Multiple choice (Inverse Trigonometry)

choose the correct option in the following question:(class 12 inverse trigonometric functions multiple choice)

Question 1: The value of $\tan^{-1}(\sqrt{3})+\cos^{-1}(-\frac{1}{2})$ corresponding to principal branches is

(a) -π/12 (b) 0

Answer (c)

Explanation: $\tan^{-1}(\sqrt{3})+\cos^{-1}(-\frac{1}{2})$

$\Rightarrow \pi/6 + (\pi - \cos^{-1}(\frac{1}{2})$

$\Rightarrow \pi/6 + (\pi - \pi/6)$

$\Rightarrow \pi/6 + \pi - \pi/6$

$\Rightarrow \pi$

Question 2: The value of $\cot(\sin^{-1}x)$ is

(a) $\frac{\sqrt{1+x^2}}{x}$ (b) $\frac{x}{\sqrt{1+x^2}}$

Answer (d)

Explanation: $\cot(\sin^{-1}x)$

Let $\sin^{-1} x = y$

$\Rightarrow \sin y = x$

$\Rightarrow \cos^2y = 1 - \sin^2 y$

$\Rightarrow \cos^2 y = 1 - x^2$

$\Rightarrow \cos y = \sqrt{1-x^2}$

$\Rightarrow \sin^{-1} x = y = \cos^{-1}\sqrt{1-x^2}$

Since, $\cot y = \dfrac{\cos y}{\sin y}$

$y= \cot^{-1}[\dfrac{\sqrt{1-x^2}}{x}]$

$\sin^{-1} x = y= \cot^{-1}[\dfrac{\sqrt{1-x^2}}{x}]$

$\cot(\sin^{-1}x) = \cot( \cot^{-1}(\dfrac{\sqrt{1-x^2}}{x}))$

$= \dfrac{\sqrt{1-x^2}}{x}$

Question 3: The value of $\sin^{-1}(\cos\frac{\pi}{9})$ is

(a) π/9 (b) 5π/9

Answer (d)

Explanation: $\sin^{-1}(\cos\frac{\pi}{9})$

$=\sin^{-1}(\sin(\frac{\pi}{2}-\frac{\pi}{9}))$

$= \dfrac{7\pi}{18}$

Question 4: Let $\theta = \sin^{-1}(\sin(-600))$ , Then value of θ is

(a) π/3 (b) π/2

Answer (a)

Explanation: $\theta = \sin^{-1}(\sin(-600))$

$\theta = \sin^{-1}(\sin(-10\pi/3))$

$\Rightarrow \theta = \sin^{-1}(\sin(-10\pi/3+4\pi))$

$\Rightarrow \theta = \sin^{-1}(\sin(2\pi/3))$

$\Rightarrow \theta = \sin^{-1}(\sin(\pi - 2\pi/3))$

$\Rightarrow \theta = \sin^{-1}(\sin(\pi/3))$

$\Rightarrow \theta = \pi/3$

Question 5: The principal value of $\tan^{-1}(\tan 3\pi/5)$ is

(a) 2π/5 (b) -2π/5

Answer: (b) -2π/5

Explanation: $\tan^{-1} (\tan 3\pi/5)$

$\tan^{-1} (\tan 3\pi/5) = \tan^{-1} (\tan[\pi - 2\pi/5])$

$= \tan^{-1} (- \tan 2\pi/5)$ {since $\tan(\pi - x) = -\tan x$ }

= – $\tan^{-1} (\tan 2\pi/2)$

= –2π/5

Question 6: $\sin[\pi/3 - \sin^{-1}(-½)]$ is equal to:

(a) 1/2 (b) 1/3

Answer: (d) 1

Explanation: $\sin[\pi/3 - \sin^{-1}(-½)]$

$= \sin[\pi/3 - \sin^{-1}[\sin (-\pi/6))]$

= sin[π/3 – (-π/6)]

= sin(π/3 + π/6)

= sin (π/2)

= 1

Question 7: The domain of $\sin^{-1}(2x)$ is

(a) [0, 1] (b) [– 1, 1]

Answer: (c) [-1/2, 1/2]

Explanation: Let, $\sin^{-1}(2x) = \theta$ .

$\Rightarrow$ 2x = sin θ.

