Question

if have a function as $f\left( x \right)=\cos \left[ 2\pi \right]x+\cos \left[ -2\pi \right]x$, where $\left[ x \right]$ stands for the greatest integer function, then This question has multiple correct options
A. $f\left( \dfrac{\pi }{2} \right)=-1$
B. $f\left( \pi \right)=+1$
C. $f\left( -\pi \right)=0$
D. $f\left( \dfrac{\pi }{4} \right)=1$

Answer 1

Hint: We will first calculate the values of $\left[ 2\pi \right],\left[ -2\pi \right]$ using the values of $\pi $ as $3.14$ and from the given data $\left[ x \right]$ stands for the greatest integer function. Now we will substitute those values in the given function. In the problem they mentioned that the question has multiple correct answers, so we will check each option for the obtained function.

Complete step-by-step solution
Given that,
$f\left( x \right)=\cos \left[ 2\pi \right]x+\cos \left[ -2\pi \right]x$
Substituting the value of $\pi =3.14$ to find the value of $\left[ 2\pi \right]$, then
$\begin{align}
  & \left[ 2\pi \right]=\left[ 2\times 3.14 \right] \\
 & =\left[ 6.28 \right]
\end{align}$
Given that $\left[ x \right]$stands for the greatest integer function, then greatest integer value of $6.28$ is $6$, hence the value of $\left[ 6.28 \right]$ is $6$. i.e.
$\left[ 2\pi \right]=6....\left( \text{i} \right)$
Substituting the value of $\pi =3.14$ to find the value of $\left[ -2\pi \right]$, then
$\begin{align}
  & \left[ -2\pi \right]=\left[ -2\times 3.14 \right] \\
 & =\left[ -6.28 \right]
\end{align}$
Given that $\left[ x \right]$stands for the greatest integer function, then greatest integer value of $-6.28$ is $-7$, hence the value of $\left[ -6.28 \right]$ is $-7$. i.e.
$\left[ -2\pi \right]=-7....\left( \text{ii} \right)$
From equations $\left( \text{i} \right)$ and $\left( \text{ii} \right)$, substituting the values of $\left[ 2\pi \right],\left[ -2\pi \right]$ in given function, then
$\begin{align}
  & f\left( x \right)=\cos \left[ 2\pi \right]x+\cos \left[ -2\pi \right]x \\
 & =\cos \left( 6 \right)x+\cos \left( -7 \right)x
\end{align}$
We know that $\cos \left( -\theta \right)=\cos \theta $, then
$f\left( x \right)=\cos 6x+\cos 7x$
Hence the simplified form of the given function is $f\left( x \right)=\cos 6x+\cos 7x$
The values of
$\begin{align}
  & f\left( \dfrac{\pi }{2} \right)=\cos 6\left( \dfrac{\pi }{2} \right)+\cos 7\left( \dfrac{\pi }{2} \right) \\
 & =\cos 3\pi +\cos \left( \dfrac{7\pi }{2} \right) \\
 & =\cos \left( 2\pi +\pi \right)+\cos \left( 5\pi +\dfrac{\pi }{2} \right) \\
 & =\cos \left( 2\pi +\pi \right)+\cos \left( 2\left( 2\pi \right)+\pi +\dfrac{\pi }{2} \right)
\end{align}$
We know that $\cos \left( 2n\pi +\theta \right)=\cos \theta $, where $n=1,2,3,4...$, then
$f\left( \dfrac{\pi }{2} \right)=\cos \pi +\cos \left( \dfrac{3\pi }{2} \right)$
We know that the values of $\cos \pi $ as $-1$ and $\cos \left( \dfrac{3\pi }{2} \right)$ as $0$, then
$\begin{align}
  & f\left( \dfrac{\pi }{2} \right)=-1+0 \\
 & f\left( \dfrac{\pi }{2} \right)=-1.....\left( \text{a} \right)
\end{align}$
Now the value of $f\left( \pi \right)$ is
$\begin{align}
  & f\left( \pi \right)=\cos 6\pi +\cos 7\pi \\
 & =\cos \left( 2\left( 3 \right)\pi \right)+\cos \left( 2\left( 3 \right)\pi +\pi \right)
\end{align}$
We know that $\cos \left( 2n\pi +\theta \right)=\cos \theta $ and $\cos \left( 2n\pi \right)=1$ for $n=1,2,3,4...$, then
$f\left( \pi \right)=1+\cos \pi $
We know that the value of $\cos \pi $ as $-1$, then
$\begin{align}
  & f\left( \pi \right)=1-1 \\
 & =0.....\left( \text{b} \right)
\end{align}$
Now the value of $f\left( -\pi \right)$ is
$f\left( -\pi \right)=\cos 6\left( -\pi \right)+\cos 7\left( -\pi \right)$
We know that $\cos \left( -\theta \right)=\cos \theta $, then
$\begin{align}
  & f\left( -\pi \right)=\cos 6\pi +\cos 7\pi \\
 & f\left( -\pi \right)=f\left( \pi \right) \\
 & f\left( -\pi \right)=0....\left( \text{c} \right)
\end{align}$
Now the value of $f\left( \dfrac{\pi }{4} \right)$ is
$\begin{align}
  & f\left( \dfrac{\pi }{4} \right)=\cos 6\left( \dfrac{\pi }{4} \right)+\cos 7\left( \dfrac{\pi }{4} \right) \\
 & =\cos \left( \dfrac{3\pi }{2} \right)+\cos \left( \pi +\dfrac{3\pi }{4} \right)
\end{align}$
Substituting the values of $\cos \left( \dfrac{3\pi }{2} \right)=0$ and $\cos \left( \dfrac{7\pi }{4} \right)=\dfrac{1}{\sqrt{2}}$ then
$\begin{align}
  & f\left( \dfrac{\pi }{4} \right)=0+\dfrac{1}{\sqrt{2}} \\
 & =\dfrac{1}{\sqrt{2}}.....\left( \text{d} \right)
\end{align}$
From the equations $\left( \text{a} \right),\left( \text{b} \right),\left( \text{c} \right),\left( \text{d} \right)$ we can say that options $\text{A,C}$ are the correct options.

Note: The students may make mistakes when they are taking the greatest integer of $-6.28$. The greatest Integer of $-6.28$ is $7$, don’t take it as $6$. If you take it as $6$, then the whole solution will be deflected from the correct way. You can also refer to the below picture to check the greatest integer values