If the functions f, g and h are defined from the set of real numbers R to R such that
`f(x)=x^(2)-1,g(x)=sqrt((x^(2)+1))`,
`h(x)={{:("0,","if",xle0),("x,","if",xge0):}`
Then find the composite function hofog and determine whether the function fog is invertible and h is the identity function.

Answer

A, B, C, D

Explanation

`f(x)=x^(2)-1`
`g(x)=sqrt(x^(2)+1),h(x)={{:("0, if",xle0),("x, if",xge0):}`
`therefore (hofog)(x)=(hof){g(x))`
`=(hof)sqrt(x^(2)+1)`
`=h{fsqrt((x^(2)+1))}`
`=h{sqrt((x^(2)+1)^(2))-1}=h(x^(2)+1-1}`
`=h(x^(2))=x^(2)" "[because x^(2)ge0]`
and `fog(x)=f{g(x))`
`=f(sqrt(x^(2)+1))=(sqrt(x^(2)+1))^(2)-1=x^(2)+1-1=x^(2)`
Let y = (fog) x = `x^(2), AA x in R`
If x = 1, then y = 1
If x = - 1, then y = 1
So, fog is not one-one, so it is not invertible `h(x)={{:("0,",xle0),("x,",xge0):}`
For x = - 1, h(-1) = 0 and for x = - 2, h(-2) = 0
`therefore` h is not identity function.