Inverse Trigonometric Functions Test

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Inverse Trigonometric Functions Test - 5

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Weekly Quiz Competition

Question 1
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The set of values of ‘a ’for which x² + ax + sin^–1 (x² –4x + 5) + cos^–1 (x² –4x+5 ) = 0 has at least one solution is

A
R²

B

C
R

D

Solution
Question 2
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The value of x for which sin [cot^–1 (1+x)] = cos(tan^–1 x)

A
1/2

B
1

C
0

D
-1/2

Solution

sin[cot⁻¹ (x+1)] = cos(tan⁻¹ x)
⇒sin[sin^−1(1/√1+(x+1)² )] = cos[ cos⁻¹ (1/√1+x² )]
⇒1/(√1+(x+1)² ) = 1/(√1+x² )
⇒1/(x² +2x+2) = 1/(√x² +1)
⇒x² +2x+2 = x² +1
⇒2x + 2 = 1
⇒2x = 1 −2
⇒2x = −1
⇒x = −1/2
Question 3
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The value of , if ∠C = 90⁰ in triangle ABC, is

A
π/4

B
π/3

C
π/2

D
π

Solution

a² + b² = c²
tan^-1 A + tan^-1 B = tan(A + B)/(1 - AB)
=>tan^-1 (a/(b+c)) + tan^-1 (b/(c+a))
tan^-1 [(a/(b+c) + b/(c+a))/(1 - ab/(b+c)(c+a))]
=>tan^-1 (ac + a² + b² + bc)/(bc + ab + c² + ac - ab)
= tan^-1 (ac + bc + a² + b² )/(ac + bc + c² )
= tan^-1 (ac + bc + c² )/(ac + bc + c² ) As we know that {a² + b² = c² }
= tan^-1 (1)
= π/4
Question 4
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The complete solution set of the inequality [cot^–1 x]² –6[cot –1 x] + 9 ≤0 is (where [ * ] denotes the greatest integer function)

A
(–∞, cot 3]

B
[cot 3, cot 2]

C
[cot 3, ∞)

D
None of these

Solution
Question 5
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A
tan^-1 (9/5 √10)

B
cos^-1 (9/5 √10)

C
tan^-1 (9/14)

D
sin^-1 (9/5 √10)

Solution

Given α= cos^-1 (⅗)
cos α= 3/5
sin α= ⅘
tan α= 4/3
Also β= tan^-1 (⅓)
So tan β= 1/3
tan (α-β) = (tan α–tan β)/(1 + tan αtan β)
= (4/3 –⅓)/(1 + 4/9)
= 1/(13/9)
= 9/13
So (α-β) = tan^-1 (9/13)
= sin^-1 (9/5 √10)
Hence option d is the answer.
Question 6
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The solution of the equation

A
x = 2

B
x = –4

C
x = 4

D
None of these

Solution

sin^-1 (tan π/4) - sin^-1 (√3/x) - /6 = 0
sin^-1 (1) - π/6 = sin^-1 (√3/x)
π/2 - π/6 = sin^-1 (√3/x)
2 π/6 = sin^-1 (√3/x)
π/3 = sin^-1 (√3/x)
⇒. sin(π/3) = (√3/x)
⇒(√3/2) = (√3/x)
⇒(x)^½ = 2
x = 4
Question 7
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The number of solution(s) of the equation, sin^–1 x + cos^–1 (1 –x) = sin^–1 (–x), is/are

A
0

B
1

C
2

D
More than 2

Solution

sin^-1 x + cos^-1 (1-x) = -sin^-1 x
cos^-1 (1-x) = -2sin^-1 x
(1-x) = cos(-2sin^-1 x) ……………………….(1)
Let 2sin^-1 x = t
x= sin t/2
Cos t = 1 - 2sin² (t/2)
⇒1 - 2x^2
From eq(1), we get ⇒1- x= 1-2x²
2x² - x = 0
x(2x - 1) = 0
x = 0,½
sin^-1 + cos^-1 (1-x) = sin^-1 (-x)
0 + 0 + 0 = 0
π/6 + π/3 ⇒π/2
sin^(-x) = -π/6 not equal to π/2
x = 0, therefore it has 1 solution.
Question 8
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If sin^-1 x + sin^-1 y + sin^{-1z =} , then

A

B
x⁵⁰ + y²⁵ + z⁵ = 0

C

D
None of these

Solution

sin^-1 x implies [-π/2, π/2]
= sin^-1 x + sin^-1 y + sin^-1 z = 3 π/2
sin^-1 x = π/2 sin^-1 y = π/2 sin^-1 z = π/2
x = 1 y = 1 z = 1
(A) = 1+1+1 -(9/(1+1+1)) = 3 - 9/3
⇒3 - 3 = 0
Question 9
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If α satisfies the inequation x² –x –2 >0, then which of the following exists

