Relations Question 20
Question: Let $ f:[ -\frac{\pi }{3},\frac{2\pi }{3} ]\to [0,4] $ be a function defined as $ f(x)=\sqrt{3}\sin x-\cos x+2 $ .Then $ {f^{-1}}(x) $ is given by
Options:
A) $ {{\sin }^{-1}}( \frac{x-2}{2} )-\frac{\pi }{6} $
B) $ {{\sin }^{-1}}( \frac{x-2}{2} )+\frac{\pi }{6} $
C) $ \frac{2\pi }{3}+{{\cos }^{-1}}( \frac{x-2}{2} ) $
D) none of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ y=f(x)=\sqrt{3}\sin x-\cos x+2 $
$ =2\sin ( x-\frac{\pi }{6} )+2 $ …(1)
Since $ f(x) $ is one-one and onto, f is invertible.
From (1), $ \sin ( x-\frac{\pi }{6} )=\frac{y-2}{2} $
Or $ x={{\sin }^{-1}}\frac{y-2}{2}+\frac{\pi }{6} $
Or $ {f^{-1}}(x)=si{n^{-1}}( \frac{x-2}{2} )+\frac{\pi }{6} $
BETA
