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CBSE Board Examination 2026 · Section A · 1 mark

Q1: Range of yy when 2cos⁡−1x=y2\cos^{-1}x = y

Board 2026Ch 2 · Inverse Trigonometric Functions
The question

If 2cos⁡−1x=y2\cos^{-1}x = y⁠, then

(A) 0≤y≤π0 \le y \le \pi

(B) −π≤y≤π-\pi \le y \le \pi

(C) 0≤y≤2π0 \le y \le 2\pi

(D) −π≤y≤0-\pi \le y \le 0

Exam answer

The principal value branch of cos⁡−1x\cos^{-1}x is [0,π][0, \pi]⁠.

Why?

cos⁡−1x\cos^{-1}x is defined for −1≤x≤1-1 \le x \le 1⁠, and its principal value always lies in [0,π][0, \pi]⁠. For example, cos⁡−11=0\cos^{-1}1 = 0 and cos⁡−1(−1)=π\cos^{-1}(-1) = \pi⁠.

0≤cos⁡−1x≤π⇒ 0≤2cos⁡−1x≤2π⇒ 0≤y≤2π\begin{aligned} &0 \le \cos^{-1}x \le \pi \\ \Rightarrow\ &0 \le 2\cos^{-1}x \le 2\pi \\ \Rightarrow\ &0 \le y \le 2\pi \end{aligned}
Why?

Multiplying every part of an inequality by a positive number keeps the signs the same way round. Here we multiply by 2, then write yy for 2cos⁡−1x2\cos^{-1}x⁠.

Both ends are reached: x=1x = 1 gives y=0y = 0⁠, and x=−1x = -1 gives y=2πy = 2\pi⁠.

(A) stops at π\pi⁠, and (B), (D) allow y<0y < 0⁠, so only (C) matches.

Why?

Watch out: (A) is the range of cos⁡−1x\cos^{-1}x itself. Doubling the angle doubles the range.

(B) and (D) allow negative values of yy⁠, but cos⁡−1x\cos^{-1}x is never negative.

Answer: (C) 0≤y≤2π0 \le y \le 2\pi

Where this comes from

Chapter 2, Inverse Trigonometric Functions: Principal value branch of cos⁻¹x.

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