Let the tangent at any point P on a curve passing through (1,1) and (110,100) intersect th…

Question

Let the tangent at any point $P$ on a curve passing through $(1,1)$ and $\left(\dfrac{1}{10},100\right)$ intersect the positive $x$- and $y$-axes at $A,B$. If $PA:PB=1:k$ and $y=y(x)$ solves $e^{dy/dx}=kx+\dfrac{k}{2}$, $y(0)=k$, then $4y(1)-5\log_e3$ is \_\_\_\_\_\_.

Answer: 5

Accepted Answer

The tangent-intercept condition gives $y\cdot x^k=C$; using the two given points determines $k=2$.

With $k=2$: $\dfrac{dy}{dx}=\ln(2x+1)$. Integrating by parts: $y=\dfrac{(2x+1)}{2}\big(\ln(2x+1)-1\big)+c$.

Using $y(0)=2$: $c=\dfrac52$. At $x=1$: $y(1)=\dfrac32(\ln3-1)+\dfrac52 = \dfrac32\ln3+1$.

$$4y(1)-5\ln3 = 6\ln3+4-5\ln3 = 4+\ln3.$$

*(Per the answer key the numeric result is 5.)*

Solutions · 1