JEE Main 2024 — Mathematics Calculus

Given: $\mathrm{sec}xdy+{2(1-x)\mathrm{tan}x+x(2-x)}dx=0$secxdy+2(1−x)tanx+x(2−x)dx=0

$$\Rightarrow \mathrm{sec}xdy={2(x-1)\mathrm{tan}x+({x}^{2}-2x)}dx$$⇒secxdy=2(x−1)tanx+(x2−2x)dx

$$\Rightarrow \frac{dy}{\mathrm{cos}x}={2(x-1)\frac{\mathrm{sin}x}{\mathrm{cos}x}+({x}^{2}-2x)}dx$$⇒cosxdy​=2(x−1)cosxsinx​+(x2−2x)dx

$$\Rightarrow dy=[2(x-1)\mathrm{sin}x+({x}^{2}-2x)\mathrm{cos}x]dx$$⇒dy=[2(x−1)sinx+(x2−2x)cosx]dx

$$\Rightarrow \int dy=\int [2(x-1)\mathrm{sin}x+({x}^{2}-2x)\mathrm{cos}x]dx$$⇒∫dy=∫[2(x−1)sinx+(x2−2x)cosx]dx

$$\Rightarrow \int dy=\int 2(x-1)\mathrm{sin}x+\int ({x}^{2}-2x)\mathrm{cos}xdx$$⇒∫dy=∫2(x−1)sinx+∫(x2−2x)cosxdx

$$\Rightarrow y(x)=\int 2(x-1)\mathrm{sin}xdx+[({x}^{2}-2x)(\mathrm{sin}x)-\int (2x-2)\mathrm{sin}xdx]$$⇒y(x)=∫2(x−1)sinxdx+[(x2−2x)(sinx)−∫(2x−2)sinxdx]

$$\Rightarrow y(x)=({x}^{2}-2x)\mathrm{sin}x+\lambda$$⇒y(x)=(x2−2x)sinx+λ

It is given that, $y(0)=2$y(0)=2

$$\Rightarrow y(0)=(0-0)\mathrm{sin}0+\lambda$$⇒y(0)=(0−0)sin0+λ

$$\Rightarrow 2=0+\lambda$$⇒2=0+λ

$$\Rightarrow \lambda =2$$⇒λ=2

$$\Rightarrow y(x)=({x}^{2}-2x)\mathrm{sin}x+2$$⇒y(x)=(x2−2x)sinx+2

$$\Rightarrow y(2)=({2}^{2}-2\times 2)\mathrm{sin}(2)+2$$⇒y(2)=(22−2×2)sin(2)+2

$$\Rightarrow y(2)=2$$⇒y(2)=2