Question

If $f(x) = [x]$, where [x]denotes the greater integers and $g(x) = \cos (\pi x)$then the range of gof is
A. \left\\{ 0 \right\\}
B. \left\\{ { - 1,1} \right\\}
C. \left\\{ { - 1,0,1} \right\\}
D. \left\\{ {x: - 1 \leqslant x < 1} \right\\}

Answer 1

If $f:A \to B\& g:C \to D$then $gof:A \to D$is defined as $gof(x) = g(f(x))$providing range(f)$\subseteq$Domain(g).
in real functions case f and g if the range of f is not a subset of the domain of f, then gof can be defined for those elements in the domain of f which have their reflection in the domain of g.

Complete step by step answer:

Let’s start with what is given to us, They have given,
$\Rightarrow f(x) = [x]$and
$\Rightarrow g(x) = \cos (\pi x)$, where x can have any integer value.

Therefore, f(x) can also have integer value for all $x \in R.$
As, $\Rightarrow f(x) = [x]$

Therefore, the Range of f(x) can have a greater integer value for all $x \in R.$

Also, $g(x) = \cos (\pi x)$
And $- 1 \leqslant \cos \theta < 1$ for all $x \in R.$

$\begin{gathered} \Rightarrow gof = g(f(x)) \\\ \Rightarrow gof = \cos (\pi f(x)) = \cos ([x]) \\\ \Rightarrow gof = \cos (\pi x) \\\ \end{gathered}$
Where $\pi x$is the integral multiple of x and can be lie between -1 and 1.
Range of gof is \left\\{ { - 1,1} \right\\}.
So, option B is the correct answer.

Note: Algorithm to solve is given step by step as follows:
Step 1: - f(x) will be given in the question and put $y = f(x)$
Step 2: - Solve the equation in terms of y, $y = f(x)$ for x.
Step 3: - For all values of x, find the values of y which are real and in the domain of f.
Step 4: - Hence, the range is the values of y obtained in step 3.

1. If Range(f)$\cap$Domain(g)$= \Phi$, then gof does not exist.
2. Both fog and gof are bijections if f and g are bijections.
3. For any two real functions f and g, then fog exists but gof may not exist, there is the possibility they may not be equal even if both exist.