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Question: Let \( f\left( x \right) = \dfrac{{\sin \left( {\left[ x \right]\pi } \right)}}{{{x^2} + 2x + 4}} \)...

Let f(x)=sin⁡([x]π)x2+2x+4f\left( x \right) = \dfrac{{\sin \left( {\left[ x \right]\pi } \right)}}{{{x^2} + 2x + 4}} where [.]=G.I.F.\left[ . \right] = G.I.F. , then which of the following in not true:
(A) f is periodic
(B) f is even
(C) f is many-one
(D) f is onto.

Explanation

Solution

Hint : The given question revolves around the concepts of trigonometry and various types of functions. We are given a complex composite function in the problem including the greatest integer function. So, we first try to simplify the function provided to us. We must know the output of the sine function for integral multiples of π\pi . Then, we classify the function and find the incorrect statement out of the four options.

Complete step by step solution:
So, we are given the function f(x)=sin⁡([x]π)x2+2x+4f\left( x \right) = \dfrac{{\sin \left( {\left[ x \right]\pi } \right)}}{{{x^2} + 2x + 4}} where [.]=G.I.F.\left[ . \right] = G.I.F.
Now, in the composite function provided to us, the expression [x]\left[ x \right] is the greatest integer function of x and thus gives only integral values. Thus, the collective angle of the sine function [x]π\left[ x \right]\pi is always an integral multiple of π\pi .
Also, we know that the sine function assumes zero value at the integral multiple of π\pi .
Hence, the complete complex composite function assumes zero as the value for any value of x.
So, f(x)=sin⁡([x]π)x2+2x+4=0f\left( x \right) = \dfrac{{\sin \left( {\left[ x \right]\pi } \right)}}{{{x^2} + 2x + 4}} = 0 for any value of x.
Thus, the given composite function is simplified as a constant function assuming zero value for every value of variable x.
Hence, the given function f(x)=sin⁡([x]π)x2+2x+4f\left( x \right) = \dfrac{{\sin \left( {\left[ x \right]\pi } \right)}}{{{x^2} + 2x + 4}} is a many-one function.
Also, the function f has the same value for all values of x as it is a constant function. So, it assumes the same value for negative and positive numbers of the same magnitude. Thus, f(−x)=f(x)f\left( { - x} \right) = f\left( x \right) .
Hence, the function f is even function.
Also, the output of the function f is repeated for every value of x. So, the function f is periodic.
But the range of the function f consists of only one number 00 . So, the function is not onto. Hence, this statement is incorrect.
Thus, option (D) is correct.
So, the correct answer is “Option D”.

Note : We must know the classification of functions and the properties of all types of functions in order to solve the problem. We also must know about some simple trigonometric formulae and identities to solve more questions of this type. One should be open to innovative solutions and should possess creative thinking to find the range of the function and classify it as a constant function.