`f(x)={(x-1",",-1 le x le 0),(x^(2)",",0le x le 1):} and g(x)=sinx`
Consider the functions `h_(1)(x)=f(|g(x)|) and h_(2)(x)=|f(g(x))|.`
Which of the following is not true about `h_(1)(x)`?

Options

A. Domain of `h_(1)(x) and h_(2)(x)" is " x in [2n pi,(2n+1)pi],n in Z.`
B. Range of `h_(1)(x) and h_(2)(x)" is " [0,1]`
C. Period of `h_(1)(x) and h_(2)(x) " is " pi`
D. None of these

Answer

C

Explanation

`|g(x)|=|sinx|, x in R`
`f(|g(x)|)={(|sinx|-1",",-1 le |sinx| lt 0),((|sinx|)^(2)",", 0le (|sinx|) le 1):}=sin^(2)x,x in R`
`f(g(x))={(sinx-1",",-1 le sinx lt 0),(sin^(2)x",", 0le sinx le 1):}`
`={(sinx-1",",(2n-1) pi lt x lt 2n pi),(sin^(2)x",", 2n pi le x le (2n+1)pi):},n in Z.`
or `|f(g(x))|={(sinx-1",",(2n-1) pi lt x lt 2n pi),(sin^(2)x",", 2n pi le x le (2n+1)pi):},n in Z.`
For `h_(1)(x)-=h_(2)(x)=sin^(2)x, x in [2n pi,(2n+1) pi], n in Z,` and has range [0, 1] for the common domain.
Also, the period is `2pi` (from the graph).