SSCE HSC Mathematics Extension 1 Inverse functions: Let $f(x) = 2x + ext{ln} x$, fo

Let $f(x) = 2x + ext{ln} x$, for $x > 0$ - HSC - SSCE Mathematics Extension 1 - Question 14 - 2023 - Paper 1

Question 14

Let $f(x) = 2x + ext{ln} x$, for $x > 0$. (i) Explain why the inverse of $f(x)$ is a function. (ii) Let $g(x) = f^{-1}(x)$. By considering the value of $f(1)$, or... show full transcript

Worked Solution & Example Answer:Let $f(x) = 2x + ext{ln} x$, for $x > 0$ - HSC - SSCE Mathematics Extension 1 - Question 14 - 2023 - Paper 1

Step 1

Explain why the inverse of $f(x)$ is a function.

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Answer

The function $f(x) = 2x + ext{ln} x$ is strictly increasing since its derivative, $f'(x) = 2 + rac{1}{x}$ , is always positive for $x > 0$ . Therefore, it passes the horizontal line test, which ensures that it has an inverse function.

Step 2

Let $g(x) = f^{-1}(x)$. By considering the value of $f(1)$, or otherwise, evaluate $g(2)$.

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Answer

To find $g(2)$ , we first evaluate $f(1)$ :

$f(1) = 2(1) + ext{ln}(1) = 2 + 0 = 2.$

Since $f(1) = 2$ , it follows that $g(2) = 1$ .

Step 3

Show that the x-coordinates of any points of intersection of the hyperbola and circle are zeros of the polynomial $P(x) = x^4 - 2cx^3 + 1$.

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Answer

To find the points of intersection, we substitute $y = rac{1}{x}$ into the equation of the circle:

$(x - c)^2 + rac{1}{x^2} = c^2.$

Expanding gives:

$x^2 - 2cx + c^2 + rac{1}{x^2} - c^2 = 0$

Multiplying through by $x^2$ gives:

$x^4 - 2cx^3 + 1 = 0,$

which shows that the x-coordinates of the intersection points are indeed the zeros of $P(x) = x^4 - 2cx^3 + 1$ .

Step 4

By considering the given graphs, or otherwise, find the exact value of $c > 0$ such that the hyperbola $y = \frac{1}{x}$ and the circle $(x - c)^2 + y^2 = c^2$ intersect at only one point.

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Answer

For the hyperbola and circle to intersect at only one point, the graphs must be tangent to each other. Thus, we need to set the derivative of $y = rac{1}{x}$ equal to the derivative of the circle equation at the point of tangency. The exact value of $c$ occurs when the discriminant of $P(x)$ equals zero, which provides the conditions for a double root. Setting conditions and solving for $c$ gives:

$c = rac{ ext{min value of } y}{x^3}.$

After analyzing both graphs, we determine that the exact value of $c$ occurs when $c = 1$ .

Let $f(x) = 2x + ext{ln} x$, for $x > 0$ - HSC - SSCE Mathematics Extension 1 - Question 14 - 2023 - Paper 1

Worked Solution & Example Answer:Let $f(x) = 2x + ext{ln} x$, for $x > 0$ - HSC - SSCE Mathematics Extension 1 - Question 14 - 2023 - Paper 1

Explain why the inverse of $f(x)$ is a function.

Let $g(x) = f^{-1}(x)$. By considering the value of $f(1)$, or otherwise, evaluate $g(2)$.

Show that the x-coordinates of any points of intersection of the hyperbola and circle are zeros of the polynomial $P(x) = x^4 - 2cx^3 + 1$.

By considering the given graphs, or otherwise, find the exact value of $c > 0$ such that the hyperbola $y = \frac{1}{x}$ and the circle $(x - c)^2 + y^2 = c^2$ intersect at only one point.

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