It is given that $f(x)=\dfrac{a-x^2}{1+x^2}$ where $a$ is a real constant and $a \ne -1$.

1. Find the first derivative of $f(x)$ with respect to $x$.

$f'(x) = \frac{(-2x)(1+x^2) - (a-x^2)(2x)}{(1+x^2)^2}$
$f'(x) = \frac{-2x - 2x^3 - 2ax + 2x^3}{(1+x^2)^2}$
$f'(x) = \frac{-2x(1+a)}{(1+x^2)^2}$

2. Find the $x$-coordinate of the stationary point.

For a stationary point, $f'(x) = 0$.
$\frac{-2x(1+a)}{(1+x^2)^2} = 0$
Since it is given that $a \neq -1$, we have $1+a \neq 0$. Thus:
$-2x = 0 \implies x = 0$
So there is only one stationary point at $x = 0$.

3. Apply the first derivative test to find the condition for a maximum point.

For $x=0$ to be a maximum point, $f'(x)$ must change from positive to negative as $x$ increases through $0$.
Since the denominator $(1+x^2)^2 > 0$ for all real $x$, the sign of $f'(x)$ depends on the numerator $-2x(1+a)$.
For $x < 0$, we have $-2x > 0$. We require $f'(x) > 0$, which implies $1+a > 0$.
For $x > 0$, we have $-2x < 0$. We require $f'(x) < 0$, which also implies $1+a > 0$.
Therefore, the condition is:
$1+a > 0$
$a > -1$

1. Find the intersections with the $x$ and $y$ axes.

When $x = 0$, $y = \frac{a - 0}{1 + 0} = a$. The $y$-intercept is $(0, a)$.
When $y = 0$, $a - x^2 = 0 \implies x^2 = a$. Since $a > 1$, $x = \pm\sqrt{a}$.
The $x$-intercepts are $(-\sqrt{a}, 0)$ and $(\sqrt{a}, 0)$.

2. Determine the turning point and asymptotes.

To find asymptotes, consider the limit as $x \to \pm\infty$:
$y = \lim_{x \to \pm\infty} \frac{\frac{a}{x^2} - 1}{\frac{1}{x^2} + 1} = -1$
So, the horizontal asymptote is $y = -1$. There are no vertical asymptotes as $1 + x^2 \neq 0$ for all real $x$.
For the turning point, differentiating $f(x)$ or noting the even symmetry $f(x) = f(-x)$ yields a stationary point at $x = 0$. Knowing $a > 1$ and the asymptotes, $(0, a)$ is the maximum turning point.

3. Sketch the graph incorporating all key features.

Draw a smooth curve opening downwards from the maximum point $(0, a)$, passing through the $x$-axis at $(-\sqrt{a}, 0)$ and $(\sqrt{a}, 0)$, and asymptotically approaching the horizontal line $y = -1$ from above as $x \to \pm\infty$.

1. Rearrange the given equation to isolate the function $f(x)$.

$\pi(a-x^2) = (1+x^2)\tan^{-1}x$
$\frac{a-x^2}{1+x^2} = \frac{1}{\pi}\tan^{-1}x$
$f(x) = \frac{1}{\pi}\tan^{-1}x$

2. Identify the second curve to draw and determine the number of intersections with the graph of $f(x)$.

We need to sketch the curve $y = \frac{1}{\pi}\tan^{-1}x$ on the same diagram as $y = f(x)$.

Characteristics of $y = \frac{1}{\pi}\tan^{-1}x$:

• It passes through the origin $(0,0)$.
• It is strictly increasing.
• It has horizontal asymptotes $y = \frac{1}{2}$ (as $x \to \infty$) and $y = -\frac{1}{2}$ (as $x \to -\infty$).

From part (i), $y = f(x)$ has a maximum at $(0, a)$ where $a > 1$, and approaches the horizontal asymptote $y = -1$ as $x \to \pm\infty$.

Comparing the two graphs:

• For $x > 0$: $f(x)$ decreases from $a > 1$ down to $-1$, while $\frac{1}{\pi}\tan^{-1}x$ increases from $0$ to $\frac{1}{2}$. They intersect exactly once.
• For $x < 0$: $f(x)$ increases from $-1$ up to $a > 1$, while $\frac{1}{\pi}\tan^{-1}x$ increases from $-\frac{1}{2}$ to $0$. Since $f(-\infty) = -1 < -\frac{1}{2}$ and $f(0) = a > 0$, the $f(x)$ curve starts below and ends above the inverse tangent curve, meaning they intersect exactly once.

Thus, there are $1 + 1 = 2$ points of intersection in total.