If $f$ is an even function from $R$ to $R$, then $f(0)$ must be equal to 0 .
$f: R \rightarrow R$ defined by $f(x)=x-[x], \forall x \in R$, where $[x]$ is the greatest integer not greater than $x$, is a periodic function
If $f: R \rightarrow R$ is an odd function, then $f(0)=0$
Number of onto functions from $\{1,2,3,4,5,6\}$ to $\{1,2\}$ is 62
Solution
(a) If $f$ is an even function from $R$ to $R$, then $f(0)$ must be equal to 0 .
$\because$ We know that if a function $f(x)$ is even, then $f(-x)=f(x)$
Now, if we assume $f(x)=\cos x$
$
\begin{aligned}
f(-x) & =\cos (-x) \\
& =\cos x \\
& =f(x)
\end{aligned} \quad\{\because \cos (-\theta)=(\cos \theta)\}
$
$\therefore f(x)=\cos x$ is an even function.
Now, $f(0)=\cos 0=1 \neq 0$
The given statement is false.
(b) $f: R \rightarrow R, f(x)=x-[x]$
$\because$ We know that, $x=[x]+\{x\}$
where $\{x\}=$ fractional part function
$
\begin{aligned}
& \Rightarrow \quad x-[x]=\{x\} \\
& \therefore \quad f(x)=\{x\} \\
&
\end{aligned}
$
where $\{x\}$ is a periodic function.
$\Rightarrow f(x)$ is periodic function.
$\therefore$ The given statement is true.
(c) $f: R \rightarrow R$ is an odd function, then $f(0)=0$
$\because$ We know that if $f(x)$ is an odd function, then
$
f(-x)=-f(x)
$
Put $x=0$, we get
$
\begin{aligned}
f(0) & =-f(0) \\
\Rightarrow \quad f(0)+f(0) & =0 \\
2 f(0) & =0 \Rightarrow f(0)=0
\end{aligned}
$
The given statement is true.
(d) Number of onto functions from $\{1,2,3,4,5,6\}$ to $\{1,2\}$ is 62 .
Let $A=\{1,2,3,4,5,6\}$ and $B=\{1,2\}$
$
\because \Rightarrow n(A)=6 \text { and } n(B)=2
$
$\because$ We know that the number of onto function from a set $A$ with $m$ number of elements to set $B$ with $n$ number of elements is
$
n^m-\left\{{ }^n C_1(n-1)^m+{ }^n C_2(n-2)^m+\ldots{ }^n C_{n-1}(1)^m\right\}
$
Here, $n=2$ and $m=6$
So, total number of onto function are
$
\begin{aligned}
& 2^6-\left\{{ }^2 C_1(2-1)^6+{ }^2 C_2(2-2)^6\right] \\
& =64-[2+0] \\
& =64-2=62
\end{aligned}
$
$\therefore$ The given statement is true