If `0lt=xlt=2pi,` then `2^cos e c^(2x)` `sqrt(1/2y^2-y+1)lt=sqrt(2)` is satisfied by exactly one value of `y` is satisfied by exactly one value of `z` is satisfied by `x` for which `cosx=0` is satisfied by `x` for which `s inx=0`

Options

A. is satisfied by exactly one value of y
B. is satisfied by exactly two value of x
C. is satisfied by x for which `cos x=0`
D. is satisfied by x for which `sin x=0`

Answer

A, B, C

Explanation

The given inequality can be written as
`2^(cosec^(2) x) sqrt((y-1)^(2)+1) le 2` ...(i)
Since `cosec^(2) x ge 1` for all real x, we have
`2^(cosec^(2) x) ge 2` ...(ii)
Also `(y-1)^(2) + 1 ge 1`
`rArr sqrt((y-1)^(2)+1) ge 1` ...(iii)
From Eqs. (ii) and (iii), we get
`2^(cosec^(2) x) sqrt((y-1)^(2)+1) ge 2` ...(iv)
Therefore, from Eqs. (i) and (iv) equality holds only when `2^(cosec^(2))=2 and sqrt((y-1)^(2)+1)=1`. Thus,
`cosec^(2) x=1 and (y-1)^(2)+1=1`
`rArr sin x = pm 1 and y=1`
`rArr x =pi/2, (3pi)/2 and y=1`
Hence, the solution of the given inequality is `x=pi/2, (3pi)/2` and `y=1`.