MARCH ANSWERS (MINE)
1) P=6
2) f(x)=2/9x^3-x^2+2
3)118 cm^2
4)dy/dx=-2(2x-1)^-2 +2
d^2y/dx^2=8(2x-1)^-3
5) x=8 in minimum and x=1 is the maximum
6) a) q= 2 or -11/3 b) 130.7
7) b) p=8 and n=42
8) 54.8 and 125.3 b)sq root 3 +tan (2x)
9) a) k=2 b) f inverse is 2- square root (x-3) and domain is f(x) >4 c) gf(x) range is < or equal to 1/3 (I confused the most here tho)
10) 117pi/4 b) (0,78/11)
11)a)8 square root 2, b) T(4,8)
Please share your answers if you solved,

I think you've confused some numbering.....
I've got some different answers.
For 4 ii) x= 0 max and x=1 min
One of the angles in 7 a) is 54.7° not 54.8° according to me...a little difference though but the answer mark will be lost
For 7 b) my a is same but the value of b is -4
For 8 iv) my range is 0<gf(x)<1
And plz tell me the ans of the very last part i.e. 10 iii) mine is (4.5, 7.5)

 

[Shemyaa]

I think you've confused some numbering.....
I've got some different answers.
For 4 ii) x= 0 max and x=1 min
One of the angles in 7 a) is 54.7° not 54.8° according to me...a little difference though but the answer mark will be lost
For 7 b) my a is same but the value of b is -4
For 8 iv) my range is 0<gf(x)<1
And plz tell me the ans of the very last part i.e. 10 iii) mine is (4.5, 7.5)

4) I wrote an 8 instead of 0 by mistake so I took the wrong values, yes your answer is correct.
7) I think I got it 54.8 roundings it to 3sf but I really don't remember tbh,
and I got b=2 If you actually try -4 you'll get y=0, but I think you took 2/3pi which is wrong because t (in my pic) value has to be on the negative side so you should take -pie/3 (you're not shifting it towards left or right ), maybe they'll accept -4 but its not in the domain range they've given us.
8) for iv I confused somewhere I understood the range of f(x) has to be greater than 1 but I can't figure how to progress so please help
and 10)iii yes its (9/2, 15/2) sorry I messed up the order

 

[A*****]

4) I wrote an 8 instead of 0 by mistake so I took the wrong values, yes your answer is correct.
7) I think I got it 54.8 roundings it to 3sf but I really don't remember tbh,
and I got b=2 View attachment 64678 If you actually try -4 you won't get y=0, I think you took 2/3pi which is wrong because your x value has to be on the negative side so you should take -pie/3 (you're not shifting it towards left or right )
8) for iv I confused somewhere I understood the range of f(x) has to be greater than 1 but I can't figure how to progress so please help
and 10)iii yes its (9/2, 15/2) sorry I messed up the order

Oh yes I've looked into it your value for b is correct
For 8 iv) first see the domain of the inner function f(x) which is x<2 and then its range for this domain which is f(x)>3
Now the range of f will be domain of g and according to this domain the range of g(x) will be the range of gf(x). From here the answer 0<gfx<1 comes
But I was a bit confused as what to use as the domain of f(x). x<2 or x<1

 

[Shemyaa]

Oh yes I've looked into it your value for b is correct
For 8 iv) first see the domain of the inner function f(x) which is x<2 and then its range for this domain which is f(x)>3
Now the range of f will be domain of g and according to this domain the range of g(x) will be the range of gf(x). From here the answer 0<gfx<1 comes
But I was a bit confused as what to use as the domain of f(x). x<2 or x<1

Yes, that's the thing, I think your answer is correct tho,
I'll ask my mom's friend to help out so we could be sure, and I'll inform you

 

[A*****]

Yes, that's the thing, I think your answer is correct tho,
I'll ask my mom's friend to help out so we could be sure, and I'll inform you

Alright thanks!!

 

[Hamza Umair]

anyone who can give me recent topical past papers for p1 and s1

 

[A*****]

anyone who can give me recent topical past papers for p1 and s1

You can find them at gceguide.com

 

[Shemyaa]

You can find them at gceguide.com

Heey I still haven't got a reply from my moms' friend but I was thinking that I think the question is pretty simple like they said domain has to be greater than 1, therefore it can't be 0, and actually we were confused with what domain to take but since they said x>1 and we know f(x) range will be the domain of g(x) so if we place f(x)>1 you won't get a value for it except when f(x)>3, therefore, its obvious to say that x can only be 3 and up to infinity, and if you try values greater than 3 you'll find it tends towards 0, so I think its right to say 0<gf(x)<1,
I gave it a thought its actually a reciprocal standard formuala (a/x-h)+k , so if we take 2/x-1 then this means the asymptotes are x=1 and y=0, and for x>1 the range has to be what I mentioned before,
does it make sense?

 

[A*****]

Heey I still haven't got a reply from my moms' friend but I was thinking that I think the question is pretty simple like they said domain has to be greater than 1, therefore it can't be 0, and actually we were confused with what domain to take but since they said x>1 and we know f(x) range will be the domain of g(x) so if we place f(x)>1 you won't get a value for it except when f(x)>3, therefore, its obvious to say that x can only be 3 and up to infinity, and if you try values greater than 3 you'll find it tends towards 0, so I think its right to say 0<gf(x)≤1,
I gave it a thought its actually a reciprocal standard formuala (a/x-h)+k , so if we take 2/x-1 then this means the asymptotes are x=1 and y=0, and for x>1 the range has to be what I mentioned before,
does it make sense?

It's less than 1 not greater...if we use x<1 then the range would be 0<gfx<2/3

 

[Shemyaa]

It's less than 1 not greater...if we use x<1 then the range would be 0<gfx<2/3

 

[A*****]

View attachment 64680

No we have to use the domain of f(x)

 

[asad khan 124746]

what are answers to feb march 2019 paper 1

 

[Shemyaa]

No we have to use the domain of f(x)

Yo they're saying the domain of g(x) has to be x>1, f(x) range is the domain of g(x), in other words, f(x)>1
f(x) can't be >1 but can be greater than 3 based on completed sq.form, therefore the range of gf(x) is 0<gf(x)<1

 

[asad khan 124746]

domain of f(x) is x<1

 

[A*****]

Yo they're saying the domain of g(x) has to be x>1, f(x) range is the domain of g(x), in other words, f(x)>1
f(x) can't be >1 but can be greater than 3 based on completed sq.form, therefore the range of gf(x) is 0<gf(x)<1

Here I've got the answers

 

[Shemyaa]

Here I've got the answers

why are they taking the domain of gf(x) to be lesser than 1?

 

[asad khan 124746]

read the first part

 

[asad khan 124746]

it says x<k

 

[asad khan 124746]

then it says replace k with 1
so x<1