The domain of cos-1(x2 - 4) isA. [3, 5]B. [-1, 1]C. [-√5, -√3] ⋃

1.	The domain of cos-1(x2 - 4) isA. [3, 5]B. [-1, 1]C. [-√5, -√3] ⋃ [√3, √5]D.[-√5, - √3] ∩[-√5, √3]
Answer» Correct answer is D.[-√5, - √3] ∩[-√5, √3] We need to find the domain of cos^-1 (x² – 4). We must understand that, the domain of definition of a function is the set of "input" or argument values for which the function is defined. We know that, domain of an inverse cosine function, cos^-1x is, x ∈ [-1, 1] Then, (x² – 4) ∈ [-1, 1] Or, -1 ≤ x 2 – 4 ≤ 1 Adding 4 on all sides of the inequality, -1 + 4 ≤ x² – 4 + 4 ≤ 1 + 4 ⇒ 3 ≤ x 2 ≤ 5 Now, since x has a power of 2, so if we take square roots on all sides of the inequality then the result would be ⇒ ±√3 ≤ x ≤ ±√5 But this obviously isn’t continuous. So, we can write as x ∈ [-√5, - √3] ∩[-√5, √3]

The domain of cos-1(x2 - 4) isA. [3, 5]B. [-1, 1]C. [-√5, -√3] ⋃ [√3, √5]D.[-√5, - √3] ∩[-√5, √3]

Answer»

Correct answer is D.[-√5, - √3] ∩[-√5, √3]

We need to find the domain of cos^-1 (x² – 4).

We must understand that, the domain of definition of a function is the set of "input" or argument values for which the function is defined.

We know that, domain of an inverse cosine function, cos^-1x is,

x ∈ [-1, 1]

Then, (x² – 4) ∈ [-1, 1]

Or, -1 ≤ x 2 – 4 ≤ 1

Adding 4 on all sides of the inequality,

-1 + 4 ≤ x² – 4 + 4 ≤ 1 + 4

⇒ 3 ≤ x 2 ≤ 5

Now, since x has a power of 2, so if we take square roots on all sides of the inequality then the result would be

⇒ ±√3 ≤ x ≤ ±√5

But this obviously isn’t continuous.

So, we can write as

x ∈ [-√5, - √3] ∩[-√5, √3]