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Q.1. The principal value of tan1(tan 3π/5) is
(a) 2π/5
(b) 2π/5
(c) 3π/5
(d) 3π/5
Correct Answer is option (b)
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
Correct Answer is option (d)
sin[π/3 – sin1(½)]
= sin[π/3 – sin1[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]
Correct Answer is option (c)
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 sin1(2x) is [½, ½].
Q.4. If sin^{–1}x + sin^{–1}y = π/2, then value of cos^{–1}x + cos^{–1}y is
(a) π/2
(b) π
(c) 0
(d) 2π/3
Correct Answer is option (a)
Given,
sin^{–1} x + sin^{–1} y = π/2
[(π/2) – cos^{1}x] + [(π/2) – cos^{1}y] = π/2
(π/2) + (π/2) – (π/2) = cos1x + cos1y
Therefore, cos^{–1}x + cos^{–1}y = π/2.
Q.5. 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}
Correct Answer is option (c)
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)
Correct Answer is option (d)
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 [sin1(√(⅔))] {by Pythagoras theorem}
= √(⅔)
Q.7. 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]
Correct Answer is option (d)
Given,
y = cos^{–1 }(x^{2} – 4 )
⇒ cos y = x^{2} – 4
As we know, –1 ≤ cos y ≤ 1
So, – 1 ≤ x^{2} – 4 ≤ 1
Adding 4 on both sides, we get;
⇒ 3 ≤ x^{2} ≤ 5
Taking square root on both sides, we get;
⇒ √3 ≤ x ≤ √5
⇒ x∈ [√5, √3] ∪ [√3, √5]
Q.8. If α ≤ 2 sin^{–1}x + cos^{–1}x ≤ 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
Correct Answer is option (b)
Given,
α ≤ 2 sin^{–1}x + cos^{–1}x ≤ β
We know that,
π/2 ≤ sin^{–1} x ≤ π/2
⇒ (π/2) + (π/2) ≤ sin^{–1}x + (π/2) ≤ (π/2) + (π/2)
⇒ 0 ≤ sin–1x + (sin^{–1}x + cos^{–1}x) ≤ π
⇒ 0 ≤ 2 sin^{–1}x + cos^{–1}x ≤ π
By comparing with α ≤ 2 sin^{–1}x + cos^{–1}x ≤ 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
Correct Answer is option (d)
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(tan1 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})
Correct Answer is option (d)
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 θ = x/√(1 + x^{2})
cos θ = 1/√(1 + x^{2})
Now,
sin(tan1 x) = sin θ = x/√(1 + x^{2}).
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