Key Concepts and Formulas

• Tangent to a curve: $Y - y = \frac{dy}{dx}(X - x)$, where $(X, Y)$ is a point on the tangent and $(x, y)$ is the point of tangency.
• Section Formula: If point $P(x, y)$ divides the line segment joining $A(x_1, y_1)$ and $B(x_2, y_2)$ in the ratio $m:n$, then $x = \frac{mx_2 + nx_1}{m+n}$ and $y = \frac{my_2 + ny_1}{m+n}$.
• Integration by Parts: $\int u \, dv = uv - \int v \, du$.

Step-by-Step Solution

Step 1: Find the x and y intercepts of the tangent line.

The tangent line at a point $P(x, y)$ on the curve is given by $Y - y = \frac{dy}{dx}(X - x)$.

• To find the x-intercept (point A), set $Y = 0$: $0 - y = \frac{dy}{dx}(X - x)$ $X = x - y \frac{dx}{dy}$ So, $A = \left(x - y \frac{dx}{dy}, 0\right)$.

• To find the y-intercept (point B), set $X = 0$: $Y - y = \frac{dy}{dx}(0 - x)$ $Y = y - x \frac{dy}{dx}$ So, $B = \left(0, y - x \frac{dy}{dx}\right)$.

Step 2: Apply the Section Formula.

Given that $PA : PB = 1 : k$, the point $P(x, y)$ divides the line segment $AB$ in the ratio $1:k$. Using the section formula: $x = \frac{1(0) + k\left(x - y \frac{dx}{dy}\right)}{1 + k} = \frac{k\left(x - y \frac{dx}{dy}\right)}{1 + k}$ $y = \frac{1\left(y - x \frac{dy}{dx}\right) + k(0)}{1 + k} = \frac{y - x \frac{dy}{dx}}{1 + k}$

Step 3: Simplify the equations from the Section Formula.

From the equation for $x$: $x(1 + k) = kx - ky \frac{dx}{dy}$ $x + kx = kx - ky \frac{dx}{dy}$ $x = -ky \frac{dx}{dy}$ $x \frac{dy}{dx} = -ky$

From the equation for $y$: $y(1 + k) = y - x \frac{dy}{dx}$ $y + ky = y - x \frac{dy}{dx}$ $ky = -x \frac{dy}{dx}$ $x \frac{dy}{dx} = -ky$ Both equations give us the same differential equation: $x \frac{dy}{dx} = -ky$.

Step 4: Solve the first differential equation.

Separate variables: $\frac{dy}{y} = -k \frac{dx}{x}$ Integrate both sides: $\int \frac{dy}{y} = -k \int \frac{dx}{x}$ $\ln|y| = -k \ln|x| + C_1$ $\ln|y| = \ln|x^{-k}| + C_1$ $y = C x^{-k} = \frac{C}{x^k}$

Step 5: Use the given points to find C and k.

The curve passes through $(1, 1)$ and $\left(\frac{1}{10}, 100\right)$.

• For $(1, 1)$: $1 = \frac{C}{1^k} \implies C = 1$ So, $y = \frac{1}{x^k}$.
• For $\left(\frac{1}{10}, 100\right)$: $100 = \frac{1}{\left(\frac{1}{10}\right)^k}$ $100 = 10^k$ $10^2 = 10^k \implies k = 2$ Thus, $y = \frac{1}{x^2}$.

Step 6: Solve the second differential equation.

The second differential equation is $e^{\frac{dy}{dx}} = kx + \frac{k}{2}$, with $y(0) = k$. Substitute $k = 2$: $e^{\frac{dy}{dx}} = 2x + \frac{2}{2} = 2x + 1$ $\frac{dy}{dx} = \ln(2x + 1)$ Integrate with respect to $x$: $y(x) = \int \ln(2x + 1) \, dx$ Use integration by parts: Let $u = \ln(2x + 1)$ and $dv = dx$. Then $du = \frac{2}{2x + 1} \, dx$ and $v = x$. $y(x) = x \ln(2x + 1) - \int x \cdot \frac{2}{2x + 1} \, dx$ $y(x) = x \ln(2x + 1) - \int \frac{2x}{2x + 1} \, dx$ $y(x) = x \ln(2x + 1) - \int \frac{2x + 1 - 1}{2x + 1} \, dx$ $y(x) = x \ln(2x + 1) - \int \left(1 - \frac{1}{2x + 1}\right) \, dx$ $y(x) = x \ln(2x + 1) - \left(x - \frac{1}{2} \ln(2x + 1)\right) + C$ $y(x) = x \ln(2x + 1) - x + \frac{1}{2} \ln(2x + 1) + C$ $y(x) = \left(x + \frac{1}{2}\right) \ln(2x + 1) - x + C$

Step 7: Apply the initial condition y(0) = k = 2.

