If `(sinalpha)x^2-2x+bgeq2,` for all real values of `xlt=1a n dalpha in (0,pi/2)uu(pi//2,pi),` then possible real value of `b` is /are `a2` b. `3` c. `4` d. `5`

Options

A. 2
B. 3
C. 4
D. 5

Answer

C, D

Explanation

Abscissa of the vertex is given by
`x=1/(sin alpha) gt 1`
The graph of `f(x)=(sin alpha) x^(@) -2x+b-2` is shown in figure.
Therefore, the minimum of `f(x)=(sin alpha) x^(2)-2x+b-2` must be greater than zero, but minimum is at `x=1`, i.e.,
`sin alpha-2+b-2 ge 0`
`rArr b ge 4-sin alpha, alpha in (0, pi)`
`rArr b ge 4` as `sin alpha gt 0` in `(0, pi)`