Which of the following functions is inverse of itself?

Question

Which of the following functions is inverse of itself? (A)  f(t) = ((1-t)/(1+t))  (B)  f(t) = ((1-t2)/(1+t2))  (C)  f(t) = 4log t  (D)  f(t) = 2t

Tardigrade Admin · Accepted Answer

Let y=f (t) ∴ t=f-1(y) Now, y=f (t) =(1-t/1+t) ⇒ y+ty=1-t ⇒ t+ty=1-y ⇒ t=(1-y/1+y) i.e., f-1(y) =(1-y/1+y) or f-1(t)=(1-t/1+t) Thus, this function is inverse of itself

Q. Which of the following functions is inverse of itself?

$f (t) = \frac{( 1 - t )}{( 1 + t )}$

$f (t) = \frac{( 1 - t ^{2} )}{( 1 + t ^{2} )}$

$f (t) = 4^{l o g t}$

$f (t) = 2^{t}$

Let $y = f (t)$
$∴ t = f^{- 1} (y)$
Now, $y = f (t) = \frac{1 - t}{1 + t}$
$\Rightarrow y + t y = 1 - t$
$\Rightarrow t + t y = 1 - y$
$\Rightarrow t = \frac{1 - y}{1 + y}$
i.e., $f^{- 1} (y) = \frac{1 - y}{1 + y}$
or $f^{- 1} (t) = \frac{1 - t}{1 + t}$
Thus, this function is inverse of itself

Q. Which of the following functions is inverse of itself?

f(t)=(1+t)(1−t)​

f(t)=(1+t2)(1−t2)​

f(t)=4logt

f(t)=2t

Let y=f(t) ∴t=f−1(y) Now, y=f(t)=1+t1−t​ ⇒y+ty=1−t ⇒t+ty=1−y ⇒t=1+y1−y​ i.e., f−1(y)=1+y1−y​ or f−1(t)=1+t1−t​ Thus, this function is inverse of itself

$f (t) = \frac{( 1 - t )}{( 1 + t )}$

$f (t) = \frac{( 1 - t ^{2} )}{( 1 + t ^{2} )}$

$f (t) = 4^{l o g t}$

$f (t) = 2^{t}$

Let $y = f (t)$
$∴ t = f^{- 1} (y)$
Now, $y = f (t) = \frac{1 - t}{1 + t}$
$\Rightarrow y + t y = 1 - t$
$\Rightarrow t + t y = 1 - y$
$\Rightarrow t = \frac{1 - y}{1 + y}$
i.e., $f^{- 1} (y) = \frac{1 - y}{1 + y}$
or $f^{- 1} (t) = \frac{1 - t}{1 + t}$
Thus, this function is inverse of itself