`f(x)=sin[x]+[s in x],0

Options

A. `{{:(0",",0ltxlt1),(1+sin1",",1lexlt(pi)/(4)):}`
B. `{{:((1)/(sqrt(2))",",0ltxlt(pi)/(4)),(1+(1)/(2)+(1)/(sqrt2)+(sqrt3)/(2)",",(pi)/(4)lexlt(pi)/(4)):}`
C. `{{:(0",",0ltxlt1),(sin1",",1lexlt(pi)/(2)):}`
D. `{{:(0",",0ltxlt(pi)/(2)),(1",",(pi)/(4)ltxlt1),(sin1",",1lexle(pi)/(4)):}`

Answer

C

Explanation

`0ltxlt(pi)/(2)`
`therefore" "[x]={{:(0,"if,0ltxlt1),(1,if,1lexlt(pi)/(2)):}`
`rArr" "sin[x]={{:(sin0=0,if,0ltxlt1),(sin1,if,1lexlt(pi)/(2)):}`
we have `0sinxlt1` when `0ltxlt(pi)/(2).`
`therefore" "[sinx]=0" for "0ltxlt(pi)/(2)`
`therefore" "sin[x]+[sinx]=({:(0,if,0ltxlt1),(sin1,if,1lexlt(pi)/(2)):}`