The minimum value of α for which the
equation 4/(sin x) + 1/(1 - sin x) = α has at least one solution in ( 0,π/2 ) is .......

Solution

Maths sir · 4 steps

Put t = sin x and find the lowest value the left side can reach; that value is your answer.

1. Change the variable. For x in (0, π/2), t = sin x takes every value in (0, 1). So the equation has a solution exactly when α lies in the range of f(t) = 4/t + 1/(1-t) on (0, 1).
2. Find the critical point. Set f'(t) = -4/t² + 1/(1-t)² = 0. This gives (1-t)² = t²/4. Both 1-t and t are positive here, so 1 - t = t/2, which gives t = 2/3.

   f(2/3) = 4/(2/3) + 1/(1/3) = 6 + 3 = 9

3. Confirm the minimum. As t → 0⁺ or t → 1⁻, f(t) → ∞. So the only critical point is the minimum, and the range of f on (0, 1) is [9, ∞).
4. So. The equation has a solution only when α ≥ 9. The minimum value of α is 9.

The idea to keep. When asked for which α the equation g(x) = α has a solution, find the range of g. Its lowest value is the minimum α. Check the open ends of the interval too.

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