Let functions are defined from set A to set B where B = $\{\alpha, \beta\}$ and $\alpha , \beta$ are the roots of the equation $t^2 - \sqrt 2 t - \pi = 0$ , then the number of functions which are
(A) discontinuous only at each even integers if $A = [0, 11]$ is $682$
(B) discontinuous only at each odd integer if $A = [0, 11]$ is $243$
(C) discontinuous only at prime numbers if $A = [0, 11]$ is $81$
(D) discontinuous only at $x = 5$k, ($k \in \mathbb Z^+$) if A = [0, 11] is $27$
Attempt:
For A,
There are only 2 methods through which $0$ can be discontinuous. Now through $0$ direction gets defined. After that there are 3 possibilities for each of $2,4,6,8,10 $ to be discontinuous. Hence the total number of functions is: $2\times 3^5$
For B ,
There are 6 ways through which function at $1$ can be discontinuous. Again after getting discontinuous at 1 direction gets defined. By direction, I mean after the point 1 which path it will follow $\alpha$ or $\beta$. Subsequently, there are again 3 ways for each of $3,5,7,9$ but only one way for $11$. Hence total number of ways = $6 \times 3^4 = 2 \times 3^5$
For C,
Again, there are $6$ ways for $2$ to be discontinuous, and 3 ways for each $3, 5,7$ and only one way for $11$. Hence number of ways = $6 \times 3^3 = 2\times 3^4$
For D,
$5$ can be discontinuous in $6$ ways and 10 in $3$ ways. Thus total number of ways = $6 \times 3 = 18$
So none of my answers match A B C or D.
But the answer given is $B,C$.
Can someone please tell me where I have gone wrong or overcounted?
This is the official solution given:
My answer matches the answer of A. But I really can't figure out my mistake in B C and D.
Your answers look correct to me.
There are two possibilities ($\alpha$ or $\beta$) for the value at $0$.
There are three possibilities for a point in $(0,11)$ to be discontinuous as follows :
$\qquad\quad$
There is only one possibility for each of the endpoints ($0$ and $11$) to be discontinuous.
For $(A)$ : The function is discontinuous only at $0,2,4,6,8,10$ where $0$ is the end point, so the the number of the functions is $$2\times 1\times 3^5=486$$
For $(B)$ : The function is discontinuous only at $1,3,5,7,9,11$ where $11$ is the end point, so the the number of the functions is $$2\times 3^5\times 1=486$$
For $(C)$ : The function is discontinuous only at $2,3,5,7,11$ where $11$ is the end point, so the the number of the functions is $$2\times 3^4\times 1=162$$
For $(D)$ : The function is discontinuous only at $5,10$, so the the number of the functions is $$2\times 3^2=18$$
The official solutions for $(B)(C)(D)$ are incorrect since the number of the functions in each case has to be, at least, even. This is because the number of the functions such that the value at $0$ is $\alpha$ is the same as the number of the functions such that the value at $0$ is $\beta$.