(A) −7/4
(B) −3/4
(C) 3/4
(D) 7/4
Correct Answer: 3/4
Given differential equation:
x dy/dx − sin 2y = x³(2 − x³)cos²y
Use identity sin 2y = 2 sin y cos y. Divide the entire equation by cos²y:
x (sec²y) dy/dx − 2 tan y = x³(2 − x³)
Let t = tan y. Then dt/dx = sec²y · dy/dx.
x dt/dx − 2t = x³(2 − x³)
This is a linear differential equation:
dt/dx − (2/x)t = x²(2 − x³)
Integrating factor:
IF = e^{∫(−2/x)dx} = x^{−2}
Multiplying throughout:
d/dx (t/x²) = 2 − x³
Integrate:
t/x² = 2x − x⁴/4 + C
So,
t = 2x³ − x⁶/4 + Cx²
Using condition y(2) = 0 ⇒ tan y(2) = 0:
0 = 2(8) − 64/4 + 4C 0 = 16 − 16 + 4C C = 0
Thus,
tan y = 2x³ − x⁶/4
At x = 1:
tan(y(1)) = 2 − 1/4 = 3/4
Hence, the required value is 3/4.
Updated for JEE Main 2026: This PYQ is important for JEE Mains, JEE Advanced and other competitive exams. Practice more questions from this chapter.