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f(x)=sin[x]+[sin x],0 < x < pi/2 (where ...

`f(x)=sin[x]+[sin x],0 < x < pi/2` (where [.] represents the greatest integer function), can also be represented as

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)):}`

Text Solution

Verified by Experts

The correct Answer is:
C

`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)):}`
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