Question

If equation `x^2+2x+2 + e^alpha - 2 sin beta = 0` has a real solution in `x,` then (where `n in ZZ`)
A. `alpha, beta in R`
B. `alpha in(0,1), beta in(pi//6,pi//2)`
C. `alpha in(0,1),beta in(2n pi+pi//6,2n pi+5pi//6)`
D. `alpha in (-oo,0],beta in (2n pi-pi//6)`

Answer 1

Correct Answer - C
`x^(2)+2x+2+e^(alpha)-2 sin beta =0` has real solution
`therefore D=(2)^(2)-4{2+e^(alpha)-2 sin beta} ge 0`
`therefore 1-2-e^(alpha)+2 sin beta ge 0`
`therefore 2sin beta ge (1)/(2)+(1)/(2)e^(alpha`
`rArr alpha le 0`
`rArr (1)/(2)le sin beta lt 1`
`rArr beta in (2n pi + pi//6, 2n pi + 5pi//6), n in Z`