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# Single Correct MCQs: Inverse Trignometric Functions - Notes | Study All Types of Questions for JEE - JEE

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Q.1. The principal value of tan-1(tan 3π/5) is
(a) 2π/5
(b) -2π/5
(c) 3π/5
(d) -3π/5

tan-1 (tan 3π/5)
This can be written as:
tan-1 (tan 3π/5) = tan-1 (tan[π – 2π/5])
= tan-1 (- tan 2π/5) {since tan(π – x) = -tan x}
= –tan-1 (tan 2π/2)
= –2π/5

Q.2. sin[π/3 – sin-1(-½)] is equal to:
(a) ½
(b) ⅓
(c) -1
(d) 1

sin[π/3 – sin-1(-½)]

= sin[π/3 – sin-1[sin (-π/6))]

sin[π/3 – (-π/6)]

= sin(π/3 + π/6)

= sin π/2

= 1

Q.3. The domain of sin–1(2x) is

(a) [0, 1]

(b) [– 1, 1]

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

(d) [–2, 2]

Let sin–1(2x) = θ.
Thus, 2x = sin θ.
As we know, – 1 ≤ sin θ ≤ 1
We can write this as– 1 ≤ 2x ≤ 1, which gives -1/2 ≤ x ≤ 1/2.
Therefore, the domain of sin-1(2x) is [-½, ½].

Q.4. If sin–1x + sin–1y = π/2, then value of cos–1x + cos–1y is
(a) π/2
(b) π
(c) 0
(d) 2π/3

Given,
sin–1 x + sin–1 y = π/2
[(π/2) – cos-1x] + [(π/2) – cos-1y] = π/2
(π/2) + (π/2) – (π/2) = cos-1x + cos-1y
Therefore, cos–1x + cos–1y = π/2.

Q.5. Which of the following is the principal value branch of cos–1x?
(a) [–π/2, π/2]
(b) (0, π)
(c) [0, π]
(d) (0, π) – {π/2}

The principal value branch of cos–1x is [0, π].

Q.6. The value of the expression sin [cot–1 (cos (tan–1 1))] is
(a) 0
(b) 1
(c) 1/√3
(d) √(2/3)

sin [cot–1 (cos (tan–1 1))]

= sin[cot-1 {cos (tan-1 (tan π/4))}] {since tan π/4 = 1}

= sin[cot-1 (cos π/4)]

= sin[cot-1(1/√2)]

= sin [sin-1(√(⅔))] {by Pythagoras theorem}

= √(⅔)

Q.7. The domain of y = cos–1 (x2 – 4) is
(a) [3, 5]
(b) [0, π]
(c) [-√5, -√3] ∩ [-√5, √3]
(d) [-√5, -√3] ∪ [√3, √5]

Given,

y = cos–1 (x2 – 4 )

⇒ cos y = x2 – 4

As we know, –1 ≤ cos y ≤ 1

So, – 1 ≤ x2 – 4 ≤ 1

Adding 4 on both sides, we get;

⇒ 3 ≤ x2 ≤ 5

Taking square root on both sides, we get;

⇒ √3 ≤ x ≤ √5

⇒ x∈ [-√5, -√3] ∪ [√3, √5]

Q.8. If α ≤ 2 sin–1x + cos–1x ≤ 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

Given,

α ≤ 2 sin–1x + cos–1x ≤ β

We know that,

-π/2 ≤ sin–1 x ≤ π/2

⇒ (-π/2) + (π/2) ≤ sin–1x + (π/2) ≤ (π/2) + (π/2)

⇒ 0 ≤ sin–1x + (sin–1x + cos–1x) ≤ π

⇒ 0 ≤ 2 sin–1x + cos–1x ≤ π

By comparing with α ≤ 2 sin–1x + cos–1x ≤ β, we get α = 0, β = π.

Q.9. The value of sin (2 tan–1 (.75)) is equal to
(a) .75
(b) 1.5
(c) .96
(d) sin 1.5

sin (2tan–1 (.75))
Let, tan–1 (.75) = θ
tan θ = 0.75
tan θ = 3/4
Thus by Pythagoras theorem, we get;
sin θ = 3/5 and cos θ = 4/5.
Now,
sin (2tan–1 (.75)) = sin 2θ {as tan-1(.75) = θ}
= 2 sin θ cos θ
= 2 × (3/5) × (4/5)
= 24/25
= 0.96
Therefore, sin (2tan–1 (.75)) = .96.

Q.10. sin(tan-1 x), where |x| < 1, is equal to:
(a) x/√(1 – x2)
(b) 1/√(1 – x2)
(c) 1/√(1 + x2)
(d) x/√(1 + x2)

Let tan-1x = θ.
So, tan θ = x = x/1
From this, we can write the sin θ and cos θ values as:
sin θ = x/√(1 + x2)
cos θ = 1/√(1 + x2)
Now,
sin(tan-1 x) = sin θ = x/√(1 + x2).

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