Which of the following is false?

(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.