Let $f (x) = x cos^{- 1} (- sin ∣ x ∣), x \in [- \frac{π}{2}, \frac{π}{2}]$ , then which of the following is true?

Question

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Answer 1

$f (x) = x cos^{- 1} (- sin (∣ x ∣))$

$= x (π - cos^{- 1} (sin ∣ x ∣))$

$= x (π - \frac{π}{2} + sin^{- 1} (sin ∣ x ∣))$

$= x (\frac{π}{2} + ∣ x ∣)$

$∴ f (x) = ⎩ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎧ x (\frac{π}{2} + x) x (\frac{π}{2} - x) \frac{π}{2} > x \geq o \frac{π}{2} < x < 0$

$∴ f^{'} (x) = ⎩ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎧ \frac{π}{2} + 2 x \frac{π}{2} - 2 x \frac{π}{2} > x \geq 0 - \frac{π}{2} \leq x < 0$

$f^{'} (x)$ is increasing in $(0, \frac{π}{2})$ and decreasing in $(- \frac{π}{2}, 0)$ .

Answer 2

f (x) = x cos^{- 1} (- sin (∣ x ∣))

= x (π - cos^{- 1} (sin ∣ x ∣))

= x (π - \frac{π}{2} + sin^{- 1} (sin ∣ x ∣))

= x (\frac{π}{2} + ∣ x ∣)

∴ f (x) = ⎩ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎧ x (\frac{π}{2} + x) x (\frac{π}{2} - x) \frac{π}{2} > x \geq o \frac{π}{2} < x < 0

∴ f^{'} (x) = ⎩ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎧ \frac{π}{2} + 2 x \frac{π}{2} - 2 x \frac{π}{2} > x \geq 0 - \frac{π}{2} \leq x < 0

f^{'} (x)

is increasing in

(0, \frac{π}{2})

and decreasing in

(- \frac{π}{2}, 0)

.