Question

Let \(f:R \to R\) be a function defined by \(f(x) = \sin (2x - 3)\), then \(f\)is
A. Injective
B. Surjective
C. Bijective
D. None of these

Answer 1

Hint: Injective means \[{x_1} \ne {x_2} \Rightarrow f({x_1}) \ne f({x_2})\]or \[{x_1} = {x_2} \Rightarrow f({x_1}) = f({x_2})\]and surjective means range and co-domain of a function are equal, and bijective means injective as well as surjective both.



Complete step by step solution: We know \[f(x) = \sin (2x - 3)\]is a periodic function. 1^st graph of \[\sin x\]and period of this is \[2\pi \]

Then graph of \[\sin (2x - 3)\]and period is \[\pi \]


Here period is \[\pi \]because we know in \[\sin x\]graph \[f(ax) = \dfrac{T}{a}\]where T is a time period , subtraction or addition does not affect graph time period. From graph \[\sin (2x - 3)\]

We see that \[{x_1} \ne {x_2}\]but we get value \[f({x_1}) = f({x_2})\] means for two values of pre-images we get one image, So clearly it is not injective function and it is also not bijective because for bijective it should be 1^st become injective. Now check the surjective condition,
We see that the range of this periodic function varies from -1 to 1 and \[x \in R\]which is a proper subset of the real number (Co-domain), So it is not a surjective also.

Thus, Option (D) is correct.

Note: Student must Know the definition and properties of injective, surjective, one to one, many to one function, etc. also a surjective range is equal to co-domain, and in a many-one function range is not equal to the co-domain.