Trigonometry
Inverse Trigonometric Functions
MCQ (Single Correct Answer)Subjective
1
WB JEE 2016
MCQ (Single Correct Answer)
+2
-0.5
General solution of $$y{{dy} \over {dx}} + b{y^2} = a\cos x,0 < x < 1$$ is
A
y2 = 2a(2b sin x + cos x) + ce$$-$$2bx
B
$$(4{b^2} + 1){y^2} = 2a(\sin x + 2b\cos x) + C{e^{ - 2bx}}$$
C
$$(4{b^2} + 1){y^2} = 2a(\sin x + 2b\cos x) + C{e^{ 2bx}}$$
D
$${y^2} = 2a(2b\sin x + \cos x) + C{e^{ - 2bx}}$$
WB JEE Subjects