Which of the following statements is false?

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(A) If $f$ is an even function from $R$ to $R$,then $f(0)$ must be equal to $0$. Since we know that if a function $f(x)$ is even,then $f(-x)=f(x)$. No

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If $f$ is an even function from $R$ to $R$,then $f(0)$ must be equal to $0$.

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(A) If $f$ is an even function from $R$ to $R$,then $f(0)$ must be equal to $0$. Since we know that if a function $f(x)$ is even,then $f(-x)=f(x)$. Now,if we assume $f(x)=\cos x$,then $f(-x)=\cos(-x)=\cos x=f(x)$. Thus,$f(x)=\cos x$ is an even function. However,$f(0)=\cos 0=1 
eq 0$. Therefore,the given statement is false. $(b)$ $f: R ightarrow R$,$f(x)=x-[x]$. Since $x=[x]+\{x\}$,where $\{x\}$ is the fractional part function,we have $f(x)=\{x\}$. Since $\{x\}$ is a periodic function,$f(x)$ is a periodic function. Therefore,the given statement is true. $(c)$ If $f: R ightarrow R$ is an odd function,then $f(0)=0$. Since $f(x)$ is an odd function,$f(-x)=-f(x)$. Putting $x=0$,we get $f(0)=-f(0)$,which implies $2f(0)=0$,so $f(0)=0$. Therefore,the given statement is true. $(d)$ The 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 $B=\{1,2\}$,so $n(A)=6$ and $n(B)=2$. The number of onto functions from a set with $m$ elements to a set with $n$ elements is $n^m - \binom{n}{1}(n-1)^m + \binom{n}{2}(n-2)^m - \dots$. For $n=2$ and $m=6$,the number of onto functions is $2^6 - \binom{2}{1}(1)^6 = 64 - 2 = 62$. Therefore,the given statement is true.

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