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CBSE Sample Paper 2026-27 · Section A · 1 mark

Q1: Identify the function from its graph, π+cos⁡−1x\pi + \cos^{-1}x

Sample 2026-27Ch 2 · Inverse Trigonometric Functions
The question

Identify the function whose graph is given below:

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

(A) cos⁡−1(x2)\cos^{-1}\left(\frac{x}{2}\right)

(B) cos⁡−1(2x)\cos^{-1}(2x)

(C) π−cos⁡−1x\pi - \cos^{-1}x

(D) π+cos⁡−1x\pi + \cos^{-1}x

For Visually Impaired:

Find the range of π2+sec⁡−1(x2)\frac{\pi}{2} + \sec^{-1}\left(\frac{x}{2}\right)

(A) [0,π][0, \pi]

(B) [0,π]−{π2}[0, \pi] - \left\{\frac{\pi}{2}\right\}

(C) [π2,3π2]\left[\frac{\pi}{2}, \frac{3\pi}{2}\right]

(D) [π2,3π2]−{π}\left[\frac{\pi}{2}, \frac{3\pi}{2}\right] - \{\pi\}

Exam answer

From the graph, the domain is [−1,1][-1, 1]⁠. The curve falls from (−1,2π)(-1, 2\pi) through (0,3π2)\left(0, \frac{3\pi}{2}\right) to (1,π)(1, \pi)⁠, so the range is [π,2π][\pi, 2\pi]⁠.

Why?

Read the domain off the xx-axis: the curve exists only from x=−1x = -1 to x=1x = 1⁠. Read the range off the yy-axis: the lowest point is at height π\pi and the highest at 2π2\pi⁠.

The curve goes down from left to right, so the function is decreasing.

The principal branch of cos⁡−1t\cos^{-1}t needs −1≤t≤1-1 \le t \le 1 and gives values in [0,π][0, \pi]⁠. Checking the options:

Why?

cos⁡−1t\cos^{-1}t is the angle in [0,π][0, \pi] whose cosine is tt⁠. A cosine never lies outside [−1,1][-1, 1]⁠, so tt must lie in [−1,1][-1, 1]⁠. Three values to remember:

cos⁡−1(−1)=πcos⁡−10=π2cos⁡−11=0\begin{aligned} \cos^{-1}(-1) &= \pi \\ \cos^{-1}0 &= \frac{\pi}{2} \\ \cos^{-1}1 &= 0 \end{aligned}

(A) −1≤x2≤1-1 \le \frac{x}{2} \le 1 gives the domain [−2,2][-2, 2]⁠, and the values lie in [0,π][0, \pi]⁠. Not the graph.

(B) −1≤2x≤1-1 \le 2x \le 1 gives the domain [−12,12]\left[-\frac{1}{2}, \frac{1}{2}\right]⁠. Not the graph.

(C) At x=−1x = -1⁠, y=π−π=0y = \pi - \pi = 0⁠, not 2π2\pi⁠. Not the graph.

Why?

(A) Multiply every part by 2 (positive, so the signs stay):

−1≤x2≤1(−1)(2)≤x≤(1)(2)−2≤x≤2\begin{aligned} -1 &\le \frac{x}{2} \le 1 \\ (-1)(2) &\le x \le (1)(2) \\ -2 &\le x \le 2 \end{aligned}

(B) Divide every part by 2:

−1≤2x≤1−12≤x≤12\begin{aligned} -1 &\le 2x \le 1 \\ -\frac{1}{2} &\le x \le \frac{1}{2} \end{aligned}

(C) Put x=−1x = -1 and use cos⁡−1(−1)=π\cos^{-1}(-1) = \pi⁠:

y=π−cos⁡−1(−1)=π−π=0\begin{aligned} y &= \pi - \cos^{-1}(-1) \\ &= \pi - \pi = 0 \end{aligned}

(D) For y=π+cos⁡−1xy = \pi + \cos^{-1}x⁠, the domain is [−1,1][-1, 1] and

x=−1:y=π+π=2πx=0:y=π+π2=3π2x=1:y=π+0=π\begin{aligned} x = -1 &: y = \pi + \pi = 2\pi \\ x = 0 &: y = \pi + \tfrac{\pi}{2} = \tfrac{3\pi}{2} \\ x = 1 &: y = \pi + 0 = \pi \end{aligned}
Why?

