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JEE Questions for Maths Inverse Trigonometric Functions Quiz 2 - MCQExams.com
JEE
Maths
Inverse Trigonometric Functions
Quiz 2
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0%
2/3√5
0%
2/3
0%
1/√5
0%
4/√5
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0%
π/2
0%
π/4
0%
π/3
0%
π
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cosβ > 0
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sinβ < 0
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cos(α + β) > 0
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cosα < 0
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0
0%
1
0%
2
0%
∞
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0%
1
0%
2
0%
3
0%
4
If a > b > 0, then the value of tan
-1
(a/b) + tan
-1
((a+b)/(a-b))
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0%
both a and b
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b but not a
0%
a but not b
0%
neither a nor b
The value of cos[2 tan
-1
(-7)] is
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49/50
0%
- (49/50)
0%
24/25
0%
- (24/25)
0%
48/49
If α and β are the roots of the equations 6χ
2
- 5χ + 1 = 0, then the value of tan
-1
α + tan
-1
β is
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0%
0
0%
π/4
0%
1
0%
π/2
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0%
1/2
0%
1/√3
0%
√3
0%
2
The equation sin
-1
χ - cos
-1
χ = cos
-1
(√3/has
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no solutoion
0%
unique solution
0%
infinite number of solutions
0%
None of the above
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tan2 (α/2)
0%
cot2 (α/2)
0%
tanα
0%
cot (α/2)
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0
0%
1
0%
tan-1 χ
0%
tan-12χ
If 4 sin
-1
χ + cos
-1
χ = π, then χ is equal to
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0%
0
0%
1/2
0%
- (1/2)
0%
1
Which one of the following is correct ?
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sin (cos-1χ) = cos (sin-1χ)
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sec (tan-1χ) = tan (sec-1χ)
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cos (tan-1χ) = tan (cos-1χ)
0%
tan(sin-1χ) = sin (tan-1χ)
If tan
-1
χ + tan
-1
y = π/4 , then
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χ + y + χy = 1
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χ + y - χy = 1
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χ + y + χy + 1 = 0
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χ + y - χy + 1 = 0
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- 1
0%
2/5
0%
1/3
0%
1
0%
1/5
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0%
0%
2)
0%
0%
none
If we consider only the principle values of the inverse trigonometric functions then the value of
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0%
0%
2)
0%
0%
If χ, y and z are in AP and tan
-1
χ, tan
-1
y and tan
-1
z are also in AP, then
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χ = y = z
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χ = y = - z
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χ = 1, y = 2, z = 3
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χ = 2, y = 4, z = 6
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χ = 2, y = 3z
For the equations cos
-1
χ + cos
-1
2χ + π = 0, the number of real solutions is
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0%
1
0%
2
0%
0
0%
∞
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one
0%
two
0%
zero
0%
None of these
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π/3
0%
π/4
0%
π/2
0%
0
If tan
-1
a + tan
-1
b = sin
-1
1 - tan
-1
c, then
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a + b + c = abc
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ab + bc + ca = abc
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1/a + 1/b + 1/c - (1/abc) = 0
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ab + bc + ca = a + b + c
The value of cos
-1
(cos- sin
-1
(sinis
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2
0%
8π - 26
0%
4π + 2
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None of these
If cos
-1
χ > sin
-1
χ , then
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0%
2)
0%
0%
χ > 0
If tan
-1
(a/χ) + tan
-1
(b/χ) = π/2, then χ is equal to
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0%
0%
2)
0%
2ab
0%
ab
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0%
0%
2)
0%
0%
none
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0%
0%
2)
0%
0%
None of these
If sin
-1
a + sin
-1
b + sin
-1
c = π, then the value of a√(1- a
2
) + b √(1 - b
2
) + c√(1 - c
2
) will be
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0%
2abc
0%
abc
0%
(1/2)abc
0%
(1/abc
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0%
1
0%
√3
0%
- 1
0%
1/√3
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