JEE Main 2021 — Mathematics Calculus

We have,

$$A=[\begin{matrix}y & \mathrm{sin}x & 1 \\ 0 & -1 & 1 \\ 2 & 0 & \frac{1}{x}\end{matrix}]$$A=[y02​sinx−10​11x1​​]

$$\Rightarrow |A|=-\frac{y}{x}+2\mathrm{sin}x+2$$⇒∣A∣=−xy​+2sinx+2

Now,

$$\frac{dy}{dx}=|A|$$dxdy​=∣A∣

$$\Rightarrow \frac{dy}{dx}=-\frac{y}{x}+2\mathrm{sin}x+2$$⇒dxdy​=−xy​+2sinx+2

$$\Rightarrow \frac{dy}{dx}+\frac{y}{x}=2\mathrm{sin}x+2$$⇒dxdy​+xy​=2sinx+2

$$\text{I.F.}={e}^{\int \frac{1}{x}dx}=x$$I.F.=e∫x1​dx=x

Solution is

$$y\times (I.F.)=\int (I.F.)\times (2\mathrm{sin}x+2)dx$$y×(I.F.)=∫(I.F.)×(2sinx+2)dx

$$\Rightarrow yx=\int x(2\mathrm{sin}x+2)dx$$⇒yx=∫x(2sinx+2)dx

$$\Rightarrow yx=2\int x(\mathrm{sin}x)dx+2\int xdx$$⇒yx=2∫x(sinx)dx+2∫xdx

$$\Rightarrow xy={x}^{2}-2x\mathrm{cos}x+2\mathrm{sin}x+c\ldots (i)$$⇒xy=x2−2xcosx+2sinx+c…(i)

Now $x=\pi ,y=\pi +2$x=π,y=π+2, hence

$$\Rightarrow \pi (\pi +2)={\pi }^{2}-2\pi (-1)+0+c$$⇒π(π+2)=π2−2π(−1)+0+c

$$\Rightarrow c=0$$⇒c=0

Now $(i)$(i) becomes

$$xy={x}^{2}-2x\mathrm{cos}x+2\mathrm{sin}x$$xy=x2−2xcosx+2sinx

Put $x=\frac{\pi }{2}$x=2π​, then

$$(\frac{\pi }{2})y={(\frac{\pi }{2})}^{2}-2\cdot \frac{\pi }{2}\mathrm{cos}(\frac{\pi }{2})+2\mathrm{sin}(\frac{\pi }{2})$$(2π​)y=(2π​)2−2⋅2π​cos(2π​)+2sin(2π​)

$$\Rightarrow (\frac{\pi }{2})y=\frac{{\pi }^{2}}{4}+2$$⇒(2π​)y=4π2​+2

$$\Rightarrow y=\frac{\pi }{2}+\frac{4}{\pi }$$⇒y=2π​+π4​