Question

Question 1

Solve the following equation

a) 2e

2x−1 + 5e

x

2

= 0

b) 2

4x + 22x−1 > 2

[8,6]

Question 2

Let f : (−∞, 2] → R be given by

f(x) = √

2 − x

a) Show that f is injective.

b) Determine im(f).

c) Find a left inverse g : R → (−∞, 2] of f.

[7,6,5]

Question 3

Let f : Z → Z be given by

f(z) = (

2z − 5, if z ≥ 0;

z + 5, if z < 0.

a) Is f an injective function?

b) Let u ∈ Z, u ≤ 5. show that u ∈ im(f).

c) Let v ∈ Z, v > 5, Show that v ∈ im(f) if and only if v + 5 is even .

[5,6,6]

Answer 1

a) $2\cdot e^{2x-1}+5\cdot e^x-2=0$

$e^x=t$ - new unknown

$\frac{2}{e}\cdot t^2+5\cdot t-2=0$

$D=25+\frac{16}{e}$

$t_1=\frac{-5-\sqrt{25+\frac{16}{e}}}{\frac{4}{e}}$

$t_2=\frac{-5+\sqrt{25+\frac{16}{e}}}{\frac{4}{e}}$

1)$e^x=t_1$ - no solutions, because $t_1<0$

2) $e^x=t_2.\space x=ln\left( \frac{-5+\sqrt{25+\frac{16}{e}}}{\frac{4}{e}} \right)$

Answer: $x=ln\left( \frac{-5+\sqrt{25+\frac{16}{e}}}{\frac{4}{e}} \right)$

b)

$2^{4x}+2^{2x-1}>2\\ 2^{2x}=t;\\ t^2+\frac{1}{2}t-2>0;$

$t_1=\frac{-1-\sqrt{33}}{4};$

$t_2=\frac{-1+\sqrt{33}}{4};$

$(t-t_1)(t-t_2)>0$

$(t<t_1)\vee(t>t_2)\\$

$t<t_1$ -impossible, because $t_1<0$

$t>t_2;\\ 2^{2x}>\frac{-1+\sqrt{33}}{4};\\ 2x>log_2\left( \frac{-1+\sqrt{33}}{4} \right)\\ x>\frac{1}{2}\cdot log_2\left( \frac{-1+\sqrt{33}}{4} \right)\\ answer:\space x\in \left( \frac{1}{2}\cdot log_2\left( \frac{-1+\sqrt{33}}{4} \right).\infty \right)$

Question 2

$f(x)=\sqrt{2-x},\space dom(f)=(-\infty,2]$

f(x) is monotone increasing because $f_1(x)=2-x$ is monotone decreasing and

$f_2(y)=\sqrt{y}$ is monotone increasing , therefore $f(x)=f_2\left(f_1(x)\right)$ is monotone decreasing as composition of monotone functions.

Monotine i funcrion f is bijection from dom(f) on im(f).

$f((-\infty,2])=[f(2), lim_{x\rarr -\infty}f(x)=[0.\infty)$

so im(f)=$[0.\infty)$

Inverse function $g:R\rarr[-\infty,2)$

has dom(g)=$[0,\infty)$ and is determined by equation

f(y)=x, $or\space\sqrt{2-y}=x$

therefore

$y=2-x^2$

We have $g(x)=2-x^2$

Question 3

$f(z) = \begin{cases} 2z-5 &{z\ge 0} \\ z+5 &{z<0} \end{cases}$

a) Function f(z) is not injective because

f(0)=f(-10)=-5 and this is counterexample to injectivity

b) Let $u\in Z.u\le5$

For proving that $u\in im(f)$ we must find solution for equation

f(z)=u

Let we search solution in region $(-\infty,0)$

In this case the last equation has form z+5=u;

From it we have z=u-5 and this solution correct if u<5

Now let us consider the case u=5, we have f(5)=$2\cdot 5-5=5$ and eqution

f(z)=u has solution anywhere if $u\le5$ .

c) Let v ∈ Z, v > 5, Show that v ∈ im(f) if and only if v + 5 is even .

We consider two possibilities:

1. z+5=v,z<0

But in this case z+5<5<v and this case impossible

2.

$2z-5=v,\\ z=\frac{v+5}{2}$

if v+5 is odd then this solution is not valid because $z\notin Z$

Contrary if v+5 is even, i.e. v+5=2k,$k\in Z,k>5$

we have z=k- is correct solution of equation f(z)=v, because z=k>5>0