$I f f : D \to R$ be the function with domain $D = {x : - \frac{π}{2} < x < \frac{π}{2}} a n d f (x) = 3 + 4 x, R$ being the set of all real, then which one of the following statement is correct ?

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f is not one – one but onto on R

f is one – one but not onto on R

f is neither one – one nor onto on R

f is one – one as well as onto on R

answer is B.

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Detailed Solution

$\begin{array}{l} f (x) = 3 + 4 x \\ L e t x_{1}, x_{2} \in D a n d f (x_{1}) = f (x_{2}) \\ \Rightarrow 3 + 4 x_{1} = 3 + 4 x_{2} \\ \Rightarrow 4 x_{1} = 4 x_{2} \\ \Rightarrow x_{1} = x_{2} \\ f : D \to R i s o n e - o n e \\ - \frac{π}{2} < x < \frac{π}{2} \Rightarrow - 2 π < 4 x < 2 π \\ \Rightarrow 3 - 2 π < 3 + 4 x < 3 + 2 π \\ R a n g e \neq C o d o m a i n \\ f i s n o t o n t o \end{array}$