JEE Main 2021 · 16 March · morning · Single correct · +4 / −1
If y = y(x) is the solution of the differential equation,
dy/dx + 2y tan x = sin x,y( π/3 ) = 0, then the maximum value of the function y(x) over R is equal to:
- (a) 1/8
- (b) 8
- (c) -15/4
- (d) 1/2
Answer and solution
Answer (a) 1/8
Solution
Maths sir · 5 steps
Use the integrating factor sec² x, then put t = cos x and maximise a quadratic.
- Integrating factor. The equation is linear with P = 2 tan x. So the integrating factor is e^(∫ 2 tan x dx) = e^(2 ln|sec x|) = sec² x.
d/dx(y sec² x) = sin x sec² x = sec x tan x
- Integrate and fix C. Integrating gives y sec² x = sec x + C. At x = π/3, y = 0 and sec π/3 = 2, so 0 = 2 + C and C = -2.
- Rewrite in cos x. Multiply by cos² x: y = cos x - 2 cos² x. Put t = cos x, where t runs over [-1, 1], so y = t - 2t².
- Maximise. This downward parabola peaks at its vertex t = 1/4, which lies inside [-1, 1]. There y = 1/4 - 2 × 1/16 = 1/4 - 1/8.
- So. The maximum value of y(x) is 1/8, which is option (a).
The idea to keep. After solving a linear DE, rewrite the answer in a single trig function like cos x. Then the max or min is just a quadratic's vertex, as long as the vertex lies in that function's range.
Answer from the NTA final answer key. Solution by PrepWiser.