JEE Main 2024 — Mathematics Calculus

Given:

$$f(x)\mathrm{sin}2x+\mathrm{sin}x-(1+{\mathrm{cos}}^{2}x){f}^{'}(x)=0$$f(x)sin2x+sinx−(1+cos2x)f′(x)=0

Now, let $f(x)=y$f(x)=y we get,

$$\Rightarrow y\mathrm{sin}2x+\mathrm{sin}x-(1+{\mathrm{cos}}^{2}x)\frac{dy}{dx}=0$$⇒ysin2x+sinx−(1+cos2x)dxdy​=0

$$\Rightarrow \frac{y\mathrm{sin}2x}{(1+{\mathrm{cos}}^{2}x)}+\frac{\mathrm{sin}x}{(1+{\mathrm{cos}}^{2}x)}-\frac{dy}{dx}=0$$⇒(1+cos2x)ysin2x​+(1+cos2x)sinx​−dxdy​=0

$$\Rightarrow \frac{dy}{dx}-(\frac{\mathrm{sin}2x}{1+{\mathrm{cos}}^{2}x})y=\frac{\mathrm{sin}x}{1+{\mathrm{cos}}^{2}x}$$⇒dxdy​−(1+cos2xsin2x​)y=1+cos2xsinx​

$$\Rightarrow I.F.={e}^{\int -(\frac{\mathrm{sin}2x}{1+{\mathrm{cos}}^{2}x})dx}$$⇒I.F.=e∫−(1+cos2xsin2x​)dx

$$\Rightarrow I.F.={e}^{\mathrm{log}(1+{\mathrm{cos}}^{2}x)}$$⇒I.F.=elog(1+cos2x)

$$\Rightarrow I.F.=1+{\mathrm{cos}}^{2}x$$⇒I.F.=1+cos2x

So, solution of the equation is given by,

$$y(1+{\mathrm{cos}}^{2}x)=\int \frac{\mathrm{sin}x}{(1+{\mathrm{cos}}^{2}x)}(1+{\mathrm{cos}}^{2}x)dx$$y(1+cos2x)=∫(1+cos2x)sinx​(1+cos2x)dx

$$\Rightarrow y(1+{\mathrm{cos}}^{2}x)=\int \mathrm{sin}xdx$$⇒y(1+cos2x)=∫sinxdx

$$\Rightarrow y(1+{\mathrm{cos}}^{2}x)=-\mathrm{cos}x+C$$⇒y(1+cos2x)=−cosx+C

It is given that, $f(0)=0$f(0)=0.

$$\Rightarrow 0(1+1)=-\mathrm{cos}0+C$$⇒0(1+1)=−cos0+C

$$\Rightarrow C=1$$⇒C=1

$$\Rightarrow y(1+{\mathrm{cos}}^{2}x)=1-\mathrm{cos}x$$⇒y(1+cos2x)=1−cosx

$$\Rightarrow y=\frac{1-\mathrm{cos}x}{1+{\mathrm{cos}}^{2}x}$$⇒y=1+cos2x1−cosx​

$$\Rightarrow y(\frac{\pi }{2})=\frac{1-\mathrm{cos}\frac{\pi }{2}}{1+{\mathrm{cos}}^{2}\frac{\pi }{2}}$$⇒y(2π​)=1+cos22π​1−cos2π​​

$$\Rightarrow y(\frac{\pi }{2})=1$$⇒y(2π​)=1