CBSE Board Examination 2025 · Section A · 1 mark

Q18: Graph of the inverse of the cosine graph

Board 2025Ch 2 · Inverse Trigonometric Functions

The question

The graph of a trigonometric function is as shown. Which of the following will represent graph of its inverse ?

(A)

(B)

(C)

(D)

Exam answer

The graph shown is $y = \cos x$ (value 1 at $x = 0$ ⁠, zero at $\pm\frac{\pi}{2}$ ⁠, value $-1$ at $\pi$ ⁠).

Why?

Match the key points: the highest point is on the y-axis ( $\cos 0 = 1$ ⁠), the curve crosses the x-axis at $\pm\frac{\pi}{2}$ ⁠, and the lowest point is at $\pi$ ( $\cos\pi = -1$ ⁠). A sine graph would pass through the origin instead.

On $[0, \pi]$ ⁠, $\cos x$ is one-one and takes every value in $[-1, 1]$ ⁠, so

\cos^{-1} : [-1, 1] \to [0, \pi]

Why?

Only a one-one and onto function has an inverse. On the whole real line cos repeats its values, so we restrict it to $[0, \pi]$ ⁠, the principal branch. There it falls steadily from 1 to $-1$ ⁠.

For the inverse, the domain and range swap places.

Its inverse $y = \cos^{-1}x$ has domain $[-1, 1]$ and range $[0, \pi]$ ⁠. It decreases and passes through $(-1, \pi)$ ⁠, $(0, \frac{\pi}{2})$ and $(1, 0)$ ⁠.

Why?

The graph of an inverse is the mirror image of the original in the line $y = x$ ⁠, so every point $(a, b)$ becomes $(b, a)$ ⁠:

\begin{aligned} (0, 1) &\to (1, 0) \\ (\tfrac{\pi}{2}, 0) &\to (0, \tfrac{\pi}{2}) \\ (\pi, -1) &\to (-1, \pi) \end{aligned}

(A) and (B) take negative values, and (D) continues beyond $x = \pm 1$ ⁠. Only (C) matches.

Answer: (C) the graph of $y = \cos^{-1}x$

Where this comes from

Chapter 2, Inverse Trigonometric Functions: Graph of cos⁻¹x.

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