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Mathematics: Post your doubts here!

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(i) : gf(x) --> U take f(x) and then U replace in g(x)
f(x) = 2x + 1
g(x) = (2x - 1)/ (x +3)

gf(x) = (2(2x+1) + 1) / ((2x + 1) +3)
= (4x + 3) / (2x + 4)

Now, gf(x) = x
that is, (4x +3) / (2x +4) = x
4x + 3 = 2x^2 + 4x
2x^2 - 3 = 0
x ^ 2 = (3/2)
x = *root* (3/2)

(ii) : To find f inverse, U let the function f(x) to be equal to y

y = f(x)
y = 2x + 1
Now make x the subject of formula

x = (y -1)/ 2
Therefore f inverse : (x - 1)/2 (U replace the y and make it become x)

For g(x)

y = g(x)
(2x - 1)/ (x +3) = y
2x -1 = xy + 3y
2x - xy = 3y + 1
(2 - y)x = 3y + 1
x = (3y +1)/(2 - y)

Therefore g inverse : (3x +1) / (2 - x)

(iii) To sketch the graphs, first U sketch the graph of f(x).. (I believe this one is quite easy)

Now for the inverse, all that we have to do is to reflect the graph in the line y = x.

The reflected part will show the graph of f inverse relative to f(x).

Note : Any calculations were done mentally so am not sure if the answers are correct but I've explained all my workings... Hope it helps.

Waaaaoooww ! Mentally ? You are good !
Thanks :)
 
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Yo can someone plz help me with this question ( show steps ty):
solve the equation : 2 sin^2(x) -cos x - cos^2(x) 0<x<360
 
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ty a lot but i missed the question...the question is same expect with the addition of a sin x at the cos x place..its like : 2 sin^2(x) - sin x cos x - cos^2(x) 0<x<360
srry again but i got a lil question..in ur answer uve sent me for the 2 sin^2(x) - cos x - cos^2(x)..uve write cos inverse of 1 is equal to 48.2 ( by sending cos inverse on the other side and 131.8 on the other side..it gives u cos inverse of 1 as 48.2...my calculator give me for cos inverse of 1...0 ..ive done something wrong? again thanks for the fast answer
 
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ty a lot but i missed the question...the question is same expect with the addition of a sin x at the cos x place..its like : 2 sin^2(x) - sin x cos x - cos^2(x) 0<x<360
srry again but i got a lil question..in ur answer uve sent me for the 2 sin^2(x) - cos x - cos^2(x)..uve write cos inverse of 1 is equal to 48.2 ( by sending cos inverse on the other side and 131.8 on the other side..it gives u cos inverse of 1 as 48.2...my calculator give me for cos inverse of 1...0 ..ive done something wrong? again thanks for the fast answer

Oh yea srry I took cos inverse of (2/3) instead of cos inverse of 1. Urs is fine. :)
 
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Anyone taking Maths AS in november? please help i wanna what level are u upto... exam in 20 days :C and im late in past papers but il catch up by this week....GoD! studying in vacations is damn hard :-/
 
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ok ty..iam looking forward to the other one ( 2 sin^2(x) - sin x cos x - cos^2(x) 0<x<360 ) ty in advance :)
Well srry for l8 rep., was busy yesterD.

=> 2 sin^2(x) - sin x cos x - cos^2(x) = 0
=> (Divide by cos^2 (x) everywhere) => 2 tan^2 (x) - tan x - 1 = 0
=> (2 tan x + 1) (tan x - 1) = 0
=> tan x = (-1/2) or tan x = 1

tan x = (-1/2) :

x = 180 + [*tan inverse* (1/2) ] = 206.6
or x = 360 + [*tan inverse* (1/2) ] = 333.4

tan x = 1 :

x = *tan inverse* (1) = 45
or x = 180 + [*tan inverse* (1) ] = 225

Hope It helps.
 
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A question from statistics 2 chapter:-sampling in statistics 2

can sum1 plzz show me the steps!!and is Continuity correction required?
minato112
whitecorp
leadingguy
leosco1995
nightrider1993
smzimran

5. The masses of kilogram bags of flour produced in a factory have a normal distribution
with mean 1.005 kg and standard deviation 0.0082 kg. A shelf in a store is loaded with
22 of these bags, assumed to be random sample.
(a) Find the probability that a randomly chosen bags has mass less than 1 kg.
(b) Find the probability that the mean mass of the 22 bags is less than 1 kg
State, giving a reason, which of the above answers would be little changed if the
distribution of masses were not normal.
 
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