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Home  >>  CBSE XII  >>  Math  >>  Relations and Functions
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Let $\; f:R \rightarrow R$ be given as $\;f(x)=\tan x $. Then $\;f^{-1}(1)$ is

\begin{array}{1 1}(A)\;\frac{\pi}{4} & (B)\;{n\pi+\frac{\pi}{4}:n\in Z}\\(C)\;does\;not\;exist & (D)\;none\; of\; these\end{array}

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • $f(x)=\tan x$
  • To find $f^{-1}(1)$ we check if for what values
  • $(fof^{-1})(1)=1=>f(f^{-1}(1))=1$
  • $or =\tan (f^{-1}(1))=1$
$f:R \to R$
 
$f(x)=\tan x$
 
since $f^{-1} $ is inverse of f
 
$(fof^{-1})(1)=1$
 
$f(f^{-1}(1))=1$
 
$\tan (f^{-1}(1))=1$
 
But $\tan(\frac{\pi}{4})=1$
 
$\tan (f^{-1}(1))=\tan \frac{\pi}{4}$
 
=>$f^{-1}(1)=n\pi+\frac{\pi}{4} \; n \in z$
 
'B' option is correct
 
If $\tan \theta =\tan \alpha$
 
$\theta= n \pi +\alpha \; n \in z$

 

answered Mar 5, 2013 by meena.p
 

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