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Find the domain of the following function
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Answers
Given function is
We know,
So, using this
We know,
So, using this
As square of number can never be negative.
Hence,
Answer:
{-1, 1}
Step-by-step explanation:
arcsin of g(x) is definable only for the real values of x which lie between -1 and 1 (including -1 and 1).
if we have sin⁻¹g(x), then
- 1 ≤ g(x) ≤ 1
In the given question, g(x) = (1 + x²)/2x
1 ≥ (1 + x²)/2x ≥ -1
Multiply and divide by (2x)² :
(2x)² ≥ 2x(1 + x²) ≥ -(2x)²
4x² ≥ 2x + 2x³ or 2x + 2x³ ≥ -4x²
0 ≥ x - 2x² + x³ or x + 2x² + x³ ≥ 0
Case 1: 0 ≥ x - 2x² + x³
0 ≥ x(1 - 2x + x²)
0 ≥ (x - 0)(x - 1)²
∴ x ∈ (-∞, 0] U {1}
As for x = 0, f(x) is undefined, x ∈ (-∞, 0) U {1} [from this case]
Case 2: x + 2x² + x³ ≥ 0
x(1 - 2x + x²) ≥ 0
(x - 0)(x + 1)² ≥ 0
x ∈ [0, ∞) U {-1} But f(x) is undefined for x = 0,
therefore, x ∈ (0, ∞) U {-1}
Combining both the cases, we get
x ∈ {1, -1}
Therefore, domain of this function is {1, -1}