The set of all possible values of θ in the interval
(0, π) for which the points (1, 2) and (sin θ, cos θ) lie
on the same side of the line x + y = 1 is :

Solution

Maths sir · 4 steps

Put both points into x + y - 1: same side means the two values have the same sign.

1. Sign test. Two points lie on the same side of ax + by + c = 0 exactly when ax + by + c has the same sign at both. Here the expression is L = x + y - 1.
2. First point. At (1, 2), L = 1 + 2 - 1 = 2, which is positive. So we need sin θ + cos θ - 1 > 0.

   sin θ + cos θ > 1 ⇔ √2 sin(θ + π/4) > 1 ⇔ sin(θ + π/4) > 1/√2

3. Solve the inequality. For θ ∈ (0, π), the angle θ + π/4 runs over (π/4, 5π/4). On this range the sine is greater than 1/√2 only for π/4 < θ + π/4 < 3π/4.
4. So. Subtracting π/4 gives 0 < θ < π/2, which is option (d). At θ = π/2 the point (1, 0) lies on the line itself, so that end is excluded.

The idea to keep. Same side of a line means the line's expression has the same sign at both points. Then write a sin θ + b cos θ as one sine and solve the inequality on the given range.

Answer from the NTA final answer key. Solution by PrepWiser.