CBSE Class 12 Mathematics Inverse Trigonometric Functions MCQs Set 08

Answer: (b) [−1, 1]

Explanation:
1. sin⁻¹x = θ means sin θ = x, and the value of sin θ always lies between −1 and 1.
2. So sin⁻¹x is defined only for −1 ≤ x ≤ 1; for example sin⁻¹2 does not exist.
3. Hence the domain of sin⁻¹x is [−1, 1].

Teacher's Note:
1. The domain of sin⁻¹x is the range of sin x, and the range of sin⁻¹x is the restricted domain of sin x.
2. Do not confuse the domain [−1, 1] with the principal range \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \).

Answer: (b) \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \)

Explanation:
1. tan x takes every real value exactly once on \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \), so this is the principal branch.
2. The end points are excluded because tan x is not defined at \( x = \pm\frac{\pi}{2} \).
3. Hence the range of tan⁻¹x is \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \).

Teacher's Note:
1. The range of tan⁻¹x is an open interval, while that of sin⁻¹x is closed; this difference is often tested.
2. As x → ∞, tan⁻¹x → \( \frac{\pi}{2} \) but never reaches it, which is why the graph has two horizontal asymptotes.

Answer: (d) 1

Explanation:
1. \( \sin^{-1}\frac{1}{2} = \frac{\pi}{6} \) and \( \cos^{-1}\frac{1}{2} = \frac{\pi}{3} \).
2. Their sum is \( \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2} \).
3. So the value is \( \sin\frac{\pi}{2} = 1 \).

Teacher's Note:
1. Use the identity \( \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \) for every x in [−1, 1]; the sum is π/2 without any calculation.
2. So any expression of the form sin(sin⁻¹x + cos⁻¹x) equals 1 for every allowed x.

Answer: (a) \( \frac{\pi}{2} \)

Explanation:
1. \( \cos\frac{5\pi}{3} = \cos\left(2\pi - \frac{\pi}{3}\right) = \cos\frac{\pi}{3} = \frac{1}{2} \).
2. \( \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} \) and \( \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6} \).
3. Sum \( = \frac{\pi}{3} + \frac{\pi}{6} = \frac{\pi}{2} \).

Teacher's Note:
1. Both inverse functions act on the same number 1/2, so the identity \( \cos^{-1}x + \sin^{-1}x = \frac{\pi}{2} \) gives the answer at once.
2. Writing \( \cos^{-1}\left(\cos\frac{5\pi}{3}\right) = \frac{5\pi}{3} \) is wrong, because 5π/3 lies outside [0, π].

Answer: (d) \( -\frac{\pi}{10} \)

Explanation:
1. \( \frac{33\pi}{5} = 6\pi + \frac{3\pi}{5} \), so \( \cos\frac{33\pi}{5} = \cos\frac{3\pi}{5} \).
2. \( \cos\frac{3\pi}{5} = \sin\left(\frac{\pi}{2} - \frac{3\pi}{5}\right) = \sin\left(-\frac{\pi}{10}\right) \).
3. Since \( -\frac{\pi}{10} \) lies in \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \), \( \sin^{-1}\left(\sin\left(-\frac{\pi}{10}\right)\right) = -\frac{\pi}{10} \).

Teacher's Note:
1. Always convert cos into sin with \( \cos\theta = \sin\left(\frac{\pi}{2} - \theta\right) \) and then bring the angle into the principal range.
2. Check the sign: 3π/5 is in the II quadrant, so its cosine is negative and the answer must be negative.

Answer: (b) 1

Explanation:
1. \( \sin^{-1}(1) = \frac{\pi}{2} \), so \( 2\sin^{-1}(1) = \pi \).
2. The equation becomes \( \sec^{-1}\left(\frac{1}{x}\right) + \pi = \pi \), i.e., \( \sec^{-1}\left(\frac{1}{x}\right) = 0 \).
3. So \( \frac{1}{x} = \sec 0 = 1 \), giving x = 1.