Since, – 1 ≤ sin θ ≤ 1

$\Rightarrow$ – 1 ≤ 2x ≤ 1\Rightarrow $-1/2 \le x \le 1/2. Therefore, the domain of$ \sin^{-1}(2x) $is [-½, ½]. Question 8: If$ \sin^{–1}x + \sin^{–1}y = \pi/2 $, then value of$ \cos^{–1}x + \cos^{–1}y $is (a) π/2 (b) π (c) 0 (d) 2π/3 Answer: (a) π/2 Explanation: Given,$ \sin^{–1} x + \sin^{–1} y = \pi/2\Rightarrow [(\pi/2) – \cos^{-1}x] + [(\pi/2) – \cos^{-1}y] = \pi/2 $(π/2) + (π/2) - (π/2) =$ \cos^{-1}x + \cos^{-1}y $Therefore,$ \cos^{-1}x + \cos^{-1}y $= π/2. Question 9: Which of the following is the principal value branch of$ \cos^{–1}x $? (a) [-π/2, π/2] (b) (0, π) (c) [0, π] (d) (0, π) - {π/2} Answer: (c) [0, π] Explanation: The principal value branch of$ \cos^{-1}x $is [0, π]. Question 10: The value of the expression$ \sin [\cot^{–1} (\cos (\tan^{–1} 1))] $is (a) 0 (b) 1 (c) 1/√3 (d) √(2/3) Answer: (d) √(2/3) Explanation:$ \sin [\cot^{–1} (\cos (\tan^{–1} 1))] $=$ \sin [\cot^{–1} (\cos (\tan^{–1} (\tan \pi/4))] ${since tan π/4 = 1}$ = \sin[\cot^{-1} (\cos \pi/4)]= \sin[\cot^{-1}(1/√2)]= \sin [\sin^{-1}(√(⅔))] $= √(2/3) Question 11: The domain of$ y = \cos^{–1} (x^2 – 4) $is (a) [3, 5] (b) [0, π] (c) [-√5, -√3] ∩ [-√5, √3] (d) [-√5, -√3] ∪ [√3, √5] Answer: (d) Explanation: Given,$ y = \cos^{–1} (x^2 – 4 )\Rightarrow \cos y = x^2 – 4 $Since, -1 \le cos y \le 1 So, - 1 \le$ x^2 – 4 $\le 1 ⇒ 3 \le$ x^2 $\le 5 ⇒ √3 \le x \le √5 ⇒ x∈ [-√5, -√3] ∪ [√3, √5] Question 12: If α \le$ 2 \sin^{–1}x + \cos^{–1}x $\le β, then (a) α = -π/2, β = π/2 (b) α = 0, β = π (c) α = -π/2, β = 3π/2 (d) α = 0, β = 2π Answer: (b) α = 0, β = π Explanation: Given, α \le$ 2 \sin^{–1}x + \cos^{–1}x $\le β We know that, -π/2 \le$ \sin^{–1} $x \le π/2 ⇒ (-π/2) + (π/2) \le$ \sin^{–1}x $+ (π/2) \le (π/2) + (π/2) ⇒ 0 \le$ \sin^{–1}x + (\sin^{–1}x + \cos^{–1}x) $\le π ⇒ 0 \le$ 2 \sin^{–1}x + \cos{–1}x $\le π By comparing with α \le$ 2 \sin^{–1}x + \cos{–1}x $\le β, we get α = 0, β = π. Question 13: The value of$ \sin (2 \tan^{–1} (.75)) $is equal to (a) .75 (b) 1.5 (c) .96 (d)$ \sin 1.5 $Answer: (c) .96 Explanation:$ \sin (2\tan^{–1} (.75)) $Let,$ \tan^{–1} (.75) $= θ$ \Rightarrow \tan \theta = 0.75\Rightarrow \tan \theta = 3/4 $Now, sin θ = 3/5 and cos θ = 4/5.$ \sin (2\tan^{–1} (.75)) = \sin 2θ\theta ${as$ tan^{-1}(.75) = \theta $} = 2 sin θ cos θ = 2 × (3/5) × (4/5) = 24/25 = 0.96 Therefore,$ \sin (2\tan^{–1} (.75)) = .96 $. Question 14:$ \sin(\tan^{-1} x) $, where |x| < 1, is equal to: (a)$ x/√(1 – x^2) $(b)$ 1/√(1 – x^2) $(c)$ 1/√(1 + x^2) $(d)$ x/√(1 + x^2) $Answer: (d)$ x/√(1 + x^2) $Explanation: Let$ \tan^{-1}x = $θ. So, tan θ = x = x/1 From this, we can write the