$y(0) = \left(0 + \frac{1}{2}\right) \ln(2(0) + 1) - 0 + C = 2$ $\frac{1}{2} \ln(1) + C = 2$ $\frac{1}{2} (0) + C = 2$ $C = 2$ So, $y(x) = \left(x + \frac{1}{2}\right) \ln(2x + 1) - x + 2$.

Step 8: Evaluate 4y(1) - 6ln(3).

$y(1) = \left(1 + \frac{1}{2}\right) \ln(2(1) + 1) - 1 + 2 = \frac{3}{2} \ln(3) + 1$ $4y(1) = 4\left(\frac{3}{2} \ln(3) + 1\right) = 6 \ln(3) + 4$ $4y(1) - 6 \ln(3) = 6 \ln(3) + 4 - 6 \ln(3) = 4$

Step 9: Check for errors

It appears there was an error in the calculation. Let's go back and check. The issue is that the correct answer is 0, but we got 4.

The error is in copying the problem. The original problem asks for $4y(1) - 6\log_e 3$.

Step 8 (Revised): Evaluate 4y(1) - 6ln(3).

$y(1) = \left(1 + \frac{1}{2}\right) \ln(2(1) + 1) - 1 + 2 = \frac{3}{2} \ln(3) + 1$ $4y(1) = 4\left(\frac{3}{2} \ln(3) + 1\right) = 6 \ln(3) + 4$ $4y(1) - 6 \ln(3) = 6 \ln(3) + 4 - 6 \ln(3) = 4$

It appears there was an error in the calculation. Let's go back and check. The issue is that the correct answer is 0, but we got 4.

We need to re-examine the differential equation setup, as there's an error somewhere.

Since the correct answer is 0, we should get $4y(1) = 6\ln 3$, implying $y(1) = \frac{3}{2} \ln 3$.

$y(x) = (x + \frac{1}{2}) \ln(2x+1) -x + C$. $y(0) = 2$, so $(\frac{1}{2}) \ln(1) - 0 + C = 2$, so $C = 2$. $y(x) = (x+\frac{1}{2}) \ln(2x+1) - x + 2$.

$y(1) = (1 + \frac{1}{2}) \ln(3) - 1 + 2 = \frac{3}{2} \ln 3 + 1$. $4 y(1) = 4(\frac{3}{2} \ln 3 + 1) = 6 \ln 3 + 4$.

$4y(1) - 6 \ln 3 = 6 \ln 3 + 4 - 6 \ln 3 = 4$.

This is not equal to 0.

Let's check the initial condition again. We have $\frac{dy}{dx} = \ln(2x+1)$. Integrating gives $y = \int \ln(2x+1) dx$. Using integration by parts with $u = \ln(2x+1)$ and $dv = dx$, we get $du = \frac{2}{2x+1}dx$ and $v = x$. So $y = x \ln(2x+1) - \int \frac{2x}{2x+1} dx = x \ln(2x+1) - \int (1 - \frac{1}{2x+1}) dx = x \ln(2x+1) - x + \frac{1}{2} \ln(2x+1) + C$. $y = (x+\frac{1}{2}) \ln(2x+1) - x + C$. Since $y(0) = k = 2$, we have $2 = (0+\frac{1}{2}) \ln(1) - 0 + C$, so $C = 2$. Thus $y(x) = (x+\frac{1}{2}) \ln(2x+1) - x + 2$.

$y(1) = (1+\frac{1}{2}) \ln(3) - 1 + 2 = \frac{3}{2} \ln 3 + 1$. $4y(1) = 4(\frac{3}{2} \ln 3 + 1) = 6 \ln 3 + 4$. $4y(1) - 6 \ln 3 = 4$.

I am unable to get the correct answer.

Common Mistakes & Tips

• Be extremely careful with the section formula and the ratio in which the point divides the line segment.
• Remember the integration by parts formula and choose appropriate $u$ and $dv$ for easier integration.
• Double-check the initial conditions and ensure they are applied correctly after integration.

Summary

We found the equation of the tangent line, used the section formula to relate the intercepts to the coordinates of the point on the curve, and derived a differential equation. Solving this differential equation and using the given points, we found the value of $k$. Then we solved the second differential equation using integration by parts and the given initial condition. Finally, we evaluated the expression $4y(1) - 6 \log _{\mathrm{e}} 3$. However, the final calculation resulted in 4, not 0. There may be an error in the problem statement.

Final Answer

The final answer is \boxed{4}.