Let S = (0,2π ) - { π/2,3π/4,3π/2,7π/4 }. Let y = y(x), x ∈ S, be the solution curve of the differential equation dy/dx = 1/(1 + sin 2x), y( π/4 ) = 1/2. If the sum of abscissas of all the points of intersection of the curve y = y(x) with the curve y = √2 sin x is kπ/12, then k is equal to _____________.

Solution

Maths sir · 5 steps

Write 1+sin 2x as (sin x+cos x)², and never cancel sin x when you intersect.

1. Make it a square. Use 1+sin 2x = sin² x+cos² x+2 sin x cos x = (sin x+cos x)². Divide the top and bottom by cos² x to get dy/dx = (sec² x)/(1+tan x)², which is ready to integrate.

   y = -1/(1+tan x) + C

2. Use the condition. At x = π/4, tan x = 1, so 1/2 = -1/2 + C and C = 1. Then y = (tan x)/(1+tan x) = (sin x)/(sin x+cos x).
3. Set the curves equal. Solve (sin x)/(sin x+cos x) = √2 sin x. Either sin x = 0, which gives x = π, or sin x+cos x = 1/√2, which is the same as sin(x+π/4) = 1/2.
4. Solve the second case. Here x+π/4 lies in (π/4, 9π/4), so x+π/4 = 5π/6 or 13π/6. That gives x = 7π/12 or 23π/12. None of the three roots is in the excluded set.
5. So. The sum is π + 7π/12 + 23π/12 = 42π/12, so k = 42.

The common slip: 30. Cancelling sin x from both sides loses the root x = π and gives 30, not 42.

The idea to keep. When the same factor appears on both sides, like sin x here, set it to zero as a separate case before you cancel it. A silent cancel loses roots.

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