Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

(C) UCLES 2023
9709/21/M/J/23
3
2 A curve has equation 
y=(2+3ln x)/(1+2x).
Find the equation of the tangent to the curve at the point 
(1,(2)/(3)). Give your answer in the form 
ax+by+c=0, where 
a,b and 
c are integers.

[5]

qquad

qquad

qquad

qquad

qquad

qquad

qquad

qquad

qquad

qquad

qquad

qquad

qquad

qquad

qquad

qquad

qquad

qquad

qquad

qquad

qquad

qquad

(C) UCLES 20232023\newline97099709/2121/M/J/2323\newline33\newline22 A curve has equation y=2+3lnx1+2x y=\frac{2+3 \ln x}{1+2 x} .\newlineFind the equation of the tangent to the curve at the point (1,23) \left(1, \frac{2}{3}\right) . Give your answer in the form ax+by+c=0 a x+b y+c=0 , where a,b a, b and c c are integers.\newline[5] [5] \newline \qquad \newline \qquad \newline \qquad \newline \qquad \newline \qquad \newline \qquad \newline \qquad \newline \qquad \newline \qquad \newline \qquad \newline \qquad \newline \qquad \newline \qquad \newline \qquad \newline \qquad \newline \qquad \newline \qquad \newline \qquad \newline \qquad \newline \qquad \newline \qquad \newline \qquad

Full solution

Q. (C) UCLES 20232023\newline97099709/2121/M/J/2323\newline33\newline22 A curve has equation y=2+3lnx1+2x y=\frac{2+3 \ln x}{1+2 x} .\newlineFind the equation of the tangent to the curve at the point (1,23) \left(1, \frac{2}{3}\right) . Give your answer in the form ax+by+c=0 a x+b y+c=0 , where a,b a, b and c c are integers.\newline[5] [5] \newline \qquad \newline \qquad \newline \qquad \newline \qquad \newline \qquad \newline \qquad \newline \qquad \newline \qquad \newline \qquad \newline \qquad \newline \qquad \newline \qquad \newline \qquad \newline \qquad \newline \qquad \newline \qquad \newline \qquad \newline \qquad \newline \qquad \newline \qquad \newline \qquad \newline \qquad
  1. Find Derivative: First, we need to find the derivative of the function y=2+3lnx1+2xy = \frac{2 + 3\ln x}{1 + 2x} to determine the slope of the tangent at the given point (1,23)(1, \frac{2}{3}).
  2. Use Point-Slope Form: Now, we use the point-slope form of the equation of a line, yy1=m(xx1)y - y_1 = m(x - x_1), where mm is the slope and (x1,y1)(x_1, y_1) is the point on the line. Here, m=59m = \frac{5}{9} and (x1,y1)=(1,23)(x_1, y_1) = (1, \frac{2}{3}).

More problems from Find equations of tangent lines using limits