A
sin^–1 α

B
cos^–1 α

C
sec^–1 α

D
None of these

Solution

α^2 - α-2 >0
α^2 - 2 α+ α-2 >0
α(α-2) +1(α-2) >0
(α-2) (α+1) >0
(-∞, -1) U (2,∞)
Therefore αexist for sec^-1 α.
Question 10
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If ƒ(x) = -2x+8, then f^-1 (1) =

A
9/2

B
7/2

C
2/9

D
0
Question 11
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If the numerical value of tan (cos^–1 (4/5) + tan^–1 (2/3)) is a/b then

A
a + b = 23

B
a –b = 10

C
3b = a + 14

D
2a = 3b

Solution

Therefore a + b = (17 + 6) = 23
Question 12
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The value of is

A

B

C

D
All are Correct

Solution

cos(−14 π/5)=cos.14 π/5=cos.4 π/5
Hence, cos.1/2cos⁻¹ (cos.4 π/5)
=cos.2 π/5
Question 13
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A
tan^-1 65/156

B
π/2

C
π

D
4 tan^-1 (5)

Solution
Question 14
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Let f(x) = sin^–1 x + cos^–1 x. Then π/2 is equal to

A

B
f(k² - 2K+3), K ∈R

C
f(–2)

D
None of these

Solution

We know that for the relation
Question 15
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If , then x may be

A
1

B
–(1/2)

C
(4/3)

D
–(3/5)
Question 16
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A

B

C

D
All are Correct
Question 17
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α, βand γare three angles given by , and γ= cos^-1 1/3. then

A

B

C

D
None of these

Solution

α= tan^-1 ((2)^½ - 1) = 2tan^-1 tan π/8
=2 * π/8
=>π/4 = cos^-1 (1/(2)^½ )
β= 3(π/4) - π/6 = 7 π/12
Therefore, β>α
⅓<1/(2)^½
=>cos^-1 (⅓) >cos^-1 (/(2)^½ ) = π/4
So, γ>α
Question 18
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For the equation 2x = tan (2 tan^–1 a) + 2 tan (tan^–1 a + tan^–1 a³ ), which of the following is invalid ?

A
a ≌ 0

B
a² + 2ax + 1 = 0

C
a² x + 2a = x

D
All are Correct

Solution

2x = tan(2tan^-1 a) + 2tan(tan^-1 a + tan^-1 a³ )
= 2 tan(tan^-1 a)/[1-{tan(tan^-1 a)}² ] + 2tan (tan^-1 (a+a³ )/(1-a*a³ ))
=2a/(1-a² ) + 2(a+a³ )/(1-a*a³ )
= [2a + 2a³ + 2a + 2a³ ]/(1-a² )(1+a² )
= 4a/(1-a² )(1+a² )
= 4a(1+a² )/(1-a² )(1+a² )
2x = 4a/(1-a² )
2x(1-a² ) = 4a
x = 2a/(1-a² )
=>a² x + 2a - x = 0
Question 19
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cos^–1 x = tan^–1 x then

A

B

C

D

Solution

Let cos⁻¹ x = tan⁻¹ x = y
⇒x=cosy and tany=x
Now,tany=x
siny/cosy = x
⇒(siny)/x=x
⇒siny=x²
siny=cos² y
⇒siny=(1 −sin² y)
⇒sin² y+siny⁻¹ = 0
By quadratic formula,
siny = −1 ±√(1)² −4 ×1 ×(−1))/(2 * 1)
⇒siny=(−1 ±√5)/2
∵siny ∈[−1,1]
∴siny=(√5 −1)/2
⇒sin(cos^−1x) = (√5 −1)/2
Question 20
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The sum is equal to

A
tan^–1 2 + tan^–1 3

B
4 tan^–1 1

C
π/2

D

Solution

Σ(n = 1) tan^-1 {4n/(n⁴ - 2n + 2)}
= tan^-1 {4n/(1 + (n² - 1)² )}
= tan^-1 {4n/(1 + (n - 1)² (n + 1)² )}
= tan^-1 {4n/( (n + 1)² - (n - 1)² )}
= Σ(n = 1) tan^-1 (n+1)² - tan(n - 1)²
for(n = 1) tan^-1 (4) - tan(0)²
for(n = 2) tan^-1 (9) - tan(1)²
for(n = 3) tan^-1 (16) - tan(2)²
for(n = n-2) tan^-1 (n-1)² - tan(n - 3)²
for(n = n-1) tan^-1 (n)² - tan(n - 2)²
for(n = n) tan^-1 (n+1)² - tan(n - 1)² we are left with = tan^-1 n + tan¹ (n+1) + tan^-1 1
= π/2 + π/2 - π/4
=>3 π/4
=>sec^-1 (-(2)^½)