Join TSR now and get all your revision questions answeredSign up now

Functions question please help

    • Thread Starter
    Offline

    3
    ReputationRep:
    Question:

    find the value of k (\neq 1) such that the quadratic function of x

    k(x+2)^2 - (x-1)(x-2)
    is equal to zero for only one value of x
    Find also, (a) th range of vaules of k for which the function possess a minimum value, (b) the range of values of k for which the value of the function never exceeds 12.5

    sketch the graph of the function for  k=\frac{1}{2} and  k = 2\frac{1}{2}


    My attempt:

    let
      f(x) =  k(x+2)^2 - (x-1)(x-2)
      k(x^2+4x+4)-(x^2-3x+2)

    i skip the a couple of steps ( please note it is only to save time)

    i got  x^2(k-1)+ x(4k+3)+4k-2

    since we need to find the value of k when x = 0
    in other for one of the root to be 0, the quadratic function has to be in the form;  x^2 + bx where c = 0

    since c = 0
     c = 4k-2
    thus  4k-2 =0
    k = 0.5


    for (a)
    using   x^2(k-1)+ x(4k+3)+4k-2
    complete the square

     (k-1)\left [ x^2 + \frac{x(4k+3)}{(k-1)} + \frac{4k-2}{k-1}\right ]

     (k-1)\left [ \left ( x +\frac{4k+3}{2(k-1)} \right )^ 2 + \frac{4k-2}{k-1} - \left ( \frac{(4k+3)}{2(k-1)} \right )^2\right ]

     (k-1) \left ( x +\frac{4k+3}{2(k-1)} \right )^ 2 + 4k-2 - \left ( \frac{(4k+3)}{2(k-1)} \right )^2

    completing the square produces;
     (k-1) \left ( x +\frac{4k+3}{2(k-1)} \right )^ 2 + 4k-2 - \left ( \frac{(4k+3)}{2(k-1)} \right )^2

    since f(x) has a minimum function a > 0, thus it has a least value of  \frac{4ac-b^2}{4a} when x =  -\frac{b}{2a}
    therefore if x =  -\frac{ 4k+3}{2(k-1)}
    then use  \left ( x + \frac{4k+3}{2(k-1)} \right ) = 0  to find k
    x =  -\frac{ 4k+3}{2(k-1)}
     -\frac{ 4k+3}{2(k-1)} = 0 k =  - \frac{3}{4}
    sub k = -(3/4) into   4k-2 - \left ( \frac{(4k+3)}{2(k-1)} \right )^2
    since  4k-2 - \left ( \frac{(4k+3)}{2(k-1)} \right )^2  =  \frac{4ac-b^2}{4a}

    therefore sub in  k = - \frac{3}{4} into  4k-2 - \left ( \frac{(4k+3)}{2(k-1)} \right )^2

    .k > 175/4


    part b

    using  (k-1) \left ( x +\frac{4k+3}{2(k-1)} \right )^ 2 + 4k-2 - \left ( \frac{(4k+3)}{2(k-1)} \right )^2

    note the maximum value = 12.5

    therefore   4k-2 -  \frac{(4k+3)^2}{4(k-1)}  =  12.5

     \frac{(4k-2)(4k-4) - (4k+3)^2}{4(k-1)} = 12.5

     (4k-2)(4k-4) - (4k+3)^2 = (4)12.5(k-1)
     (16^2 -24k + 8) - (16k^2 + 24k + 9) = 50k - 50
     -48k -1 = 50k -50
     49 = 98k
     k = \frac{1}{2}



    In the bostock book 'The core Course for A-level' The answer was k = -(1/48) , a) k > 1, b)  k \leq  \frac{1}{2}


    please help me, point out where i went wrong
    Online

    3
    ReputationRep:
    (Original post by bigmansouf)
    Question:

    find the value of k (\neq 1) such that the quadratic function of x

    k(x+2)^2 - (x-1)(x-2)
    is equal to zero for only one value of x
    Find also, (a) th range of vaules of k for which the function possess a minimum value, (b) the range of values of k for which the value of the function never exceeds 12.5

    sketch the graph of the function for  k=\frac{1}{2} and  k = 2\frac{1}{2}

    please help me, point out where i went wrong
     (k-1)\left [ \left ( x +\frac{4k+3}{2(k-1)} \right )^ 2 + \frac{4k-2}{k-1} - \left ( \frac{(4k+3)}{2(k-1)} \right )^2\right ]

    (k-1) \left ( x +\frac{4k+3}{2(k-1)} \right )^ 2 + 4k-2 - \left ( \frac{(4k+3)}{2(k-1)} \right )^2

    Are you sure you expanded the square bracket correctly?
    Offline

    2
    ReputationRep:
    X doesn't equal 0. That's your mistake, I think. (This seems to have a carry on effect for the rest of the question)
    I think the first part is asking for a value for k such that the equation has only one solution.

    EDIT: On doing the maths on paper, this is what the question is asking for. I get k=-1/48, which matches your book
    Online

    3
    ReputationRep:
    (Original post by an_atheist)
    X doesn't equal 0. That's your mistake, I think. (This seems to have a carry on effect for the rest of the question)
    I think the first part is asking for a value for k such that the equation has only one solution.

    EDIT: On doing the maths on paper, this is what the question is asking for. I get k=-1/48, which matches your book
    Good spot.

    At first glance I interpreted the question in the same way as the OP.
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by an_atheist)
    X doesn't equal 0. That's your mistake, I think. (This seems to have a carry on effect for the rest of the question)
    I think the first part is asking for a value for k such that the equation has only one solution.

    EDIT: On doing the maths on paper, this is what the question is asking for. I get k=-1/48, which matches your book
    thanks for correcting me
    • Thread Starter
    Offline

    3
    ReputationRep:
    (Original post by notnek)
     (k-1)\left [ \left ( x +\frac{4k+3}{2(k-1)} \right )^ 2 + \frac{4k-2}{k-1} - \left ( \frac{(4k+3)}{2(k-1)} \right )^2\right ]

    (k-1) \left ( x +\frac{4k+3}{2(k-1)} \right )^ 2 + 4k-2 - \left ( \frac{(4k+3)}{2(k-1)} \right )^2

    Are you sure you expanded the square bracket correctly?
     (k-1)\left [ \left ( x +\frac{4k+3}{2(k-1)} \right )^ 2 + \frac{4k-2}{k-1} - \left ( \frac{(4k+3)}{2(k-1)} \right )^2\right ]

    (k-1) \left ( x +\frac{4k+3}{2(k-1)} \right )^ 2 + 4k-2 - \left ( \frac{(4k+3)^2}{4(k-1)} \right )

    im sorry but i am stuck on this question
    normally to find the range of value of k you use the examples stated in Chandler's book as a guide to find it

    I understand why it k > 1. We know that a quadratic function hasa a minimum value if a >0. therefore since for part a it has a minimum value for the quadratic function k(x+2)^2 - (x-1)(x-2)
    rearranging the function gives

     (k-1)x^2 + (4k+3)x +4k-2
    since a = k-1 and a > 0 thus k must be k> 1 in order to have minimum value

    The problem is now although i understand the question and have the answer how do i write a mathematically structured answer to prove that k > 1
 
 
 
Poll
Which party will you be voting for in the General Election 2017?
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

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