KCET · Maths · Functions
Let \(f: R \rightarrow R\) be given \(f(x)=\tan x\). Then, \(f^{-1}(1)\) is
- A \(\frac{\pi}{4}\)
- B \(\left\{n \pi+\frac{\pi}{4}: n \in Z\right\}\)
- C \(\frac{\pi}{3}\)
- D \(\left\{n \pi+\frac{\pi}{3}: n \in Z\right\}\)
Answer & Solution
Correct Answer
(A) \(\frac{\pi}{4}\)
Step-by-step Solution
Detailed explanation
\(\because f(x)=\tan x\)
Since, \(f^{-1}\) is inverse of \(f\)
\(\Rightarrow \quad f\left(f^{-1}(l)\right)=1\)
\(\Rightarrow \quad \tan \left(f^{-1}(1)\right)=1\)
But \(\quad \tan \left(\frac{\pi}{4}\right)=1\)
\(\Rightarrow \quad \tan \left(f^{-1}(1)\right)=\tan \left(\frac{\pi}{4}\right)\)
\(\therefore \quad f^{-1}(\mathrm{l})=\frac{\pi}{4}\)
Since, \(f^{-1}\) is inverse of \(f\)
\(\Rightarrow \quad f\left(f^{-1}(l)\right)=1\)
\(\Rightarrow \quad \tan \left(f^{-1}(1)\right)=1\)
But \(\quad \tan \left(\frac{\pi}{4}\right)=1\)
\(\Rightarrow \quad \tan \left(f^{-1}(1)\right)=\tan \left(\frac{\pi}{4}\right)\)
\(\therefore \quad f^{-1}(\mathrm{l})=\frac{\pi}{4}\)
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