JEE MAIN - Mathematics (2018 - 15th April Evening Slot - No. 6)
Let f : A $$ \to $$ B be a function defined as f(x) = $${{x - 1} \over {x - 2}},$$ Where A = R $$-$$ {2} and B = R $$-$$ {1}. Then f is :
invertible and $${f^{ - 1}}(y) = $$ $${{3y - 1} \over {y - 1}}$$
invertible and $${f^{ - 1}}\left( y \right) = {{2y - 1} \over {y - 1}}$$
invertible and $${f^{ - 1}}\left( y \right) = {{2y + 1} \over {y - 1}}$$
not invertible
Explanation
Assume,
y = f(x)
$$ \Rightarrow $$ y = $${{x - 1} \over {x - 2}}$$
$$ \Rightarrow $$ yx - 2y = x - 1
$$ \Rightarrow $$ (y - 1)x = 2y - 1
$$ \Rightarrow $$ x = $${{2y - 1} \over {y - 1}}$$ = f -1(y)
As on the given domain the function is invertible and its inverse can be computed as shown above.
y = f(x)
$$ \Rightarrow $$ y = $${{x - 1} \over {x - 2}}$$
$$ \Rightarrow $$ yx - 2y = x - 1
$$ \Rightarrow $$ (y - 1)x = 2y - 1
$$ \Rightarrow $$ x = $${{2y - 1} \over {y - 1}}$$ = f -1(y)
As on the given domain the function is invertible and its inverse can be computed as shown above.
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