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CBSE Board Examination 2023 · Section A · 1 mark

Q19: Range of 2sin⁡−1x+3π22\sin^{-1}x + \frac{3\pi}{2}

Board 2023Ch 2 · Inverse Trigonometric Functions
The question

Select the correct answer from the codes (a), (b), (c) and (d) as given below.

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

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

(c) Assertion (A) is true and Reason (R) is false.

(d) Assertion (A) is false and Reason (R) is true.

Assertion (A) : The range of the function f(x)=2sin⁡−1x+3π2f(x) = 2\sin^{-1}x + \frac{3\pi}{2}⁠, where x∈[−1,1]x \in [-1, 1]⁠, is [π2,5π2]\left[\frac{\pi}{2}, \frac{5\pi}{2}\right]⁠.

Reason (R) : The range of the principal value branch of sin⁡−1(x)\sin^{-1}(x) is [0,π][0, \pi]⁠.

Exam answer
−π2≤sin⁡−1x≤π2⇒ −π≤2sin⁡−1x≤π⇒ π2≤f(x)≤5π2\begin{aligned} &-\frac{\pi}{2} \le \sin^{-1}x \le \frac{\pi}{2} \\ \Rightarrow\ &-\pi \le 2\sin^{-1}x \le \pi \\ \Rightarrow\ &\frac{\pi}{2} \le f(x) \le \frac{5\pi}{2} \end{aligned}
Why?

Start from the principal range of sin⁡−1x\sin^{-1}x⁠. Multiplying by 2, a positive number, keeps the inequality signs the same way.

2×(−π2)=−π2×π2=π\begin{aligned} 2 \times \left(-\frac{\pi}{2}\right) &= -\pi \\ 2 \times \frac{\pi}{2} &= \pi \end{aligned}

Then add 3π2\frac{3\pi}{2} to every part. Write π=2π2\pi = \frac{2\pi}{2} to get a common denominator:

−π+3π2=−2π2+3π2=π2π+3π2=2π2+3π2=5π2\begin{aligned} -\pi + \frac{3\pi}{2} &= -\frac{2\pi}{2} + \frac{3\pi}{2} \\ &= \frac{\pi}{2} \\[4pt] \pi + \frac{3\pi}{2} &= \frac{2\pi}{2} + \frac{3\pi}{2} \\ &= \frac{5\pi}{2} \end{aligned}

As xx runs over [−1,1][-1, 1]⁠, sin⁡−1x\sin^{-1}x takes every value in [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠, so the range of f is [π2,5π2]\left[\frac{\pi}{2}, \frac{5\pi}{2}\right]⁠. So Assertion (A) is true.

Why?

The inequalities alone only give bounds. The range is the whole interval because sin⁡−1x\sin^{-1}x is continuous and increasing on [−1,1][-1, 1]⁠, so it reaches every value between −π2-\frac{\pi}{2} and π2\frac{\pi}{2}⁠. The end values come from f(−1)=π2f(-1) = \frac{\pi}{2} and f(1)=5π2f(1) = \frac{5\pi}{2}⁠.

Reason (R) is false, since the principal value branch of sin⁡−1x\sin^{-1}x has range [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠.

Why?

[0,π][0, \pi] is the principal range of cos⁡−1x\cos^{-1}x⁠, not of sin⁡−1x\sin^{-1}x⁠. Mixing up the two is a common slip.

Since the Reason is false, it cannot explain the Assertion. With A true and R false, the code is (c).

Answer: (c) Assertion (A) is true and Reason (R) is false.

Where this comes from

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

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