Question

Range of ƒ(x) = cot^–1(loge (1 – x^2 )) is

Answer 1

Answer:

Let θ=cot

−1

(2x−x

2

), where θ∈(0,π)

⇒cotθ=2x−x

2

, where θ∈(0,π)

=1−(1−2x+x

2

), where θ∈(0,π)

=1−(1−x)

2

, where θ∈(0,π)

⇒cotθ≤1, where θ∈(0,π)⇒

4

π

≤θ<π⇒Range of f(x} is [

4

π

,π)

Answer 2

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