Direction: Question Based on the following paragraph. |
Let \[f:A\to B\]be a function defined by y=f(x) such that f is both one-one (Injective) and onto Surjective), then there exists a unique function \[g:B\to A\]such that \[f(x)=y\leftrightarrow g\] \[(y)=x,\forall x\in A\]and \[y\in B,\]then g is said to be inverse off Thus,\[g={{f}^{-1}}:B\to A=[\{f(x),x\}:\{x,f(x)\}\in {{f}^{-1}}]\] If no branch of an inverse trigonometric function is mentioned, then it means the principal value branch of that function. |
A) x
B) x-2x
C) \[2\pi -x\]
D) -x
Correct Answer: B
Solution :
Given, \[\frac{3\pi }{2}\le x\le \frac{5\pi }{2}\] \[\therefore \]\[{{\sin }^{-1}}(\sin x)=x-2\pi \]You need to login to perform this action.
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