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CBSE Sample Paper 2024-25 · Section A · 1 mark

Q15: Identify the graph of y=cos⁡−1xy = \cos^{-1}x

Sample 2024-25Ch 2 · Inverse Trigonometric Functions
The question

The graph drawn below depicts

−1 1 0 π/4 π/2 3π/4 π X Y
The given graph, redrawn.

(A) y=sin⁡−1xy = \sin^{-1}x

(B) y=cos⁡−1xy = \cos^{-1}x

(C) y=cosec⁡−1xy = \operatorname{cosec}^{-1}x

(D) y=cot⁡−1xy = \cot^{-1}x

Exam answer

The graph has domain [−1,1][-1, 1] and range [0,π][0, \pi]⁠. It passes through (−1,π)(-1, \pi)⁠, (0,π2)\left(0, \frac{\pi}{2}\right) and (1,0)(1, 0) and is decreasing.

Why?

Read the graph along each axis. The curve exists only for xx from −1-1 to 1 (the domain), and its height runs from 0 up to π\pi (the range). Its left end is the highest point and its right end the lowest, so it falls from left to right.

Check: cos⁡−1(−1)=π\cos^{-1}(-1) = \pi⁠, cos⁡−10=π2\cos^{-1}0 = \frac{\pi}{2} and cos⁡−11=0\cos^{-1}1 = 0⁠.

Why?

cos⁡π=−1\cos\pi = -1⁠, cos⁡π2=0\cos\frac{\pi}{2} = 0 and cos⁡0=1\cos 0 = 1⁠, and the angles π\pi⁠, π2\frac{\pi}{2}⁠, 0 all lie in [0,π][0, \pi]⁠, the principal range of cos⁡−1x\cos^{-1}x⁠.

(A) sin⁡−1x\sin^{-1}x has range [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] and is increasing. (C) cosec⁡−1x\operatorname{cosec}^{-1}x has domain R−(−1,1)R - (-1, 1)⁠. (D) cot⁡−1x\cot^{-1}x has domain RR⁠.

Why?

Each inverse trig function has its own principal branch. cot⁡−1x\cot^{-1}x also decreases with values in (0,π)(0, \pi)⁠, but it is defined for every real xx and never reaches 0 or π\pi⁠. cosec⁡−1x\operatorname{cosec}^{-1}x is not defined between −1-1 and 1 at all.

So it is the principal value branch of y=cos⁡−1xy = \cos^{-1}x⁠.

Answer: (B) y=cos⁡−1xy = \cos^{-1}x

Where this comes from

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

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