Q7:

Write the following functions in the simplest form: $\tan^{-1} (\frac{3a^2x - x^3}{a^3 - 3ax^2})$, $a > 0$; $\frac{-a}{\sqrt{3}} < x < \frac{a}{\sqrt{3}}$

Step 1: Substitution and Variable Transformation

To simplify the expression $f(x) = \tan^{-1} \left( \frac{3a^2x - x^3}{a^3 - 3ax^2} \right)$, we recognize the structural similarity between the argument of the function and the expansion formula for $\tan(3\theta)$.

Let $x = a \tan \theta$.
Then, $\frac{x}{a} = \tan \theta \implies \theta = \tan^{-1} \left( \frac{x}{a} \right)$.

Given the domain constraint $\frac{-a}{\sqrt{3}} < x < \frac{a}{\sqrt{3}}$, we divide by $a$ ($a > 0$):
$\frac{-1}{\sqrt{3}} < \frac{x}{a} < \frac{1}{\sqrt{3}}$
$\tan \left( -\frac{\pi}{6} \right) < \tan \theta < \tan \left( \frac{\pi}{6} \right)$
Thus, $-\frac{\pi}{6} < \theta < \frac{\pi}{6}$. [Constraint for validity of inverse function].

Step 2: Substitution into the Expression

Substitute $x = a \tan \theta$ into the function:

Factor out $a^3$ from the numerator and denominator:

Step 3: Applying Trigonometric Identities

By the triple angle identity for tangent:
$\tan(3\theta) = \frac{3 \tan \theta - \tan^3 \theta}{1 - 3 \tan^2 \theta}$ [Trigonometric Identity].

Therefore, the expression simplifies to:

Step 4: Range Analysis and Final Simplification

Since $-\frac{\pi}{6} < \theta < \frac{\pi}{6}$, we multiply by $3$:
$3 \times \left( -\frac{\pi}{6} \right) < 3\theta < 3 \times \left( \frac{\pi}{6} \right)$
$-\frac{\pi}{2} < 3\theta < \frac{\pi}{2}$.

Because $3\theta$ falls within the principal branch of the inverse tangent function $(-\frac{\pi}{2}, \frac{\pi}{2})$, we have:

Substitute $\theta = \tan^{-1} \left( \frac{x}{a} \right)$ back into the result:

Final Solution: 3 \tan^{-1} \left( \frac{x}{a} \right)