Question

If \[f(x) = \dfrac{{\sin ([x]\pi )}}{{{x^2} + x + 1}}\] where [.] denotes the greatest integer function, then
A. \[{\rm{f}}\] is one-one
B. \[{\rm{f}}\] is not one-one and non-constant
C. \[{\rm{f}}\] is a constant function
D. None of these

Answer 1

Hint: The constant function f is defined by the equation: \[f(x) = \dfrac{{\sin ([x]\pi )}}{{{x^2} + x + 1}}\]. This means that for every \[x\] value, \[f\] will always be equal to \[ - {\rm{ }}sin\left( {\left[ x \right]\pi } \right)\] and thus have a value of \[1\]. A function with a single element in its range is said to be a constant function. In other words, the function's output value is constant regardless of the input value inside its domain.

Complete step by step solution: We have \[F(x) = \dfrac{{\sin ([x]\pi )}}{{{x^2} + x + 1}}\]
We have give \[[x]\pi \], we know that the greatest integer \[x\] value always be an integer.
So, we can write \[[x]\pi \] as \[n\pi \].
\[ \Rightarrow f(x) = \dfrac{{\sin (n\pi )}}{{{x^2} + x + 1}}\]
Use the trivial identity:
\[\sin (\pi ) = 0\]
\[ = 0\]
Hence, \[\sin (\pi )\]is always zero.
\[\sin ([x]\pi ) = 0\]
\[\therefore f(x) = 0\quad [x]\] is an integer
Implies \[f(x)\] is a constant function and also \[f(x)\] is a zero function.
\[\therefore f(x) = 0\]
\[\therefore f\] is a constant function.

So, option C is correct.

Note: The students are most likely confused because the If statement does not appear to be a function. In order for a function to exist, there must be an input and output. When evaluating this equation, there is no apparent input or output so the statement cannot be classified as a function. To find a constant function using calculus rather than geometry. Calculus is an extremely powerful tool that can be used for many purposes, but it's not always suitable for solving problems in geometry. In fact, most constants found in calculus – such as pi and e – were discovered through methods developed for mathematical physics or differential equations.