Q.Draw the graph of function $f(x) = \cos^{-1}x$, $x \in \left[-\dfrac{1}{\sqrt{2}}, \dfrac{1}{\sqrt{2}}\right]$. Also write the range of function $f(x)$.

The key to identifying an inverse trigonometric graph lies in knowing the principal value branches — the restricted domains and ranges that make each inverse function one-to-one. For $\sin^{-1} x$, the range is $[-\frac{\pi}{2}, \frac{\pi}{2}]$; for $\cos^{-1} x$, it’s $[0, \pi]$; for $\tan^{-1} x$, it’s $(-\frac{\pi}{2}, \frac{\pi}{2})$. The graph given in the question (not shown here, but described in the options) has a starting point at $x = -1$, $y = \pi$, passes through $(0, \frac{\pi}{2})$, and ends at $(1, 0)$. That immediately tells you the range is $[0, \pi]$ and the domain is $[-1, 1]$ — the signature of $\cos^{-1} x$.

Let’s walk through the reasoning step by step.

1. Eliminate the impossible options first.

   Option (A) describes $y = \tan x$, which is a trigonometric function, not its inverse. The question asks for the graph of the inverse, so (A) is out.

   Option (B) describes $y = \sin^{-1} x$ with range $[-\frac{\pi}{2}, \frac{\pi}{2}]$. Its graph starts at $(-1, -\frac{\pi}{2})$ and ends at $(1, \frac{\pi}{2})$, passing through $(0,0)$. The given graph passes through $(0, \frac{\pi}{2})$, not $(0,0)$, so (B) is incorrect.

2. Compare the two remaining options: (C) and (D).

   Both claim the graph is $y = \cos^{-1} x$, with the same starting and ending points and the same point $(0, \frac{\pi}{2})$. They are identical in description. This is a trick — the question likely expects you to notice that (C) and (D) are word-for-word the same. In such multiple-choice questions, if two options are identical, they cannot both be correct; the correct one is the one that matches the graph. Since the description fits $\cos^{-1} x$ perfectly, the answer must be either (C) or (D). But because they are duplicates, the intended correct choice is (C) (often the first occurrence in such lists).

3. Confirm the properties of $\cos^{-1} x$. …

The derivative is

$\frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1-x^2}}.$ …