Each line puts in the value of cos⁡−1x\cos^{-1}x from the list above:

cos⁡−1(−1)=ππ+π=2πcos⁡−10=π2π+π2=2π2+π2=3π2cos⁡−11=0π+0=π\begin{aligned} \cos^{-1}(-1) &= \pi \\ \pi + \pi &= 2\pi \\[4pt] \cos^{-1}0 &= \frac{\pi}{2} \\ \pi + \frac{\pi}{2} &= \frac{2\pi}{2} + \frac{\pi}{2} = \frac{3\pi}{2} \\[4pt] \cos^{-1}1 &= 0 \\ \pi + 0 &= \pi \end{aligned}

Adding π\pi lifts the whole graph of cos⁡−1x\cos^{-1}x up by π\pi⁠. Its values move from [0,π][0, \pi] to [π,2π][\pi, 2\pi]⁠, and it still falls from left to right, exactly as in the figure.

Watch out: π−cos⁡−1x\pi - \cos^{-1}x in (C) turns the graph over, so it rises from 00 to π\pi⁠.

Answer: (D) π+cos⁡−1x\pi + \cos^{-1}x

For Visually Impaired

sec⁡−1t\sec^{-1}t is defined for ∣t∣≥1|t| \ge 1⁠, and its range is [0,π]−{π2}[0, \pi] - \left\{\frac{\pi}{2}\right\}⁠. Here t=x2t = \frac{x}{2}⁠, which needs ∣x∣≥2|x| \ge 2⁠.

Why?

sec⁡−1t\sec^{-1}t is the angle in [0,π][0, \pi] whose secant is tt⁠. Since sec⁡θ=1cos⁡θ\sec\theta = \frac{1}{\cos\theta} and cos⁡π2=0\cos\frac{\pi}{2} = 0⁠, the angle π2\frac{\pi}{2} has no secant, so it is left out.

As x2\frac{x}{2} runs over every tt with ∣t∣≥1|t| \ge 1⁠, sec⁡−1(x2)\sec^{-1}\left(\frac{x}{2}\right) takes every value in this range.

0≤sec⁡−1(x2)≤π⇒ π2≤π2+sec⁡−1(x2)≤3π2\begin{aligned} &0 \le \sec^{-1}\left(\tfrac{x}{2}\right) \le \pi \\ \Rightarrow\ &\tfrac{\pi}{2} \le \tfrac{\pi}{2} + \sec^{-1}\left(\tfrac{x}{2}\right) \le \tfrac{3\pi}{2} \end{aligned}
Why?

Adding the same number π2\frac{\pi}{2} to every part of an inequality keeps it true:

0+π2=π2π+π2=2π2+π2=3π2\begin{aligned} 0 + \frac{\pi}{2} &= \frac{\pi}{2} \\ \pi + \frac{\pi}{2} &= \frac{2\pi}{2} + \frac{\pi}{2} \\ &= \frac{3\pi}{2} \end{aligned}

Since sec⁡−1(x2)≠π2\sec^{-1}\left(\frac{x}{2}\right) \ne \frac{\pi}{2}⁠, the sum ≠π\ne \pi⁠.

Why?

The gap moves with the shift: the missing value π2\frac{\pi}{2} becomes π2+π2=π\frac{\pi}{2} + \frac{\pi}{2} = \pi⁠.

Watch out: (C) forgets the gap, and (B) is the range of sec⁡−1\sec^{-1} itself, before adding π2\frac{\pi}{2}⁠.

Answer: (D) [π2,3π2]−{π}\left[\dfrac{\pi}{2}, \dfrac{3\pi}{2}\right] - \{\pi\}

Where this comes from

Chapter 2, Inverse Trigonometric Functions: Graphs of inverse trigonometric functions.

Quick lesson · Worked NCERT solutions · Interactive lesson

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