Question

The rule f(x)= x^2 is a bijection if the domain and the co-domain are given by
1) R,R
(2) R, (0, infinity)
(3) (0,infinity), R.
4)[0,infinity),[0,infinity)​

Answer 1

Step-by-step explanation:

A cylinder of mass m and radius r is rolling on horizontal surface. Work done by force of friction, if centre is displaced by x, is

Answer 2

Answer:

Option (4) is correct.

Step-by-step explanation:

Consider the function as follows:

$f(x)=x^2$

1) For domain = $R$ and co-domain = $R$.

One to one: Let $x_1,x_2$ ∈ domain $R$ such that $f(x_1)=f(x_2)$, then $x_1=x_2$.

The function $f(x)=x^2$ in not one to one.

Since $f(1)=f(-1)=1$ but $1\neq -1$.

⇒ $f(x)=x^2$ is not a bijective function.

Thus, Option (1) is incorrect.

2) For domain = $R$ and co-domain = $(0,\infty)$.

Again, the function $f(x)=x^2$ in not one to one.

Since $f(2)=f(-2)=4$ but $2\neq -2$.

⇒ $f(x)=x^2$ is not a bijective function.

Thus, Option (2) is incorrect.

3) For domain = $(0,\infty)$ and co-domain = $R$.

Onto: For every y ∈ $R$, there exist an element x ∈ $(0, \infty)$ such that f(x) = y.

The function $f(x)=x^2$ in not onto.

If suppose $-1$ ∈ $R$, then there doesn't exist any rational $x$ ∈$(0, \infty)$ such that $x^2=-1$

⇒ $f(x)=x^2$ is not a bijective function.

Thus, Option (3) is incorrect.

4) For domain = $[0,\infty)$ and co-domain = $[0,\infty)$.

The function $f:[0, \infty) \to [0, \infty)$ is both one-one and onto.

This implies the function $f(x)=x^2$ is bijective.

Thus, option (4) is correct.

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