Teacher's Note:
1. \( \sec^{-1}y = 0 \) only when y = 1, since sec⁻¹ has range [0, π] without π/2.
2. You can also write \( \sec^{-1}\left(\frac{1}{x}\right) = \cos^{-1}x \) and solve \( \cos^{-1}x = 0 \) to get x = 1 directly.

Answer: (d) 30°

Explanation:
1. \( \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3} = 60^\circ \).
2. \( \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6} = 30^\circ \).
3. Difference = 60° − 30° = 30°.

Teacher's Note:
1. Learn the principal values of sin⁻¹ for 0, 1/2, 1/√2, √3/2 and 1 by heart; they give 0°, 30°, 45°, 60° and 90°.
2. Here the answer is asked in degrees, so convert π/6 to 30° before matching the options.

Answer: (d) −cos 2

Explanation:
1. 2 radians ≈ 114.6°, which lies outside \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \).
2. Since tan has period π, tan 2 = tan(2 − π) and 2 − π ≈ −1.14 lies in \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \); so \( \tan^{-1}(\tan 2) = 2 - \pi \).
3. Then cos(2 − π) = cos(π − 2) = −cos 2.

Teacher's Note:
1. tan⁻¹(tan x) = x only for x in (−π/2, π/2); for x in (π/2, 3π/2) it equals x − π.
2. Check: cos 2 is negative (2 rad is in the II quadrant), so −cos 2 is positive, which agrees with the cosine of an angle in (−π/2, 0).

Answer: (d) None of these

Explanation:
1. Let \( \theta = \cos^{-1}\frac{7}{25} \), so \( \cos\theta = \frac{7}{25} \) with θ in the I quadrant.
2. \( \sin\theta = \sqrt{1 - \frac{49}{625}} = \frac{24}{25} \).
3. \( \cot\theta = \frac{\cos\theta}{\sin\theta} = \frac{7}{24} \), which is not among (a), (b), (c); so the answer is None of these.

Teacher's Note:
1. Draw a right triangle with base 7 and hypotenuse 25; the third side is 24 by the 7-24-25 triple.
2. A common slip is to write 24/7 (that is tan θ); cot is base over perpendicular.

Answer: (c) \( \frac{9}{25} \)

Explanation:
1. Let \( \theta = \tan^{-1}\frac{3}{4} \); then in a right triangle the perpendicular is 3 and the base is 4.
2. Hypotenuse = \( \sqrt{3^2 + 4^2} = 5 \), so \( \sin\theta = \frac{3}{5} \).
3. \( \sin^2\theta = \frac{9}{25} \).

Teacher's Note:
1. For θ = tan⁻¹x with x > 0, \( \sin\theta = \frac{x}{\sqrt{1 + x^2}} \); with x = 3/4 this gives 3/5.
2. The question asks for the square, so do not stop at 3/5, which is also an option.

Answer: (c) 15

Explanation:
1. Let α = tan⁻¹2, so tan α = 2 and \( \sec^2\alpha = 1 + \tan^2\alpha = 5 \).
2. Let β = cot⁻¹3, so cot β = 3 and \( \text{cosec}^2\beta = 1 + \cot^2\beta = 10 \).
3. Sum = 5 + 10 = 15.

Teacher's Note:
1. The Pythagorean identities sec²θ = 1 + tan²θ and cosec²θ = 1 + cot²θ avoid drawing triangles here.
2. So \( \sec^2(\tan^{-1}x) = 1 + x^2 \) and \( \text{cosec}^2(\cot^{-1}x) = 1 + x^2 \) for every real x.

Answer: (c) \( \sec^{-1} x \)

Explanation:
1. t² − t − 2 = (t − 2)(t + 1) > 0 gives t < −1 or t > 2.
2. So x lies in (−∞, −1) ∪ (2, ∞), and every such x has |x| > 1.
3. sin⁻¹x and cos⁻¹x need |x| ≤ 1, so they are not defined; sec⁻¹x needs |x| ≥ 1, so it exists. Answer: sec⁻¹x.

Teacher's Note:
1. Sign scheme for a quadratic with roots −1 and 2: it is positive outside the roots and negative between them.
2. sec⁻¹ and cosec⁻¹ are defined exactly where sin⁻¹ and cos⁻¹ are not (apart from x = ±1).

