
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
163.5k+ views
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. 1st 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 1st 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.
Complete step by step solution: We know \[f(x) = \sin (2x - 3)\]is a periodic function. 1st 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 1st 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.
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