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## Here's the Solution to this Question

The function $f$ is a one-to-one correspondence, or a bijection, if it is both one-to-one and onto. We also say that such a function is bijective.

a) $f(x)=-3x+4$

Let $x_1,x_2 \in\R$ and let us assume $f(x_1)=f(x_2)$

So,

$f(x_1)=f(x_2)\\ \Rightarrow-3x_1+4=-3x_2+4\\ \Rightarrow x _1=x_2$

Hence, we have $f(x_1)=f(x_2)$ implies $x_1=x_2$ .

So, $f$ is one-one (injective).

Also we know

$-\infin

So, we clearly observe the Co-Domain is the same as the Range, so $f(x)$ is surjective.

Therefore $f(x)=-3x+4$ is a bijection.

b) $f(x)=-3x^2+7$

$f(-1)=-3(-1)^2+7=4$

$f(1)=-3(1)^2+7=4$

We have $f(-1)=4=f(1),$ but $-1\not=4.$

So, $f$ is not one-one (injective).

Therefore $f(x)=-3x^2+7$ is not a bijection.

c) $f(x)=\dfrac{x+1}{x+2}$

The function $f(x)$ is not defined at $x=-2.$

Therefore $f(x)=\dfrac{x+1}{x+2}$ is not a bijection from $\R$ to $\R.$

d) $f(x)=x^5+1$

Let $x_1,x_2 \in\R$ and let us assume $f(x_1)=f(x_2)$

So,

$f(x_1)=f(x_2), x_1\in\R, x_2\in \R\\ \Rightarrow x_1^5+1=x_2^5+1\\ \Rightarrow x _1=x_2$

Hence, we have $f(x_1)=f(x_2)$ implies $x_1=x_2$ .

So, $f$ is one-one (injective).

Also we know

$-\infin

So, we clearly observe the Co-Domain is the same as the Range, so $f(x)$ is surjective.

Therefore $f(x)=x^5+1$ is a bijection.