Let the tangent at any point P on a curve passing through the points (1,1) and ( 1 10,100), intersect positive x-axis and y-axis at the points A and…

Given,

The tangent at any point $P$P on a curve passing through the points $(1,1)$(1,1) and $(\frac{1}{10},100)$(101​,100), intersect positive $x$x-axis and $y$y-axis at the points $A$A and $B$B respectively,

And $PA:PB=1:k$PA:PB=1:k and $y=y(x)$y=y(x) is the solution of the differential equation ${e}^{\frac{dy}{dx}}=kx+\frac{k}{2},y(0)=k$edxdy​=kx+2k​,y(0)=k,

Now on plotting the diagram we get,

Equation of tangent at $P(x,y)$P(x,y) is :

$$Y-y=\frac{\mathrm{dy}}{\mathrm{dx}}(X-x)$$Y−y=dxdy​(X−x)

Coordinate of $A=(x-y\frac{dx}{dy},0)$A=(x−ydydx​,0)

Coordinate of $B=(0,y-x\frac{dy}{dx})$B=(0,y−xdxdy​)

$$\therefore (x,y)=(\frac{kx-ky\frac{dx}{dy}}{k+1},\frac{y-x\frac{dy}{dx}}{k+1})$$∴(x,y)=(k+1kx−kydydx​​,k+1y−xdxdy​​)

So, $y=\frac{y-x\frac{dy}{dx}}{k+1}$y=k+1y−xdxdy​​

$$\Rightarrow y(k+1)=y-x\frac{dy}{dx}$$⇒y(k+1)=y−xdxdy​

$$\Rightarrow ky=-x\frac{dy}{dx}$$⇒ky=−xdxdy​

$$\Rightarrow k\frac{dx}{x}=-\frac{dy}{y}$$⇒kxdx​=−ydy​

Now integrating both side, we get

$$k\mathrm{ln}|x|+\mathrm{ln}|y|=\mathrm{ln}C......(i)$$kln∣x∣+ln∣y∣=lnC......(i)

Now given equation $(i)$(i) passes through $(1,1)$(1,1) and $(\frac{1}{10},100)$(101​,100)

So, $C=1$C=1 and $k=2$k=2

Now putting the value of $k$k, we get

$${e}^{\frac{dy}{dx}}=kx+\frac{k}{2}$$edxdy​=kx+2k​

$$\Rightarrow {e}^{\frac{dy}{dx}}=2x+1{\text{as}k=2}$$⇒edxdy​=2x+1ask=2

$$\Rightarrow \frac{dy}{dx}=\mathrm{ln}(2x+1)$$⇒dxdy​=ln(2x+1)

Integrating the above equation we get,

$$y=\frac{1}{2}(2x+1)(\mathrm{ln}|2x+1|-1)+C$$y=21​(2x+1)(ln∣2x+1∣−1)+C

And the above equation passes through $(0,2)$(0,2)

$$\Rightarrow C=\frac{5}{2}$$⇒C=25​

So, $2y=(2x+1)(\mathrm{ln}|2x+1|-1)+5$2y=(2x+1)(ln∣2x+1∣−1)+5

$$\Rightarrow 2y(1)=3(\mathrm{ln}3-1)+5$$⇒2y(1)=3(ln3−1)+5

$$\Rightarrow 2y(1)=3\mathrm{ln}(3)+2$$⇒2y(1)=3ln(3)+2

Hence, $4y(1)-5{\mathrm{log}}_{e}3=2[3\mathrm{ln}3+2]-5\mathrm{ln}3$4y(1)−5loge​3=2[3ln3+2]−5ln3

$$\Rightarrow 4y(1)-5{\mathrm{log}}_{e}3=6\mathrm{ln}3+4-5\mathrm{ln}3$$⇒4y(1)−5loge​3=6ln3+4−5ln3

$$\Rightarrow 4y(1)-5{\mathrm{log}}_{e}3=4+\mathrm{ln}3\approx 5$$⇒4y(1)−5loge​3=4+ln3≈5

Note: This question was bonus in Jee Mains 2023 April session.