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प्रश्न
If \[f : R \to R is given by f\left( x \right) = 3x - 5, then f^{- 1} \left( x \right)\]
विकल्प
is given by \[\frac{1}{3x - 5}\]
is given by \[\frac{x + 5}{3}\]
does not exist because f is not one-one
does not exist because f is not onto
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उत्तर
Clearly, f is a bijection.
So, f -1 exists.
\[Let f^{- 1} \left( x \right) = y . . . \left( 1 \right)\] .....(1)
\[ \Rightarrow f\left( y \right) = x\]
\[ \Rightarrow 3y - 5 = x\]
\[ \Rightarrow 3y = x + 5\]
\[ \Rightarrow y = \frac{x + 5}{3}\]
\[ \Rightarrow f^{- 1} \left( x \right) = \frac{x + 5}{3} [\text{from}\left( 1 \right)]\]
So, the answer is (b).
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