Show that the function f:R→{x∈R:−1<x<1} defined by f(x)=x1+|x|, x∈R is one one and onto function.

To prove that a function is one-one (injective) and onto (surjective), we need to show that for every element in the domain there is a unique element in the co-domain, and every element in the co-domain is the image of some element in the domain.

1. Proving one-one (injectivity):

Assume that f(x1) = f(x2) for some x1, x2 in R. Then we have:

x1/(1+|x1|) = x2/(1+|x2|)

Cross multiplying gives:

x1(1+|x2|) = x2(1+|x1|)

This simplifies to:

x1 + x1|x2| = x2 + x2|x1|

Rearranging terms gives:

x1 - x2 = x2|x1| - x1|x2|

If x1 = x2, then the equation holds. If x1 ≠ x2, then the equation does not hold because the right side will be positive and the left side will be negative. Therefore, x1 must equal x2, which means the function is one-one.

2. Proving onto (surjectivity):

We need to show that for every y in the co-domain, there exists an x in the domain such that f(x) = y.

Let y = x/(1+|x|). We need to solve for x:

x = y(1+|x|)

This gives two cases:

If x ≥ 0, then x = y(1+x), which simplifies to x = y + yx. Solving for x gives x = y/(1-y) for y ≠ 1.

If x < 0, then x = y(1-x), which simplifies to x = y - yx. Solving for x gives x = -y/(1+y) for y ≠ -1.

Since y is in the interval (-1,1), there will always be a corresponding x in R for every y. Therefore, the function is onto.

In conclusion, the function f(x) = x/(1+|x|) is both one-one and onto.