Let the tangent at any point P on a curve passing through the points (1,1) and ( 1 10 , 100 ) , intersect the positive x -axis and y -axis at the points A and B respectively. If P A : P B = 1 : k and y = y ( x ) is the solution of the differential

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Solution

Q.
Let the tangent at any point P on a curve passing through the points (1,1) and $(\frac{1}{10}, 100),$ intersect the positive $x$ -axis and $y$ -axis at the points A and B respectively. If $P A : P B = 1 : k$ and $y = y (x)$ is the solution of the differential equation $e^{\frac{d y}{d x}} = k x + \frac{k}{2},$ $y (0) = k$ , then $4 y (1) - 5 \log_{e} 3$ is equal to ________ . [2023]

Ans.
(5)

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

$Y - y = \frac{d y}{d x} (X - x)$

$Y = 0 at point A, then X = \frac{- y d x}{d y} + x$

Point $P (x, y)$ divides $A B$ in $k : 1$ ratio,

$∴$ $\frac{k α + 0}{k + 1} = x \Rightarrow α = \frac{k + 1}{k} x$

$x + \frac{x}{k} = - y \frac{d x}{d y} + x$

$\Rightarrow \frac{d y}{d x} + \frac{k}{x} y = 0$

$\Rightarrow y \cdot x^{k} = C$

$\Rightarrow C = 1 [∵ Tangent passing through (1, 1)]$

From point $(\frac{1}{10}, 100),$

$100 {(\frac{1}{10})}^{k} = 1 \Rightarrow k = 2$

$\frac{d y}{d x} = \ln (2 x + 1) [Given, e^{\frac{d y}{d x}} = k x + \frac{k}{2}]$

$\Rightarrow y = \frac{(2 x + 1)}{2} (\ln (2 x + 1) - 1) + C$

Now, $2 = \frac{1}{2} (0 - 1) + C [∵ y (0) = k and k = 2]$

$C = 2 + \frac{1}{2} = \frac{5}{2}$

Now, $y (1) = \frac{3}{2} (\ln 3 - 1) + \frac{5}{2} = \frac{3}{2} \ln 3 + 1 \Rightarrow 4 y (1) = 6 \ln 3 + 4$

$∴$ $4 y (1) - 5 \ln 3 = 6 \ln 3 + 4 - 5 \ln 3 = \ln 3 + 4 \approx 5.09 \approx 5$ (approx.)

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