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HELP C3 Modulus of cos graph watch

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    Sketch the graph of f(x) = lcosxl, for 0<x<360 and hence solve the inequality lcosxl > 3^1/2 /2 (square root of 3 over 2)

    So I drew the graph correctly and found there to be 4 points of intersection however when I solved for x I got 30 degrees. so X<30 and x>330 - the answers at the back say it should be 0<x<60 or 300<x<360

    I have no idea where they got 60 from and I know I only got 2 solutions out of the 4 but Im not sure how to find the other 2 it is the -lcosl (modulus)...Any help is appreciated
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    (Original post by MrToodles4)
    Sketch the graph of f(x) = lcosxl, for 0<x<360 and hence solve the inequality lcosxl > 3^1/2 /2 (square root of 3 over 2)

    So I drew the graph correctly and found there to be 4 points of intersection however when I solved for x I got 30 degrees. so X<30 and x>330 - the answers at the back say it should be 0<x<60 or 300<x<360

    I have no idea where they got 60 from and I know I only got 2 solutions out of the 4 but Im not sure how to find the other 2 it is the -lcosl (modulus)...Any help is appreciated
    Is this by any chance question 19 on page 13 of the OUP AQA C3 C4 text book? Because if so, I have some memory that they have the wrong answer in the back of the book.
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    (Original post by Pangol)
    Is this by any chance question 19 on page 13 of the OUP AQA C3 C4 text book? Because if so, I have some memory that they have the wrong answer in the back of the book.
    THANK YOU SO MUCH, Im glad to hear it. Do you know what the answer is
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    (Original post by MrToodles4)
    I have no idea where they got 60 from and I know I only got 2 solutions out of the 4 but Im not sure how to find the other 2 it is the -lcosl (modulus)...Any help is appreciated
    But to answer your point about the other two solutions, think of this example. If | x | = 4, then either x = 4 or x = -4. What you have done is equivalent to only solving x = 4 - you need to consider the other case as well. Having said that, if you have drawn a good sketch in the first part, you can probably see the other solutions by using the symmetry of cosine.

    (And I've just checked this question in the book I mention and they do have the answers wrong. I think they were answering a version that has 1/2 in place of root(3)/2.)
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    (Original post by Pangol)
    Is this by any chance question 19 on page 13 of the OUP AQA C3 C4 text book? Because if so, I have some memory that they have the wrong answer in the back of the book.
    I'm assuming you had to check the question and page number before posting this? If not then I'm very impressed / scared.
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    (Original post by Notnek)
    I'm assuming you had to check the question and page number before posting this? If not then I'm very impressed / scared.
    Of course I didn't* have to check!

    * Did
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    (Original post by MrToodles4)
    THANK YOU SO MUCH, Im glad to hear it. Do you know what the answer is
    If you can let us know what you think it is now that we've talked about how to find the other solutions, I'll confirm/deny.
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    (Original post by Pangol)
    But to answer your point about the other two solutions, think of this example. If | x | = 4, then either x = 4 or x = -4. What you have done is equivalent to only solving x = 4 - you need to consider the other case as well. Having said that, if you have drawn a good sketch in the first part, you can probably see the other solutions by using the symmetry of cosine.

    (And I've just checked this question in the book I mention and they do have the answers wrong. I think they were answering a version that has 1/2 in place of root(3)/2.)
    I understand - can you quickly help me with this other question please:

    "On one set of axes sketch the graphs of the functions f(x)= lxl-4 and g(x) = 1/2x. Hence solve the inequality lxl-4 <1/2x so I'v solved for positive intersect which is 8. Then the reflection of the modulus intersects the graph so -(lxl-4)=1/2x and I solved this which gives 8/3.... But the answer was supposed to be negative as it intersects in the negative x axis?? I don't understand why it doesn't come out as negative.
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    (Original post by MrToodles4)
    I understand - can you quickly help me with this other question please:

    "On one set of axes sketch the graphs of the functions f(x)= lxl-4 and g(x) = 1/2x. Hence solve the inequality lxl-4 <1/2x so I'v solved for positive intersect which is 8. Then the reflection of the modulus intersects the graph so -(lxl-4)=1/2x and I solved this which gives 8/3.... But the answer was supposed to be negative as it intersects in the negative x axis?? I don't understand why it doesn't come out as negative.
    The modulus is only around the x, not x - 4. Rearrange to make the equation you have | x | = ... before you solve by removing the modulus.
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    (Original post by Pangol)
    If you can let us know what you think it is now that we've talked about how to find the other solutions, I'll confirm/deny.
    so I got x>30 x<330 and 120<x<210 are these right??
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    (Original post by Pangol)
    Is this by any chance question 19 on page 13 of the OUP AQA C3 C4 text book? Because if so, I have some memory that they have the wrong answer in the back of the book.
    And how in the lord's name did you know that?
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    (Original post by Pangol)
    The modulus is only around the x, not x - 4. Rearrange to make the equation you have | x | = ... before you solve by removing the modulus.
    Ahh I understand it now thank you so much.

    And the last thing bothering me about C3 are composite functions - like I can work out the composite function of any 2 functions without a problem but when it comes to find the domain and range of the new functions I have no idea how to do it. Especially if the original 2 functions did not have domain so x is any real number...
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    (Original post by Texxers)
    And how in the lord's name did you know that?
    Do you really need an honest answer to that? It'd be nice to be able to maintain my mystique...
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    (Original post by Pangol)
    Do you really need an honest answer to that? It'd be nice to be able to maintain my mystique...
    You wrote the book?
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    (Original post by MrToodles4)
    so I got x>30 x<330 and 120<x<210 are these right??
    I agree with 30 and 330 being two of the critical values you need (although I don't agree with the directions of your inequalities). I don't agree with your other two critical values.
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    (Original post by Pangol)
    Do you really need an honest answer to that? It'd be nice to be able to maintain my mystique...
    Unless you've literally done that question a few minutes ago, I'd not be in shock... Reveal your secret!
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    (Original post by Texxers)
    Unless you've literally done that question a few minutes ago, I'd not be in shock... Reveal your secret!
    I'm a teacher. It's not very long since we were doing this question, and I remember one of the students in the class worrying that they had done something wrong with this question, doing it collectively as a class, and then realising that the answers in the book were wrong. Of course, I had to grab the book from my shelf to find the question and page number, but I knew roughly where to find it.
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    (Original post by Pangol)
    If you can let us know what you think it is now that we've talked about how to find the other solutions, I'll confirm/deny.
    Did you do further maths? Do you have a maths degree?

    Extremely curious.
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    (Original post by MajorFader)
    Did you do further maths? Do you have a maths degree?

    Extremely curious.
    I teach further maths, and yes, I have a maths degree!
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    (Original post by Pangol)
    I'm a teacher. It's not very long since we were doing this question, and I remember one of the students in the class worrying that they had done something wrong with this question, doing it collectively as a class, and then realising that the answers in the book were wrong. Of course, I had to grab the book from my shelf to find the question and page number, but I knew roughly where to find it.
    Oh lol... I find it so weird TSR is shared by teacher as well...
 
 
 
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