Let `A=[-1,1]dot` Then, discuss whether the following functions from A to itself are one-one onto or bijective: `f(x)=x/2` (ii) `g(x)=|x|` (iii) `h(x)=x^2`

Answer

N/a

Explanation

Given that, ` A=[-1,1]`
(i) `f(x) =(x)/(2)`
Let `f(x_(1))=f(x_(2))`
`implies (x_(1))/(2)=(x_(2))/(2)implies x_(1)=x_(2)`
So, f(x) is one-one.
Now,let ` y=(x)/(2)
`implies `x=2y notin A, AA y in A`
As for `y=1 in A, x=2 notin A`
So, f(x) is not bijective as it not onto.
(ii) `g(x)=|x|`
Let `g(x_(1))=g(x_(2))`
`implies |x_(1)|=|x_(2)|implies x_(1)= +- x_(2) `
So, g(x) is not one-one.
Now, `y=|x|implies x= +- y notin A, AA y in A`
So, g(x) is not onto, also, g(x) is not bijective.
(iii) `h(x)=x|x|`
Let `h(x_(1))=h(x_(2))`
`implies x_(1)|x_(1)|=x_(2)|x_(2)|implies x_(1)= x_(2) `
So, h(x) is one-one.
Now, let `y=x|x|`
`implies y=x^(2) in A, AA x in A`
So, h(x) is onto also, h(x) is a bijective.
(iv) `k(x)=x^(2)`
Let `k(x_(1))=k(x_(2))`
`implies x_(1)^(2)=x_(2)^(2)implies x_(1)= +- x_(2) `
Thus, k(x) is not one-one.
Now, let `y=x^(2)`
`implies x=sqrt(y) notin A, AA y in A`
As for `y= -1, x= sqrt(-1) notin A`
Hence,k(x) is neither one-one nor onto.