The diagram shows the graph of $y = c \ln x$, where $c > 0$. (a) Show that the equation of the tangent to $y = c \ln x$ at $x = p$, where $p > 0$, is $$y = \frac{c}{p}(x - p) + c \ln p.$$ (b) Find the value of $c$ such that the tangent from part (a) has a gradient of 1 and passes through the origin.

SSCE HSC Mathematics Advanced Absolute value functions: The diagram shows the graph of $

The diagram shows the graph of $y = c \ln x$, where $c > 0$ - HSC - SSCE Mathematics Advanced - Question 29 - 2020 - Paper 1

Question 29

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The diagram shows the graph of $y = c \ln x$, where $c > 0$. (a) Show that the equation of the tangent to $y = c \ln x$ at $x = p$, where $p > 0$, is $$y = \frac{c}... show full transcript

Worked Solution & Example Answer:The diagram shows the graph of $y = c \ln x$, where $c > 0$ - HSC - SSCE Mathematics Advanced - Question 29 - 2020 - Paper 1

Step 1

Show that the equation of the tangent to $y = c \ln x$ at $x = p$ is $y = \frac{c}{p}(x - p) + c \ln p$

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Answer

To find the equation of the tangent line at $x = p$ , we first compute the derivative of $y = c \ln x$ :

$\frac{dy}{dx} = \frac{c}{x}.$

Now, substituting $x = p$ , we find the gradient of the tangent:

$\frac{dy}{dx}\bigg|_{x=p} = \frac{c}{p}.$

At $x = p$ , we also have the point on the curve:

$y = c \ln p.$

Using the point-gradient form for the equation of a line, we get:

$y - c \ln p = \frac{c}{p}(x - p).$

Rearranging gives us:

$y = \frac{c}{p}(x - p) + c \ln p,$ which completes the proof.

Step 2

Find the value of $c$ such that the tangent from part (a) has a gradient of 1 and passes through the origin

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Answer

From part (a), the gradient of the tangent is $\frac{c}{p}$ . To satisfy the condition that this gradient equals 1, we set:

$\frac{c}{p} = 1 \Rightarrow c = p.$

Next, we need the tangent line to pass through the origin (0,0). Substituting (0,0) into the tangent equation, we get:

$0 = \frac{c}{p}(0 - p) + c \ln p.$

Simplifying this results in:

$-c + c \ln p = 0 \Rightarrow c(\ln p - 1) = 0.$

Since $c > 0$ , we must have:

$\ln p - 1 = 0 \Rightarrow \ln p = 1 \Rightarrow p = e.$

Thus, substituting back, we find:

$c = p = e.$

The diagram shows the graph of $y = c \ln x$, where $c > 0$ - HSC - SSCE Mathematics Advanced - Question 29 - 2020 - Paper 1

Worked Solution & Example Answer:The diagram shows the graph of $y = c \ln x$, where $c > 0$ - HSC - SSCE Mathematics Advanced - Question 29 - 2020 - Paper 1

Show that the equation of the tangent to $y = c \ln x$ at $x = p$ is $y = \frac{c}{p}(x - p) + c \ln p$

Find the value of $c$ such that the tangent from part (a) has a gradient of 1 and passes through the origin

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