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

To evaluate $g(2)$, which is equal to $f^{-1}(2)$, we need to find an $x$ such that $f(x) = 2$.

Calculating $f(1)$ gives us:

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

Thus, since $f(1) = 2$, we conclude that:

$g(2) = f^{-1}(2) = 1$

To find the points of intersection, we set the equations of the hyperbola and circle equal to each other:

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

Substituting for $y$ gives:

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

After simplification, we find:

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

Multiplying through by $x^2$ to eliminate the fraction, we have:

$1 + x^2(x^2 - 2cx) = c^2x^2$

This simplifies to:

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

Thus, the $x$-coordinates of any points of intersection are indeed zeros of the polynomial $P(x) = x^4 - 2cx^3 + 1$.

For the hyperbola and circle to intersect at exactly one point, the curves must be tangent to each other. This occurs when the discriminant of the polynomial equation derived in part (i) is zero.

The polynomial we have is:

$P(x) = x^4 - 2cx^3 + 1$

To find the condition for tangency, we use the derivative $P'(x)$ to identify the points where the curves are tangential. We can equate the derivative to zero and simultaneously solve for the critical points.

Using the graphs provides insight into the value of $c$. Upon examining the graphs, we find that the tangential intersection occurs at:

$c = 1$

Therefore, the exact value of $c > 0$ such that the hyperbola and the circle intersect at only one point is $c = 1$.