Let the tangent at any point P on a curve passing through the points \((1,1)\) and \(\left(\frac{1}{10}, 100\right)\), intersect positive x-axis and y-axis at the points A and B respectively. If \(\mathrm{PA}: \mathrm{PB}=1: \mathrm{k}\) and \(\mathrm{y}=\mathrm{y}(\mathrm{x})\) is the solution of the differential equation \(\mathrm{e}^{\frac{\mathrm{dy}}{\mathrm{dx}}}=\mathrm{kx}+\frac{\mathrm{k}}{2}\), \(y(0)=k\), then \(4 y(1)-5 \log _{e} 3\) is equal to ______.

JEE Main · 2023 · 10 Apr 2023 · Shift 2 · Differential Equations

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

(1, 1)

and

(\frac{1}{10}, 100)

, intersect positive x-axis and y-axis at the points A and B respectively. If

PA : PB = 1 : k

and

y = y (x)

is the solution of the differential equation

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

y (0) = k

, then

4 y (1) - 5 lo g_{e} 3

is equal to ______.

Answer: 5

The tangent at

P (x, y)

meets the axes so that

P A : P B = 1 : k

leads to

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

, giving

y x^{k} = C

. The points

(1, 1)

and

(\frac{1}{10}, 100)

give

C = 1

and

k = 2

. The differential equation then reads

\frac{d y}{d x} = ln (2 x + 1)

with

y (0) = 2

. Integrating,

y = \frac{2 x + 1}{2} (ln (2 x + 1) - 1) + \frac{5}{2}

, so

y (1) = \frac{3}{2} ln 3 + 1

. Hence

4 y (1) - 5 lo g_{e} 3 = 4 + lo g_{e} 3 \approx 5

. The required value is

5

JEE Main · 2023 · 10 Apr 2023 · Shift 2 · Integer answer