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## Here's the Solution to this Question

1.

i)

$f(x)=12+\log(1-x^2)$

$1-x^2>0=>-1

Domain: $(-1, 1)$

$0<1-x^2\le1$

Range: $(-\infin, 12].$

ii)

$f(x)=x^2+\dfrac{1}{x^2}+5$

$x\not=0$

Domain: $(-\infin, 0)\cup (0, \infin)$

$x^2+\dfrac{1}{x^2}\ge2, x\in\R$

Range: $[7, \infin).$

2)

i)

$f(x)=x^5+5$

$f(x)$ is strictly increasing on $\R.$ So, if $n, m\in \R, n\not=m,$ then $n>m$ or $n

Without loss of generality, assume that $n>m.$

Then we have $n^5+5>m^5+5.$ So if $n\not=m,$ then $f(n)\not=f(m).$

Therefore the function $f(x)=x^5+5$ is injective.

$f(x)$ is strictly increasing on $\R.$ For every $\forall y\in \R, \exist x\in \R$ such that

$f(x)=y$

$x^5+5=y=>x=(y-5)^{1/5}$

Therefore the function $f(x)=x^5+5$ is surjective.

$f(x)$ is strictly increasing on $\R.$ For every $\forall y\in \R, \exist! x\in \R$ such that

$f(x)=y$

$x^5+5=y=>x=(y-5)^{1/5}$

Therefore the function $f(x)=x^5+5$ is bijective.

ii)

$𝑓(𝑥) = 𝑒^{4𝑥 }$

$f(x)$ is strictly increasing on $\R.$ So, if $n, m\in \R, n\not=m,$ then $n>m$ or $n

Without loss of generality, assume that $n>m.$

Then we have $e^{4n}>e^{4m}$ . So if $n\not=m,$ then $f(n)\not=f(m).$

Therefore the function $f(x)=e^{4x}$ is injective.

If $y\le0,$ we cannot find $x\in \R$ such that $f(x)=y.$

Therefore the function $f(x)=e^{4x}$ is not surjective and is not bijective.