# True/False : 2 Questions

• Jun 4th 2010, 12:58 PM
Miss
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!
• Jun 4th 2010, 01:26 PM
Miss
Its ok for number1.
Its true
sorry, I made mistake when i draw
its one-to-one

• Jun 4th 2010, 01:37 PM
undefined
Quote:

Originally Posted by Miss
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.
• Jun 4th 2010, 01:40 PM
skeeter
Quote:

Originally Posted by Miss
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.
• Jun 4th 2010, 02:10 PM
Soroban
Hello, Miss!

Quote:

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

. . $\begin{Bmatrix}x &=& 2\sin^2\!t \\
y &=& 3\cos^2\!t\end{Bmatrix} \quad\text{ where } t \in [0,\tfrac{\pi}{2}]$
. . 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).