Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    20
    ReputationRep:
    My question
    Name:  today q .jpg
Views: 162
Size:  272.9 KB

    My answer to part a
     

\displaystyle f(x)=\frac{ax+b } {cx+d}

f(x)=f(-x)

\displaystyle =\frac{ax+b} {cx+d} 

\displaystyle = \frac{-ax+b} {-cx+d}

(ax+b)(-cx+d)=(-ax+b)(cx+d)



2adx = 2bcx

ad = bc
    But for the part about f(x) = k
    My working was:
     

\displaystyle =\frac{ax+b} {cx+d}

\displaystyle  =\frac{d(ax+b)} {b(cx+d)}

\displaystyle  =\frac{adx+bd} {bcx+bd}

\because ad=bc

\displaystyle  =\frac{bcx+bd} {bcx+bd} = 1

1 = constant 

\therfore f(x)= k
    Please can you tell me if my method is right way to show that f(x) = k

    b)
    my answer:

     



\displaystyle  f(x)=\frac {ax+b} {cx+d}

\displaystyle  =\frac {(ax+b)} {(cx+d)}

\displaystyle  = \frac{-ax+b} {-cx+d}

(ax+b)(-cx+d)=(-ax+b)(cx+d)



2bd = 2acx^2

db = acx^2

\displaystyle  \frac{bd} {ac}=x^2

\displaystyle  x = \pm \sqrt{\frac{bd} {ac}}

\displaystyle  \because a= d = 0 or  b = c = 0

if  a = d = 0

x = undefined



\displaystyle  \ therefore f(x) = \frac{k} {x}

    Please show me a better way, I try to use my result to show that f(x) =k/x but i dont think i am right. Also how to i show that f(x) = kx



    c)i did part c

    please help with part a and b

    thank you
    Offline

    15
    ReputationRep:
    (Original post by bigmansouf)
    My question
    Name:  today q .jpg
Views: 162
Size:  272.9 KB

    My answer to part a
     

\displaystyle f(x)=\frac{ax+b } {cx+d}

f(x)=f(-x)

\displaystyle =\frac{ax+b} {cx+d} 

\displaystyle = \frac{-ax+b} {-cx+d}

(ax+b)(-cx+d)=(-ax+b)(cx+d)



2adx = 2bcx

ad = bc
    But for the part about f(x) = k
    My working was:
     

\displaystyle =\frac{ax+b} {cx+d}

\displaystyle  =\frac{d(ax+b)} {b(cx+d)}

\displaystyle  =\frac{adx+bd} {bcx+bd}

\because ad=bc

\displaystyle  =\frac{bcx+bd} {bcx+bd} = 1

1 = constant 

\therfore f(x)= k
    Please can you tell me if my method is right way to show that f(x) = k

    b)
    my answer:

     



\displaystyle  f(x)=\frac {ax+b} {cx+d}

\displaystyle  =\frac {(ax+b)} {(cx+d)}

\displaystyle  = \frac{-ax+b} {-cx+d}

(ax+b)(-cx+d)=(-ax+b)(cx+d)



2bd = 2acx^2

db = acx^2

\displaystyle  \frac{bd} {ac}=x^2

\displaystyle  x = \pm \sqrt{\frac{bd} {ac}}

\displaystyle  \because a= d = 0 or  b = c = 0

if  a = d = 0

x = undefined



\displaystyle  \ therefore f(x) = \frac{k} {x}

    Please show me a better way, I try to use my result to show that f(x) =k/x but i dont think i am right. Also how to i show that f(x) = kx



    c)i did part c

    please help with part a and b

    thank you
    For the first part you have  f(x)=\frac{d(ax+b)}{d(cx+d)}= \frac{adx+bd}{cdx+d^2} . You know that  ad=bc . Using this you should be able to take out a factor on top and bottom containing an expression in x.
    • Thread Starter
    Offline

    20
    ReputationRep:
    (Original post by B_9710)
    For the first part you have  f(x)=\frac{d(ax+b)}{d(cx+d)}= \frac{adx+bd}{cdx+d^2} . You know that  ad=bc . Using this you should be able to take out a factor on top and bottom containing an expression in x.
    for part a (ii) f(x0 = k I did what you recommended whereby I found b/d. the Problem is that b/d is not a constant to my understanding sine b/d can be any number doesn't that make it a variable
    Offline

    15
    ReputationRep:
    (Original post by bigmansouf)
    for part a (ii) f(x0 = k I did what you recommended whereby I found b/d. the Problem is that b/d is not a constant to my understanding sine b/d can be any number doesn't that make it a variable
    Everything is constant except the variable  x .
    • Thread Starter
    Offline

    20
    ReputationRep:
    (Original post by B_9710)
    Everything is constant except the variable  x .
    Thank you
 
 
 
Poll
Do you agree with the PM's proposal to cut tuition fees for some courses?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.