`f(x)={(x-1, -1lexlt0),(x^2, 0ltxle1):}` and g(x)=sin x. Then find h(x)=f(|g(x)|)+|f(g(x))|

Options

A. It is a periodic function with period `pi`.
B. The range is [0, 1].
C. The domain is R.
D. None of these

Answer

D

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.`
Clearly, `h_(1)(x)=f(|g(x)|)=sin^(2)x` has period `pi`, range [0, 1], and domain R`.