The correct option is B 2
Given: sin2θ=x2−4x+5
Now, x2−4x+5 has complex roots as it's discriminant D=16−20=−4<0
Also, coefficient of x2 is positive.
⇒x2−4x+5 is upward opening curve.
Now, it's minimum point is given by the relation:
(−b2a,−D4a)
Equating the values of a,b,c, we get the coordinates of minimum point as: (−(−4)2,−(−4)4)≡(2,1)
Hence, at x=2, the quadratic equation is minimum and have a value 1.
Also, we know −1≤sinθ≤1
Or 0≤sin2θ≤1
Thus, for sin2θ=x2−4x+5
This is possible only when x2−4x+5=1 or at x=2.