Question

a) Which of the following functions are 1-1 and which are onto? Justify your answer.
i) f : R ! R
0
given by f (x) = x
2
where R
0
is the set fx 2 Rjx 0g.
ii) f : R ! R given by f (x) = x
2
+ x + 1. (3)
b) If we consider the expression
1
2 3x
as a function on R, what will be its domain and
range? Will it have an inverse? Justify your answer. (2)
c) Let a =
(
1
2
p
2
;
p
3
2
p
2
;
1
p
2
)
and b =
(
1
p
2
; 0;
1
p
2
)
.
i) Find the direct cosines of a and b.
ii) Find the angle between a and b. (2)
d) Check that the vectors u =
(
3
5
;
4
5
; 0
)
, v =
(
4
5
;
3
5
; 0
)
and w = (0; 0; 1) are orthonormal.
Further, write the vector a = (1; 1; 2) as a linear combination of the vectors.

Answer 1

Answer on Question #45759, Engineering, Other

Task:

a) Which of the following functions are 1-1 and which are onto? Justify your answer.

i) $f: R \to R_0$ given by $f(x) = x^2$ where $R^0$ is the set $fx \ 2Rjx \ 0g$.

ii) $f: R \to R$ given by $f(x) = x^2 + x + 1$.

b) Let $a = (1 + 2p^2; p^3 + 2p^2; 1 + p^2)$

and $b = (1 + p^2; 0; 1 + p^2)$.

i) Find the direct cosines of $a$ and $b$.

ii) Find the angle between $a$ and $b$.

c) Check that the vectors $\overline{u} = (35; 45; 0)\overline{v} = (45; 35; 0)$

and $\overline{w} = (0; 0; 1)$ are orthonormal. Further, write the vector $a = (1; 1; 2)$ as a linear combination of the vectors.

Solution:

a)

The function is injective or 1 to 1 if every element of the function's codomain is the image of at most one element of its domain.

The function $f$ from a set $X$ to a set $Y$ is surjective (or onto), or a surjection, if every element $y$ in $Y$ has a corresponding element $x$ in $X$ such that $f(x) = y$.

i) $f: R \to R_0$ given by $f(x) = x^2$

if $R_0 = R \setminus \{0\}$ then $f$ is not onto because for $y = -2$ we can't find a $x$ such that $f(x) = x^2$ and it is not 1 to 1, because $f(1) = f(-1) = 1$. If $R_0 = [0; +\infty]$ then $f$ is onto, because for every $y$ in $[0; +\infty)$ exists $x = \sqrt{y}$ such that $y = f(x)$ and it is not 1 to 1, because $f(-1) = f(1)$.

ii) $f: R \to R$ given by $f(x) = x^2 + x + 1$.

This function is not onto, because for $y = 0$ we can't find $x$ such that $0 = x^2 + x + 1$. And it is not 1 to 1 because $f(1) = f(-2) = 3$.

b) Let $a = (1 + 2p^2; p^3 + 2p^2; 1 + p^2)$

and $b = (1 + p^2; 0; 1 + p^2)$.

i) Find the direct cosines of $a$ and $b$.

ii) Find the angle between $a$ and $b$.

$$\cos \angle (\overline{a}, \overline{b}) = \frac{\overline{a} \cdot \overline{b}}{|\overline{a}| \cdot |\overline{b}|} = \frac{(1 + p^2)(1 + 2p^2) + 0(p^3 + 2p^2) + (1 + p^2)^2}{\sqrt{(1 + p^2)^2 + (1 + p^2)^2} \sqrt{(1 + 2p^2)^2 + (p^3 + 2p^2)^2 + (1 + p^2)^2}} = \frac{3p^4 + 5p^2 + 2}{(1 + p^2) \sqrt{2} \sqrt{p^6 + 4p^5 + 9p^4 + 6p^2 + 2}} \Rightarrow \angle (\overline{a}, \overline{b}) = \arccos\left(\frac{3p^4 + 5p^2 + 2}{(1 + p^2) \sqrt{2} \sqrt{p^6 + 4p^5 + 9p^4 + 6p^2 + 2}}\right)$$

c) Check that the vectors $\overline{u} = (35;45;0)\overline{\nu} = (45;35;0)$

and $\overline{w} = (0;0;1)$ are orthonormal. Further, write the vector $a = (1; 1; 2)$ as a linear combination of the vectors.

$\overline{u} = (35;45;0)\overline{v} = (45;35;0)$ and $\overline{w} = (0;0;1)$ are orthonormal if $\overline{u}\cdot \overline{v}\cdot \overline{w} = 0$

so $\overline{u} \cdot \overline{v} \cdot \overline{w} = 35 * 45 * 0 + 45 * 35 * 0 + 0 * 0 * 1 = 0$ they are orthonormal.

Further, write the vector $\mathbf{a} = (1; 1; 2)$ as a linear combination of the vectors

$\overline{u} = (35;45;0)\overline{v} = (45;35;0)$ and $\overline{w} = (0;0;1)$

$\overline{a} = x\overline{u} +y\overline{v} +z\overline{w}$

$(1;1;2) = x(35;45;0) + y(45;35;0) + z(0;0;1)$

$(1;1;2) = (35x + 45y;45x + 35y;z)$

$$\left\{ \begin{array}{l} 35x + 45y = 1 \\ 45x + 35y = 1 \\ z = 2 \end{array} \right.$$ $$\left\{ \begin{array}{l} x = \frac{1}{80} \\ y = \frac{1}{80} \\ z = 2 \end{array} \right.$$

So linear combination of the vectors is (1/80;1/80;2)

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