If α ≤ 2 sin-1x+ cos-1x ≤ β, then

Question

If α ≤ 2 sin-1x+ cos-1x ≤ β, then (A) α=(-π/2) β=(π/2) (B) α=(-π/2) β=(3 π/2) (C) α=0 β=π (D) α=0 β=2 π

Tardigrade Admin · Accepted Answer

We know that , -(π/2) ≤ sin -1 x ≤ (π/2) ⇒ 0 ≤ (π/2)+ sin -1 x ≤ π ⇒ 0 ≤ sin -1 x+ cos -1 x+ sin -1 x ≤ π ⇒ 0 ≤ 2 sin -1 x+ cos -1 x ≤ π On comparing. we get α=0 and β=π

Q. If $α \leq 2 sin^{- 1} x + cos^{- 1} x \leq β$ , then

$α = \frac{- π}{2} β = \frac{π}{2}$

$α = \frac{- π}{2} β = \frac{3 π}{2}$

$α = 0 β = π$

$α = 0 β = 2 π$

We know that , $- \frac{π}{2} \leq sin^{- 1} x \leq \frac{π}{2}$
$\Rightarrow 0 \leq \frac{π}{2} + sin^{- 1} x \leq π$
$\Rightarrow 0 \leq sin^{- 1} x + cos^{- 1} x + sin^{- 1} x \leq π$
$\Rightarrow 0 \leq 2 sin^{- 1} x + cos^{- 1} x \leq π$
On comparing. we get
$α = 0$ and $β = π$

Q. If α≤2sin−1x+cos−1x≤β, then

α=2−π​β=2π​

α=2−π​β=23π​

α=0β=π

α=0β=2π