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Math Help - True/False : 2 Questions

  1. #1
    Member Miss's Avatar
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    True/False : 2 Questions

    1- The following function is one-to-one

    f(x)= -x if x belongs to [-1,0]
    f(x)=3x+2 if x belongs to [0,1]

    2- The following parametric curve represents a line segment from (0,3) to (2,0) :

    x(t)=2sin^2(t) and y(t)=3cos^2(t) where t \in [0,\frac{\pi}{2}]

    For (1): I think its false, if we draw the function on the interval [-1,1] we will get a graph which is similar to the graph of f(x)=|x| (the v-shape). clearly there are horizontal lines intersect the curve of f in more than one point, so its not 1-1 function, so the statement is false.

    For (2): I do not have any idea!
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  2. #2
    Member Miss's Avatar
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    Its ok for number1.
    Its true
    sorry, I made mistake when i draw
    its one-to-one

    I need number 2 please.
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  3. #3
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    Quote Originally Posted by Miss View Post
    1- The following function is one-to-one

    f(x)= -x if x belongs to [-1,0]
    f(x)=3x+2 if x belongs to [0,1]

    2- The following parametric curve represents a line segment from (0,3) to (2,0) :

    x(t)=2sin^2(t) and y(t)=3cos^2(t) where t \in [0,\frac{\pi}{2}]

    For (1): I think its false, if we draw the function on the interval [-1,1] we will get a graph which is similar to the graph of f(x)=|x| (the v-shape). clearly there are horizontal lines intersect the curve of f in more than one point, so its not 1-1 function, so the statement is false.

    For (2): I do not have any idea!
    I see you already figured out (1), but try to be careful about writing intervals like [a,b] or (a,b) or (a,b] or [a,b). The f(x) you wrote above is not a function because the first part gives f(0) = 0 and the second part gives f(0) = 2.

    For (2), first check start and end points.

    x(0) = 0 and y(0) = 3

    x\left(\frac{\pi}{2}\right)=2 and y\left(\frac{\pi}{2}\right)=0.

    So we can't conclude whether it's true or false yet.

    Now express y in terms of x:

    y = 3\cos^2t=3(1-sin^2t)=3\left(1-\frac{x}{2}\right)=3-\frac{3}{2}x

    This is an equation of a line, so the answer to (2) is True.
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  4. #4
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    Quote Originally Posted by Miss View Post
    1- The following function is one-to-one

    f(x)= -x if x belongs to [-1,0]
    f(x)=3x+2 if x belongs to [0,1]

    2- The following parametric curve represents a line segment from (0,3) to (2,0) :

    x(t)=2sin^2(t) and y(t)=3cos^2(t) where t \in [0,\frac{\pi}{2}]

    For (1): I think its false, if we draw the function on the interval [-1,1] we will get a graph which is similar to the graph of f(x)=|x| (the v-shape). clearly there are horizontal lines intersect the curve of f in more than one point, so its not 1-1 function, so the statement is false.

    For (2): I do not have any idea!
    (1) agree

    (2) x = 2sin^2{t} = 2(1 - \cos^2{t}) = 2 - 2\cos^2{t}

    \cos^2{t} = \frac{2-x}{2}

    y = 3\cos^2{t} = \frac{3(2-x)}{2} = 3 - \frac{3}{2}x

    now calculate and ATQ.



    btw ... not to cramp your style, but the pink type is difficult to read.
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  5. #5
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    Hello, Miss!

    2. The following parametric curve represents a line segment from (0,3) to (2,0)

    . . \begin{Bmatrix}x &=& 2\sin^2\!t \\ <br />
y &=& 3\cos^2\!t\end{Bmatrix} \quad\text{ where } t \in [0,\tfrac{\pi}{2}] . . This is true.

    We have: . \begin{array}{ccc}\dfrac{x}{2} &=& \sin^2\!t \\ \\[-3mm]  \dfrac{y}{3} &=& \cos^2\!t \end{array}


    \text{Add the equations: }\;\frac{x}{2} + \frac{y}{3} \:=\:\underbrace{\sin^2t + \cos^2t}_{\text{This is 1}}

    So we have: . \frac{x}{2} + \frac{y}{3} \:=\:1 \quad\Rightarrow\quad 3x + 2y \:=\:6

    The graph would be a straight line (infinitely long).


    However, t ranges from 0 to \tfrac{\pi}{2}

    Then: . \begin{array}{ccccc}x \:=\:2\sin^2t\text{ ranges from 0 to 2} \\ y \:=\:3\cos^2t\text{ ranges from 3 to 0} \end{array}


    Therefore, the graph is the line segment from (0.3) to (2,0).

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