Inverse Trigonometric Functions Test

Result

Inverse Trigonometric Functions Test - 7

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Weekly Quiz Competition

Question 1
1 / -0.25

cos2 θis not equal to

A
1 −2sin² θ

B
2cos² θ−1

C

D

Solution

cos 2 θis equal to cos² θ- sin² θ= 2cos² θ- 1
Question 2
1 / -0.25

is equal to

A
π/2

B
1

C
0

D
none of these.

Solution

sin^-1 √3/5 = A
Sin A = √3/5 , cos A = √22/5
Therefore Cos^-1 √3/5 = B
Cos B = √3/5 , sin B = √22/5
sin(A+B) = sinA cosB + cosA sinB
= √3/5 * √3/5 + √22/5 * √22/5
= 3/25 * 22/25
= 25/25
= 1
Question 3
1 / -0.25

2cos⁻¹ x = cos⁻¹ (2x² −1)holds true for all

A
|x| ≤ 1/2

B
|x|⩽1

C
0 ⩽x ⩽1

D
none of these.

Solution

This is true for all real values of x ∈[0,1].
Question 4
1 / -0.25

A
25/7

B
24/25

C
25/24

D
none of these

Solution

none of these : 7/25
Question 5
1 / -0.25

If sin A + cos A = 1, then sin 2A is equal to

A
2

B
1/2

C
0

D
1

Solution

(Sin A + Cos A)² = sin² A + cos² A + 2 sinAcosA
1 = 1 + Sin 2A
so, Sin 2A = 0
Hence A = 0
Question 6
1 / -0.25

When x = π/2, then tan x, is

A
∞or −∞

B
−∞−∞

C
∞

D
not defined.

Solution

tan π/2 = n.d.i.e. not defined.
Question 7
1 / -0.25

The value of the expression sin θ+cos θlies between

A
−√2 and √2 both inclusive

B
0 and √2 both inclusive

C
0 and 2 both inclusive.

D
-2 and 2 both inclusive

Solution

Minimum value
and maximum value =
Question 8
1 / -0.25

cos⁻¹ (cosx) = x is satisfied by ,

A
x ∈[0,1]

B
x ∈[−1,1]

C
x ∈[0,π]

D
none of these.

Solution

cos⁻¹ (cosx) = x if,
0 ⩽x ⩽πi.e. if , x ∈[0,π]
Question 9
1 / -0.25

A
π/4

B
π

C
π/2

D
π/3

Solution
Question 10
1 / -0.25

If ƒ(x) = tan(x), then f^-1 (1/√(3)) =

A
π/6

B
π/4

C
π/3

D
π

Solution

Let F(x) = tanx = y
f^(-1) = tany = x
f^(-1) (1/√3), tany = 1/√3
y = tan^-(1/√3)
=π/6