Question

The period of the function $ f(x) = [6x + 7] + \cos \pi x - 6x $ where $ [.] $ denotes the greatest integer function, is
A.3
B. $ 2\pi $
C.2
D.None of these

Answer 1

Hint: In this question, we are given a function containing a greatest integer function and a trigonometric function. A greatest integer function is defined as a function that gives the rounded off value of any function as the answer. For example $ [7.1] = 8,\,[ - 4.9] = - 5 $ , it increases the magnitude of the given value by 1 while keeping the sign the same. We can write a greatest integer function as the sum of the input value and the fractional part, that is, $ [x] = x + \{ x\} $ . For example, $ [7.1] = 7.1 + 0.9,\,[ - 4.9] = - 4.9 + ( - 0.1) $ . Using this information, we can simplify the given equation and then find the period of the whole function.

Complete step by step solution:
We know that the greatest integer function of the sum of a decimal/unknown quantity and an integer is equal to the sum of the greatest integer function of the decimal/unknown quantity and the integer. For example –
 $
  [2.3 + 1] = [3.3] = 3 \\
  [2.3] + 1 = 2 = 3 \;
  $
So, $ [6x + 7] = [6x] + 7 $
We get –
 $
  f(x) = [6x] + 7 + \cos \pi x - 6x \\
   \Rightarrow f(x) = [6x] - 6x + \cos \pi x + 7 \;
  $
We know that –
 $
  [x] = x + \{ x\} \\
   \Rightarrow [x] - x = \{ x\} \;
  $
So, we get –
 $ f(x) = \{ 6x\} + \cos \pi x + 7 $
The period of $ \{ x\} = 1 $
So the period of $ \{ 6x\} = \dfrac{1}{6} $
The period of $ \cos x = 2\pi $
So the period of \[\cos \pi x = \dfrac{{2\pi }}{\pi } = 2\]
7 is a constant, so it doesn’t have any period.
Now, the period of the whole function is equal to the LCM of the above two functions, so period of the above function is 2.

So, the correct answer is “Option C”.

Note: The period of a function is defined as the length of the smallest interval that is being repeated in the graph of that function continuously. We find the LCM of the period of the functions to find the period of the whole function. The LCM is found by first finding the LCM of the numerators of the periods and then finding the HCF of the denominators, $ LCM = \dfrac{{LCM\,of\,numerators}}{{HCF\,of\,denominators}} $ , that’s why the LCM of the period of the two functions in above solution is 2.