Let f(x)= sin -1( sin x), g(x)= cos -1( cos x), then which one of the following incorrect?

Question

Let f(x)= sin -1( sin x), g(x)= cos -1( cos x), then which one of the following incorrect? (A) f(x)=g(x) ∀ x ∈(0, (π/4)) (B) f ( x )< g ( x ) ∀ x ∈((π/2), (3 π/4)) (C) f ( x )< g ( x ) ∀ x ∈(π, (5 π/4)) (D) f(x)>g(x) ∀ x ∈((3 π/2), 2 π)

Tardigrade Admin · Accepted Answer

f(x)= sin -1( sin x)= begincasesx, (-π/2) ≤ x ≤ (π/2) π-x, (π/2)< x ≤ (3 π/2) x-2 π, (3 π/2)< x ≤ (5 π/2) endcases g ( x )= cos -1( cos x )= begincases x , 0 ≤ x ≤ π 2 π- x , π< x ≤ 2 π endcases Now, verify alternatives.

Q. Let $f (x) = sin^{- 1} (sin x), g (x) = cos^{- 1} (cos x)$ , then which one of the following incorrect?

$f (x) = g (x) \forall x \in (0, \frac{π}{4})$

$f (x) < g (x) \forall x \in (\frac{π}{2}, \frac{3 π}{4})$

$f (x) < g (x) \forall x \in (π, \frac{5 π}{4})$

$f (x) > g (x) \forall x \in (\frac{3 π}{2}, 2 π)$

$f (x) = sin^{- 1} (sin x) = ⎩ ⎨ ⎧ x, π - x, x - 2 π, \frac{- π}{2} \leq x \leq \frac{π}{2} \frac{π}{2} < x \leq \frac{3 π}{2} \frac{3 π}{2} < x \leq \frac{5 π}{2}$
$g (x) = cos^{- 1} (cos x) = {x, 2 π - x, 0 \leq x \leq π π < x \leq 2 π$
Now, verify alternatives.

Q. Let f(x)=sin−1(sinx),g(x)=cos−1(cosx), then which one of the following incorrect?

f(x)=g(x)∀x∈(0,4π​)

f(x)<g(x)∀x∈(2π​,43π​)

f(x)<g(x)∀x∈(π,45π​)

f(x)>g(x)∀x∈(23π​,2π)