Question

If `sinx+cosx=sqrt(y+1/y)` for `x in [0,pi]` , then `x=pi/4` (b) `y=0` `y=1` (d) `x=(3pi)/4`

Answer 1

`sinx+cos=sqrt(y+1/y)`
`sqrt(2/3)(sinx+cosx)=sqrt((sqrty)^2+(1/sqrty)^2)`
`sqrt2(1/sqrt2sinx+1/sqrt2cosx)=sqrt((sqrty-1/sqrty)^2+2sqrty*1/sqrty)`
`sqrt2(sinpi/4sinx+sinpi/4cosx)=sqrt((sqrty-1/sqrty)^2+2`
`sqrt2sin(x+pi/4)=sqrt((sqrty-1/sqrty)^2+2`
LHS
`sqrt2sin(x+pi/4)`
`-1<=sin(x+pi/4)<=1`<br> `-sqrt2<=sqrt2sin(x+pi/4)<=sqrt2`<br> RHS
`sqrt((sqrty-1/sqrty)^2+2`
`0<=(sqrty-1/sqrty)^2ltoo`<br> `2<=(sqrty-1/sqrty)^2+2ltoo`<br> `sqrt2<=sqrt((sqrty-1/sqrty)^2+2)ltoo`<br> `LHS=RHS=sqrt2`
`sqrt2sin(x+pi/4)=sqrt2`
`sin(x+pi/4)=sinpi/2`
`x+pi/4=pi/2`
`x=pi/4`
`0<=x<=pi`<br> `RHS=sqrt((sqrty-1/sqrty)^2+2)=sqrt2`
`(sqrty-1/sqrty)^2+2=2`
`sqrty-1/sqrty=0`
`y=1`
Option A and C is correct.