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Solve the following.
is a curve. The tangent to the curve at the point meets x-axis at A and y-axis at B. The normal to…
$y = \ln(x+1) - \ln x$ is a curve. The tangent to the curve at the point $P(1, \ln 2)$ meets x-axis at A and y-axis at B. The normal to the curve at P meets the x-axis at C and y-axis at D.
(a)[1.0]
Find the slope of tangent at P and find the slope of normal at P.
(b)[1.0]
Find the equation of tangent at P.
(c)[1.0]
Find the equation of normal at P.
(d)[1.0]
Find the co-ordinates of A and C in terms of $\ln 2$.
Answer
Answer (a)
Official answer key$y = \ln\left(\frac{x+1}{x}\right)$, $\frac{dy}{dx} = \frac{x}{x+1}\times\frac{x\times1 - (x+1)}{x^2} = \frac{-1}{x(x+1)}$
slope of tangent at P $m_1 = \frac{-1}{2}$
slope of normal at P $m_2 = 2$
Final answer: $m_1 = -\frac{1}{2},\ m_2 = 2$
Answer (b)
Official answer keyEquation of tangent at P: $y - \ln 2 = \frac{-1}{2}(x - 1)$
$2y - 2\ln 2 = -x + 1$
$x + 2y = 2\ln 2 + 1$
Final answer: $x + 2y = 2\ln 2 + 1$
Answer (c)
AIEquation of normal at $P(1, \ln 2)$ with slope $2$: $y - \ln 2 = 2(x - 1)$
$y - \ln 2 = 2x - 2$
$2x - y = 2 - \ln 2$
Final answer: $2x - y = 2 - \ln 2$
Answer (d)
AIThe tangent $x + 2y = 2\ln 2 + 1$ meets the x-axis ($y = 0$) at $x = 2\ln 2 + 1$, so $A(2\ln 2 + 1, 0)$.
The normal $2x - y = 2 - \ln 2$ meets the x-axis at $2x = 2 - \ln 2$, $x = \frac{2 - \ln 2}{2}$, so $C\left(\frac{2 - \ln 2}{2}, 0\right)$.
Final answer: $A(2\ln 2 + 1, 0),\ C\left(\frac{2-\ln 2}{2}, 0\right)$
From ISC 2025 Practice Mathematics, question 104.