Question

Statement-1 All points of intersection of $y=f\left(x\right)$ and $y=f^{-1}\left(x\right)$ lies on $y=x$ only
Statement-2 if point P(lies on $y=f\left(x\right)$, then $Q(\alpha ,\beta )$ lies on $y=f^{-1}\left(x\right)$
Statement-3 Inverse of invertible functin is unique and its range is equal to the function domain.
Which of the following option is correct for above statement in order

• A. $TTF$
• B. $FTT$
• C. $TTT$
• D. $TFT$

Answer 1

The correct option is D $TFT$

$\text{Taking of statement - I}$

$\text{All point of intersection of}f\left(x\right)\text{and}$

$f^{\prime }\left(x\right)\text{lies on}y=x\text{line.}$

Therefore statement I is true.

$\text{Statement - II}$

$\text{It is not necessary for any point}$

$Q(\alpha ,\beta )\text{to lie on}{f}^{\prime }\left(x\right)\text{.}$

Therefore statement II is false.

$\text{Statement - III}$

$\text{If}{f}^{-1}\left(x\right)\text{is the inverse of}f\left(x\right)\text{,}$

Then,

Domain of $f\left(x\right)$ is range of $f^{-1}\left(x\right)$

and Range of $f\left(x\right)$ is domain of $f^{-1}\left(x\right)$

Therefore statement III is true.

Option D = TFT