Question

If the function $ f $ and $ g $ are given by $ f(x) = x - [x] $ , and $ g(x) = \dfrac{1}{2}\sin [x] $ where $ [x] $ denotes the integral part of $ x $ and the the range of $ gof $ is $ A $ , the range of $ fog $ is $ B $ , then
A. $ A \subset B $
B. $ B \subset A $
C. $ A = B $
D. $ A \cap B = \varphi $

Answer 1

Hint: The given functions are $ f{\text{ and }}g $ . We have to find the range of the composite functions. $ gof $ and $ fog $ . The function $ gof $ will be achieved by putting the function f inside the function g whereas the function $ fog $ will be found by putting the function g inside the function f , we have to compare their respective ranges and then compare with the given options.

Complete step-by-step answer:
The function are given as,
 $ f(x) = x - [x] $
 $ g(x) = \dfrac{1}{2}\sin [x] $
The composite functions will be given by,
 $ gof = \dfrac{1}{2}\sin [x - [x]] $
 $ fog = \dfrac{1}{2}\sin [x] - [\dfrac{1}{2}\sin [x]] $
(remember that these composite functions are made by just substituting one function as a variable for another function)
The range $ A $ of the function $ gof $ will be $ [0] $ . Because when $ [x - [x]] $ will be solved it's only value can be $ 0 $ .
But the range $ B $ of the function $ fog $ will be both positive numbers and negative real numbers . We can say this because when in the function
 $ fog = \dfrac{1}{2}\sin [x] - [\dfrac{1}{2}\sin [x]] $ is solved it will have two scenarios:
First situation: The value of $ x $ is positive in that case the function $ fog $ will return either zero or return a positive number.
Second Situation: The value of $ x $ is negative, in this case the second term will give zero and the first term will give a negative number.
Hence the range $ B $ is of both positive and negative numbers. Since the range of $ B $ is bigger the option (a) is correct, which is that the first range is a subset of the second range.
So, the correct answer is “Option A”.

Note: The given question although asked us to find the range of the given functions, the range would always had come out to be of discrete nature rather than a continuous one, and for that as well we had to remember the value of radian such as $ \sin [1] $ , where the angle is in radian rather than the degree and hence we have only calculated its rough nature rather than the exact one.