MCQ
If $f(x) = 3x - 5$, then ${f^{ - 1}}(x)$
  • A
    Is given by $\frac{1}{{3x - 5}}$
  • 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

Answer

Correct option: B.
Is given by $\frac{{x + 5}}{3}$
b
(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 ${f^{ - 1}}(x) = \frac{{x + 5}}{3}$.

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