JEE Main 2022 — Mathematics Calculus

Given,

$$\frac{dy}{dx}-y=2-{e}^{-x}$$dxdy​−y=2−e−x

Linear differential equation,

So, $\displaystyle I.F.={e}^{-\int \mathrm{dx}}={e}^{-x}$I.F.=e−∫dx=e−x

Now solution of differential equation is given by

$$y\times IF=\int (2-{e}^{-x})\times IFdx$$y×IF=∫(2−e−x)×IFdx

$$\Rightarrow y{e}^{-x}=\int (2-{e}^{-x}){e}^{-x}dx$$⇒ye−x=∫(2−e−x)e−xdx

$$\Rightarrow y{e}^{-x}=\int (2{e}^{-x}-{e}^{-2x})dx$$⇒ye−x=∫(2e−x−e−2x)dx

$$\Rightarrow y{e}^{-x}=-2{e}^{-x}+\frac{{e}^{-2x}}{2}+C$$⇒ye−x=−2e−x+2e−2x​+C

$$\Rightarrow y=-2+\frac{{e}^{-x}}{2}+C{e}^{x}$$⇒y=−2+2e−x​+Cex

Given $\underset{x\rightarrow \infty }{\mathrm{lim}}y$x→∞lim​y is finite

So, $\underset{x\rightarrow \infty }{\mathrm{lim}}(-2+\frac{{e}^{-x}}{2}+C\cdot {e}^{x})\rightarrow$x→∞lim​(−2+2e−x​+C⋅ex)→finite

This is possible only when $C=0$C=0

Hence$y=y(x)=-2+\frac{{e}^{-x}}{2}$y=y(x)=−2+2e−x​

Now finding the slope $\frac{dy}{dx}=-\frac{1}{2}{e}^{-x}$dxdy​=−21​e−x

$${\frac{dy}{dx}}_{x=0}=-\frac{1}{2}=m,y(0)=-2+\frac{1}{2}=\frac{-3}{2}$$dxdy​x=0​=−21​=m,y(0)=−2+21​=2−3​

Now equation of tangent will be,

$$y+\frac{3}{2}=-\frac{1}{2}(x-0)$$y+23​=−21​(x−0)

$$\Rightarrow x+2y=-3$$⇒x+2y=−3

x-intercept $a=-3,$a=−3, y-intercept $b=\frac{-3}{2}$b=2−3​

So, $a-4b=-3+6=3$a−4b=−3+6=3