Question

The domain of the function f(x) = `sin^-1((|x| + 5)/(x^2 + 1))` is (–∞, –a] ≈ [a, ∞). Then a is equal to ______.

Options

• `(sqrt(17) - 1)/2`

• `sqrt(17)/2`

• `(sqrt(17) + 1)/2`

• `sqrt(17)/2 + 1`

Answer 1

The domain of the function f(x) = `sin^-1((|x| + 5)/(x^2 + 1))` is (–∞, –a] ≈ [a, ∞). Then a is equal to `underlinebb((sqrt(17) + 1)/2)`.

Explanation:

f(x) = `sin^-1((|x| + 5)/(x^2 + 1))`

`-1 ≤ (|x| + 5)/(x^2 + 1) ≤ 1`

–(x² + 1) ≤ |x| + 5 ≤ x² + 1

–x² – 1 – 5 ≤ |x| ≤ x² + 1 – 5

–x² – 6 ≤ |x| ≤ x² – 4

(i) x < 0, |x| = –x

–x² – 6 ≤ –x ≤ x² ≤ – 4

–x² ≤ 6 + x ≤ 0        x² + x – 4 ≥ 0

x² – x + 6 ≥ 0          x = `(-1 +- sqrt(17))/2`

Complex for all x   `x ≤ (-1 - sqrt(17))/2` or
but x < 0        `{:x ≥ (-1 + sqrt(17))/2:}}`common

⇒ x < 0         For x < 0

⇒  `x ≤ (-1 - sqrt(17))/2`

(ii) x > 0, |x| = x

–x² – 6 ≤ x² – 4

–x² – 6 – x ≤ 0             x² – x – 4 ≥ 0

`{:(x^2 + x + 6 ≥ 0),(x > 0):}}`     x = `(1 +- sqrt(17))/2`

again x > 0 `{{:(x ≤  (1 - sqrt(17))/2 or x ≥ (1 + sqrt(17))/2),(x > 0):}`

⇒ `x ≥ (1 + sqrt(17))/2`

So (–∞, –a] ≈ [a, ∞) `(-∞, (-1 - sqrt(17))/2] ∪ [(1 + sqrt(17))/2, ∞)`

⇒ a = `(1 + sqrt(17))/2`