Question

If \(f(x) = {\sin ^{ - 1}}\left[ {\frac{{\sqrt 3 }}{2}x - \frac{1}{2}\sqrt {1 - {x^2}} } \right]\), \(x \in \left[ { - \frac{1}{2},1} \right]\), then f(x) will be
1. \({\sin ^{ - 1}}\left( {x - \frac{\pi }{2}} \right)\)
2. \({\sin ^{ - 1}}\left( {x - \frac{\pi }{3}} \right)\)
3. \({\sin ^{ - 1}}x ~-\frac{\pi}{6}\)
4. \({\sin ^{ - 1}}\left( {x - \frac{\pi }{4}} \right)\)

Answer 1

Correct Answer - Option 3 : \({\sin ^{ - 1}}x ~-\frac{\pi}{6}\)

Let, x = sinθ 

∵ -0.5 ≤ x ≤ 1, ∴ -π/6 ≤ θ ≤ π/2

\(f(x) = {\sin ^{ - 1}}\left[ {\frac{{\sqrt 3 }}{2}sinθ - \frac{1}{2}\sqrt {1 - {sin^2θ }} } \right]\)

\(f(x) = {\sin ^{ - 1}}\left[ {\frac{{\sqrt 3 }}{2}sinθ - \frac{1}{2}cosθ } \right]\)

\(f(x) = {\sin ^{ - 1}}\left[ {cos(π/6) sinθ - cosθ sin(π/6) } \right]\)

\(f(x) = {\sin ^{ - 1}}\left[ sin(θ - π/6) \right]\)

f(x) = θ - π/6

putting the value of θ we get,

f(x) = sin^-1x - π/6