A function y = f(x) satisfies f ( x ) sin 2 x + sin x - ( 1 + cos 2 x ) f ' ( x ) = 0 with condition f(0) = 0. Then f (2) is equal to [2024]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
A function y = f(x) satisfies $f (x) \sin 2 x + \sin x - (1 + \cos^{2} x) f' (x) = 0$ with condition f(0) = 0. Then $f (\frac{π}{2})$ is equal to [2024]

1 2

2 0

3 1

4 – 1

Ans.
(3)

We have, $y \sin 2 x + \sin x - (1 + \cos^{2} x) \frac{d y}{d x} = 0$

$\Rightarrow \frac{d y}{d x} (1 + \cos^{2} x) - y \sin 2 x = \sin x$

$\Rightarrow \frac{d y}{d x} - (\frac{\sin 2 x}{1 + \cos^{2} x}) y = \frac{\sin x}{1 + \cos^{2} x}$

$IF = e^{- \int \frac{\sin 2 x}{1 + \cos^{2} x} d x} = e^{\log (1 + \cos^{2} x)} = 1 + \cos^{2} x$

So, solution is given by

$y \cdot (1 + \cos^{2} x) = \int \frac{\sin x}{(1 + \cos^{2} x)} (1 + \cos^{2} x) d x$

$\Rightarrow y (1 + \cos^{2} x) = \int \sin x d x \Rightarrow y (1 + \cos^{2} x) = - \cos x + c$ ... (i)

Now, y(0) = 0

$\Rightarrow 0 = - \cos (0) + c \Rightarrow c = 1$

From (i), $y (1 + \cos^{2} x) = - \cos x + 1 \Rightarrow y = \frac{1 - \cos x}{1 + \cos^{2} x}$

$∴ y (\frac{π}{2}) = \frac{1 - 0}{1 + 0} = 1$

Want Us to Email you About Special Offers & Updates?