sin θ and cos θ values as:$ \sin \theta = x/√(1 + x^2)\cos \theta = 1/√(1 + x^2) $Now,$ \sin(\tan^{-1} x) = \sin \theta = x/√(1 + x^2) $. Question 15: The value of the expression$ 2\sec^{-1}2+\sin^{-1}(\frac{1}{2}) $is (a) π/6 (b) 5π/6 (c) 7π/6 (d) 1 Answer (b) Explanation: $ 2\sec^{-1}2+\sin^{-1}(\frac{1}{2})= \frac{2\pi}{3} + \frac{\pi}{6}=\frac{5\pi}{6} $Question 16: The value of$ \tan^2(\sec^{-1}2) +\cot^2(\opratorname{cosec}^{-1}3) $is (a) 5 (b) 11 (c) 13 (d) 15 Answer (b) Question 17: The domain of the function defined by$ f(x) =\sin^{-1}\sqrt{1-x} $is (a) [1, 2] (b) [-1, 1] (c) [0, 1] (d) None of these Answer (a) Question 18: The value of$ \cot[\cos^{-1}(\frac{7}{25})] $is (a) 25/24 (b) 25/7 (c) 24/25 (d) 7/24 Answer (d) Question 19: $ \sin(\tan^{-1}x), |x|<1 $is equal to (a)$ \frac{x}{\sqrt{1-x^2}} $(b)$ \frac{1}{\sqrt{1-x^2}} $(c)$ \frac{1}{\sqrt{1+x^2}} $(d)$ \frac{x}{\sqrt{1+x^2}} $Answer (d) Question 20: If$ \cos^{-1}\alpha+\cos^{-1}\beta+\cos^{-1}\gamma = 3\pi $then$ \alpha(\beta +\gamma)+\beta(\gamma+\alpha)+\gamma(\alpha+\beta) $equals (a) 0 (b) 1 (c) 6 (d) 12 Answer (c) Question 21: Which of the following is the principal value branch of$ \cos^{-1}x $? (a) [-π/2, π/2] (b) (0, π) (c) [0, π] (d) (o, π) - {π/2} Answer (c) Question 22: The value of $ \sin^{-1}[\cos(33π/5)] $is (a) 3π/5 (b) -7π/5 (c) π/10 (d) -π/10 Answer (d) Question 23: The domain of the function$ \cos^{-1}(2x-1) $is (a) [0, 1] (b) [-1, 1] (c) (-1, 1) (c) [0, π] Answer (a) Question 24:Which of the following is the principal value branch of$ \operatorname{cosec}^{-1}x $? (a) ( -π/2, π/2) (b) (0, π) - { π/2} (c) {- π/2, π/2} (d) [- π/2, π/2] - {0} Answer (d) Question 25: If$ \cos(\sin^{-1}\frac{3}{5}+\cos^{-1})= 0 $, Then x is equal to (a) 1/5 (b) 3/5 (c) 0 (d) 1 Answer (b) Question 26: The value of$ \cos^{-1}(2x^2-1),0 \leq x\lea 1 $is equal to (a)$ 2\cos^{-1}x $(b)$ \sin^{-1}x $(c)$ \pi – 2\cos^{-1}x $(d)$ \pi + 2\cos^{-1}x $Answer (a) Question 27: The number of real solutions of the equation$ \sqrt{1+\cos 2x} = \sqrt{2}\cos^{-1}(\cos x) $in [ π/2, π] is (a) 0 (b) 1 (c) 2 (d) ∞ Answer (a) Question 28: The value of$ 2\cos^{-1}(\frac{-1}{2})+2\sin^{-1}(\frac{-1}{2})-\cos^{-1}(-1) $is (a) 0 (b) π/2 (c) π (d) 2π Answer (a) Question 29: The value of$ \cot[\frac{1}{2}\sin^{-1}\frac{\sqrt{3}}{2}] $is (a) 1 (b) 1/√3 (c) √3 (d) 0 Answer (c) Question 30:$ \sin(\cot^{-1}x) $is equal to (a)$ \sqrt{1+x^2} $(b) x (c)$ (1+x^2)^{-3/2} $(d)$ (1+x^2)^{-1/2}$

Answer (d)

Class 12 matrix multiple choice question

Multiple choice (Inverse Trigonometry)

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