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1.Relation and Function
easy
If $f(x) = 3x - 5$, then ${f^{ - 1}}(x)$
A
Is given by $\frac{1}{{3x - 5}}$
B
Is given by $\frac{{x + 5}}{3}$
C
Does not exist because $f$ is not one-one
D
Does not exist because $f$ is not onto
(IIT-1998)
Solution
(b) Let $f(x) = y\,\, \Rightarrow \,\,x = {f^{ – 1}}(y).$
Hence$f(x) = y = 3x – 5\,\, $
$\Rightarrow \,\,x = \frac{{y + 5}}{3}\, \Rightarrow \,{f^{ – 1}}(y) = x = \frac{{y + 5}}{3}$
$\therefore \,\,{f^{ – 1}}(x) = \frac{{x + 5}}{3}$
Also $f$ is one-one and onto,
so ${f^{ – 1}}$ exists and is given by ${f^{ – 1}}(x) = \frac{{x + 5}}{3}$.
Standard 12
Mathematics
