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CBSE Sample Paper 2022-23 · Section A · 1 mark

Q19: Domain of sec⁡−12x\sec^{-1}2x and the value of sec⁡−1(−2)\sec^{-1}(-2)

Sample 2022-23Ch 2 · Inverse Trigonometric Functions
The question

In the following questions, a statement of assertion (A) is followed by a statement of Reason (R). Choose the correct answer out of the following choices.

(a) Both A and R are true and R is the correct explanation of A.

(b) Both A and R are true but R is not the correct explanation of A.

(c) A is true but R is false.

(d) A is false but R is true.

Assertion (A): The domain of the function sec⁡−12x\sec^{-1}2x is (−∞,−12]∪[12,∞)\left(-\infty, -\frac{1}{2}\right] \cup \left[\frac{1}{2}, \infty\right)

Reason (R): sec⁡−1(−2)=−π4\sec^{-1}(-2) = -\frac{\pi}{4}

Exam answer

sec⁡−1x\sec^{-1}x is defined if x≤−1x \le -1 or x≥1x \ge 1⁠.

Why?

sec⁡θ=1cos⁡θ\sec\theta = \frac{1}{\cos\theta} and ∣cos⁡θ∣≤1|\cos\theta| \le 1⁠, so ∣sec⁡θ∣≥1|\sec\theta| \ge 1⁠. The values of sec never lie strictly between −1-1 and 1, so those inputs are not allowed for sec⁡−1\sec^{-1}⁠. Its domain is R−(−1,1)\mathbb{R} - (-1, 1)⁠.

So sec⁡−12x\sec^{-1}2x is defined if 2x≤−12x \le -1 or 2x≥12x \ge 1⁠, that is x≤−12x \le -\frac{1}{2} or x≥12x \ge \frac{1}{2}⁠. So Assertion (A) is true.

Why?

Replace the input xx by 2x2x⁠, then divide each inequality by 2. Dividing by a positive number keeps the signs the same way.

2x≤−1⇒x≤−122x≥1⇒x≥12\begin{aligned} 2x \le -1 &\Rightarrow x \le -\frac{1}{2} \\ 2x \ge 1 &\Rightarrow x \ge \frac{1}{2} \end{aligned}

In interval form this is (−∞,−12]∪[12,∞)\left(-\infty, -\frac{1}{2}\right] \cup \left[\frac{1}{2}, \infty\right)⁠, exactly the Assertion.

The range of sec⁡−1x\sec^{-1}x is [0,π]−{π2}[0, \pi] - \left\{\frac{\pi}{2}\right\}⁠, which does not contain −π4-\frac{\pi}{4}⁠.

Why?

The principal value of sec⁡−1\sec^{-1} is always an angle from 0 to π\pi⁠, never π2\frac{\pi}{2}⁠, where cos⁡\cos is 0. A negative angle such as −π4-\frac{\pi}{4} can never be a value of sec⁡−1\sec^{-1}⁠.

In fact sec⁡−1(−2)=π−π3=2π3\sec^{-1}(-2) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}⁠. So Reason (R) is false.

Why?

Use the result sec⁡−1(−x)=π−sec⁡−1x\sec^{-1}(-x) = \pi - \sec^{-1}x⁠, with sec⁡−12=π3\sec^{-1}2 = \frac{\pi}{3} because cos⁡π3=12\cos\frac{\pi}{3} = \frac{1}{2}⁠.

sec⁡−1(−2)=π−sec⁡−12=π−π3=3π3−π3=2π3\begin{aligned} \sec^{-1}(-2) &= \pi - \sec^{-1}2 \\ &= \pi - \frac{\pi}{3} \\ &= \frac{3\pi}{3} - \frac{\pi}{3} \\ &= \frac{2\pi}{3} \end{aligned}

Check: cos⁡2π3=−12\cos\frac{2\pi}{3} = -\frac{1}{2}⁠, so sec⁡2π3=−2\sec\frac{2\pi}{3} = -2⁠.

A is true and R is false, so the answer is (c). A false Reason cannot explain anything, so codes (a) and (b) are out.

Answer: (c) A is true but R is false.

Where this comes from

Chapter 2, Inverse Trigonometric Functions: Domain and range of sec⁻¹x.

Question type: A negative number in cos⁡−1\cos^{-1}⁠, sec⁡−1\sec^{-1} or cot⁡−1\cot^{-1}

How to spot it: click to see

Type 3. A negative number in cos⁡−1\cos^{-1}⁠, sec⁡−1\sec^{-1} or cot⁡−1\cot^{-1}⁠: take it away from π\pi

Spot it: a negative number inside cos⁡−1\cos^{-1}⁠, sec⁡−1\sec^{-1} or cot⁡−1\cot^{-1}⁠.

Here: In the Reason, −2-2 is inside sec⁡−1\sec^{-1}⁠: sec⁡−12=π3\sec^{-1}2 = \frac{\pi}{3}⁠, so sec⁡−1(−2)=π−π3=2π3\sec^{-1}(-2) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}⁠, not −π4-\frac{\pi}{4}⁠. So R is false.

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Inverse Trigonometric Functions in past papers: question 16 of 18 · see them all