Answer: (c) \( x \in (-\infty, -1] \cup [1, +\infty) \)

Explanation:
1. tan⁻¹x is defined for every real x.
2. sec⁻¹x is defined only for |x| ≥ 1, i.e., x ≤ −1 or x ≥ 1.
3. f(x) is real where both are defined: \( x \in (-\infty, -1] \cup [1, +\infty) \).

Teacher's Note:
1. The domain of a sum is the intersection of the domains of the parts.
2. Remember x = ±1 are included, because sec⁻¹1 = 0 and sec⁻¹(−1) = π.

Answer: (b) 2n

Explanation:
1. Each \( \sin^{-1}x_r \leq \frac{\pi}{2} \), so the sum of the 2n terms is at most \( 2n \cdot \frac{\pi}{2} = n\pi \).
2. The sum equals nπ only if every term takes its maximum: \( \sin^{-1}x_r = \frac{\pi}{2} \), i.e., \( x_r = 1 \) for all r.
3. Hence \( \sum_{r=1}^{2n} x_r = 2n \times 1 = 2n \).

Teacher's Note:
1. When a sum of bounded terms equals its greatest possible value, each term must be at its maximum; this idea solves many such questions.
2. Similarly, if a sum of 2n cos⁻¹ terms equals 2nπ, every x is −1.

Answer: (b) \( \sin^{-1}\left(\sin\frac{7\pi}{5}\right) \)

Explanation:
1. \( \frac{7\pi}{5} = \pi + \frac{2\pi}{5} \), so \( \sin\frac{7\pi}{5} = -\sin\frac{2\pi}{5} = \sin\left(-\frac{2\pi}{5}\right) \).
2. Since \( -\frac{2\pi}{5} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \), \( \sin^{-1}\left(\sin\frac{7\pi}{5}\right) = -\frac{2\pi}{5} \).
3. cos⁻¹ and sec⁻¹ give values in [0, π] (here 3π/5), which can never be negative. Answer: (b).

Teacher's Note:
1. Only sin⁻¹, tan⁻¹ and cosec⁻¹ can take negative values, so a negative principal value rules out cos⁻¹, sec⁻¹ and cot⁻¹ at once.
2. For θ in \( \left[\frac{\pi}{2}, \frac{3\pi}{2}\right] \), \( \sin^{-1}(\sin\theta) = \pi - \theta \); here π − 7π/5 = −2π/5.

Answer: (c) 3

Explanation:
1. y = cos⁻¹(cos x) ≥ 0, so |y| = y and the system becomes sin x = cos⁻¹(cos x), which needs sin x ≥ 0.
2. On [0, π], cos⁻¹(cos x) = x and sin x = x only at x = 0. On [−2π, −π], cos⁻¹(cos x) = 2π + x and sin x = 2π + x only at x = −2π.
3. On (−π, 0) sin x < 0, and on (π, 2π) sin x < 0, so no solution there except the end point x = 2π, where both sides are 0.
4. Solutions: (−2π, 0), (0, 0), (2π, 0), i.e., 3 solutions.

Teacher's Note:
1. The graph of cos⁻¹(cos x) is a zig-zag of slopes ±1 touching 0 at even multiples of π; sketching it with sin x shows the meeting points quickly.
2. Since sin x < x for x > 0, the line and the sine curve meet only at the start of each tooth.

Answer: (a) \( \phi \)

Explanation:
1. 4 rad lies in \( \left(\frac{\pi}{2}, \frac{3\pi}{2}\right) \), so \( \sin^{-1}(\sin 4) = \pi - 4 \), which is negative (≈ −0.86).
2. x² − kx + (π − 4) > 0 for all real x needs discriminant < 0: k² − 4(π − 4) < 0, i.e., k² < 4(π − 4).
3. The right side is negative, so no real k satisfies it. The set is φ.

Teacher's Note:
1. A quadratic with positive leading coefficient is positive for all x only if its discriminant is negative, which needs the constant term to be positive.
2. Never write sin⁻¹(sin 4) = 4; first bring 4 into the principal range using π − 4.