Question

find the range of the function 1 / 2 cos x - 1.​

Answer 1

EXPLANATION.

Range of the function.

⇒ 1/(2cosx - 1).

As we know that,

Range of cos x = [-1,1].

Range of 2cosx = [-2,2].

⇒ - 1 ≤ cos x ≤ 1.

⇒ - 2 ≤ 2cosx ≤ 2.

⇒ - 2 - 1 ≤ 2cosx - 1 ≤ 2 - 1.

⇒ - 3 ≤ 2cosx - 1 ≤ 1.

⇒ (-1/3) ≤ 1/(2cosx - 1) ≤ (1).

Now, we get a range.

Range = (-∞, -1/3] ∪ [1,∞).

                                                                                                                 

MORE INFORMATION.

Domain and range of inverse trigonometric functions.

(1) = sin⁻¹x

Domain = [-1,1]

Range = (-π/2,π/2).

(2) = cos⁻¹x.

Domain = [-1,1]

Range = [0,π].

(3) = tan⁻¹x.

Domain = (-∞,∞).

Range = (-π/2,π/2).

(4) = cot⁻¹x.

Domain = (-∞,∞).

Range = (0,π).

(5) = sec⁻¹x.

Domain = (-∞,-1] ∪ [1,∞).

Range = [0,π/2) ∪ (π/2,π].

(6) = cosec⁻¹x.

Domain = (-∞,-1] ∪ [1,∞).

Range = [-π/2,0) ∪ (0,π/2].

Answer 2

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EXPLANATION.

Range of the function.

⇒ 1/(2cosx - 1).

As we know that,

Range of cos x = [-1,1].

Range of 2cosx = [-2,2].

⇒ - 1 ≤ cos x ≤ 1.

⇒ - 2 ≤ 2cosx ≤ 2.

⇒ - 2 - 1 ≤ 2cosx - 1 ≤ 2 - 1.

⇒ - 3 ≤ 2cosx - 1 ≤ 1.

⇒ (-1/3) ≤ 1/(2cosx - 1) ≤ (1).

Now, we get a range.

Range = (-∞, -1/3] ∪ [1,∞).

                                                                                                                 

MORE INFORMATION.

Domain and range of inverse trigonometric functions.

(1) = sin⁻¹x

Domain = [-1,1]

Range = (-π/2,π/2).

(2) = cos⁻¹x.

Domain = [-1,1]

Range = [0,π].

(3) = tan⁻¹x.

Domain = (-∞,∞).

Range = (-π/2,π/2).

(4) = cot⁻¹x.

Domain = (-∞,∞).

Range = (0,π).

(5) = sec⁻¹x.

Domain = (-∞,-1] ∪ [1,∞).

Range = [0,π/2) ∪ (π/2,π].

(6) = cosec⁻¹x.

Domain = (-∞,-1] ∪ [1,∞).

Range = [-π/2,0) ∪ (0,π/2].