Animated Solution for Mathematics - Differentiation: Match the statements given in Column-I with the intervals/union of intervals given in Column-II.

We begin with the expression $\text{Re}\left(\frac{2iz}{1-z^2}\right)$ where $|z|=1$. Many students immediately try to substitute $z=x+iy$. While that is a valid path, it is a path through a dense forest of algebra.

Instead, let us use the elegance of Euler's form. Since $z$ lies on the unit circle, we can write $z=e^{i\varphi }$. Substituting this into our expression, we get:

Now, watch the magic. If we factor out $e^{i\varphi }$ from the denominator, we are left with $e^{i\varphi }(e^{-i\varphi }-e^{i\varphi })$. The $e^{i\varphi }$ terms cancel out, leaving us with:

Recall that $e^{i\varphi }-e^{-i\varphi }=2i\sin \varphi$. Thus, our expression simplifies to $-\frac{1}{\sin \varphi }$. Since $\sin \varphi$ ranges between $[-1,1]$ (excluding zero), the reciprocal ranges from $(-\infty ,-1]\cup [1,\infty )$. The imaginary part vanishes, leaving us with a purely real range.

Next, we face $f(x)={\sin }^{-1}\left(\frac{8(3{)}^{x-2}}{1-3^{2(x-1)}}\right)$. The inverse sine function is a gatekeeper; it only accepts inputs in the interval $[-1,1]$. We must ensure that the argument satisfies:

This looks terrifying, but let us simplify it. Let $t=3^{x-1}$. Then $3^{x-2}=\frac{t}{3}$ and $3^{2(x-1)}=t^2$. The inequality becomes:

By splitting this into cases where $t<1$ and $t>1$, we solve the quadratic inequalities. We find that $t\le \frac{1}{3}$ or $t\ge 3$. Translating back to $x$, we get $3^{x-1}\le 3^{-1}⟹x\le 0$ and $3^{x-1}\ge 3^1⟹x\ge 2$.

The domain is $(-\infty ,0]\cup [2,\infty )$. It is a perfect example of how substitution can tame even the most intimidating exponential expressions.

Now, we encounter the determinant:

Do not panic at the sight of a $3\times 3$ matrix. Expand it along the first row. The calculation yields $(1+{\tan }^2\theta )-\tan \theta (0)+1({\tan }^2\theta +1)$.

This simplifies to $2(1+{\tan }^2\theta )$. Using the identity $1+{\tan }^2\theta ={\sec }^2\theta$, we get $f(\theta )=2{\sec }^2\theta$. For $0\le \theta <\frac{\pi }{2}$, $\sec \theta$ ranges from $[1,\infty )$, so $2{\sec }^2\theta$ ranges from $[2,\infty )$.

Finally, we analyze $f(x)=x^{3/2}(3x-10)$ for $x\ge 0$. To find where it is increasing, we need $f^{\prime }(x)\ge 0$. Applying the product rule:

Factoring out $3x^{1/2}$, we get:

Since $\sqrt{x}$ is always non-negative, the sign depends on $5x-10$. Thus, $f(x)$ increases when $5x-10\ge 0$, or $x\ge 2$. The interval is $