Let the tangent at any point P on a curve passing through the...

Given,

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

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

Now on plotting the diagram we get,

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

$Y - y = \dfrac{dy}{dx}\left(X - x\right)$

Coordinate of $A = \left(x - y\dfrac{dx}{dy}, 0\right)$

Coordinate of $B = \left(0, y - x\dfrac{dy}{dx}\right)$

$∴ \left(x, y\right) = \left(\dfrac{kx - ky\dfrac{dx}{dy}}{k + 1}, \dfrac{y - x\dfrac{dy}{dx}}{k + 1}\right)$

So, $y = \dfrac{y - x\dfrac{dy}{dx}}{k + 1}$

$⇒ y\left(k + 1\right) = y - x\dfrac{dy}{dx}$

$⇒ ky = - x\dfrac{dy}{dx}$

$⇒ k\dfrac{dx}{x} = - \dfrac{dy}{y}$

Now integrating both side, we get

$kln | x | + ln | y | = lnC . . . . . . \left(i\right)$

Now given equation \left(i\right)$passes through \left(1, 1\right)$ and \left(\dfrac{1}{10}, 100\right)$</p><p>So,$C = 1$and$k = 2$</p><p>Now putting the value of$k$, we get</p><p>$e\dfrac{dy}{dx} = kx + \dfrac{k}{2}$</p><p>$⇒ e\dfrac{dy}{dx} = 2x + 1\left\{\text{as}k = 2\right\}$</p><p>$⇒ \dfrac{dy}{dx} = \ln \left(2x + 1\right)$</p><p>Integrating the above equation we get,</p><p>$y = \dfrac{1}{2}\left(2x + 1\right)\left(\ln \left|2x + 1\right| - 1\right) + C$</p><p>And the above equation passes through \left(0, 2\right)$

$⇒ C = \dfrac{5}{2}$

So, $2y = \left(2x + 1\right)\left(\ln \left|2x + 1\right| - 1\right) + 5$

$⇒ 2y\left(1\right) = 3\left(\ln 3 - 1\right) + 5$

$⇒ 2y\left(1\right) = 3\ln \left(3\right) + 2$

Hence, $4y\left(1\right) - 5\log_{e}3 = 2\left[3\ln 3 + 2\right] - 5\ln 3$

$⇒ 4y\left(1\right) - 5\log_{e}3 = 6\ln 3 + 4 - 5\ln 3$

⇒ 4y\left(1\right) - 5\log_{e}3 = 4 + \ln 3 \approx 5

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