QJEEM22Mab689: JEE Main 2022 Mathematics - Class 11 JEEM22Mab689 Solution: The number of one-one functions f : {a, b, c, d} {0, 1, 2,…

Given one-one function

$f:\{ a,b,c,d\} \to \{ 0,1,2,\,\,....\,\,10\}$

and $2f(a) - f(b) + 3f(c) + f(d) = 0$

$\Rightarrow 3f(c) + 2f(a) + f(d) = f(b)$

Case I:

(1) Now let $f(c) = 0$ and $f(a) = 1$ then

$3 \times 0 + 2 \times 1 + f(d) = f(b)$

$\Rightarrow 2 + f(d) = f(b)$

Now possible value of $f(d) = 2,3,4,5,6,7,$ and $8$ .

f(d) can't be 9 and 10 as if $f(d) = 9$ or 10 then $f(b) = 2 + 9 = 11$ or $f(b) = 2 + 10 = 12$ , which is not possible as here any function's maximum value can be 10.

$\therefore$ Total possible functions when $f(c) = 0$ and $f(a) = 1$ are = 7

(2) When $f(c) = 0$ and $f(a) = 2$ then

$3 \times 0 + 2 \times 2 + f(d) = f(b)$

$\Rightarrow 4 + f(d) = f(b)$

$\therefore$ possible value of $f(d) = 1,3,4,5,6$

$\therefore$ Total possible functions in this case = 5

(3) When $f(c) = 0$ and $f(a) = 3$ then

$3 \times 0 + 2 \times 3 + f(d) = f(b)$

$\Rightarrow 6 + f(d) = f(b)$

$\therefore$ Possible value of $f(d) = 1,2,4$

$\therefore$ Total possible functions in this case = 3

(4) When $f(c) = 0$ and $f(a) = 4$ then

$3 \times 0 + 2 \times 4 + f(d) = f(b)$

$\Rightarrow 8 + f(d) = f(b)$

$\therefore$ Possible value of $f(d) = 1,2$

$\therefore$ Total possible functions in this case = 2

(5) When $f(c) = 0$ and $f(a) = 5$ then

$3 \times 0 + 2 \times 5 + f(d) = f(b)$

$\Rightarrow 10 + f(d) = f(b)$

Possible value of f(d) can be 0 but f(c) is already zero. So, no value to f(d) can satisfy.

$\therefore$ No function is possible in this case.

$\therefore$ Total possible functions when $f(c) = 0$ and $f(a) = 1,2,3$ and $4$ are $= 7 + 5 + 3 + 2 = 17$

Case II:

(1) When $f(c) = 1$ and $f(a) = 0$ then

$3 \times 1 + 2 \times 0 + f(d) = f(b)$

$\Rightarrow 3 + f(d) = f(b)$

$\therefore$ Possible value of $f(d) = 2,3,4,5,6,7$

$\therefore$ Total possible functions in this case = 6

(2) When $f(c) = 1$ and $f(a) = 2$ then

$3 \times 1 + 2 \times 2 + f(d) = f(b)$

$\Rightarrow 7 + f(d) = f(b)$

$\therefore$ Possible value of $f(d) = 0,3$

$\therefore$ Total possible functions in this case = 2

(3) When $f(c) = 1$ and $f(a) = 3$ then

$3 \times 1 + 2 \times 3 + f(d) = f(b)$

$\Rightarrow 9 + f(d) = f(b)$

$\therefore$ Possible value of $f(d) = 0$

$\therefore$ Total possible functions in this case = 1

$\therefore$ Total possible functions when $f(c) = 1$ and $f(a) = 0,2$ and $3$ are

$= 6 + 2 + 1 = 9$

Case III:

(1) When $f(c) = 2$ and $f(a) = 0$ then

$3 \times 2 + 2 \times 0 + f(d) = f(b)$

$\Rightarrow 6 + f(d) = f(b)$

$\therefore$ Possible values of $f(d) = 1,3,4$

$\therefore$ Total possible functions in this case = 3

(2) When $f(c) = 2$ and $f(a) = 1$ then,

$3 \times 2 + 2 \times 1 + f(d) = f(b)$

$\Rightarrow 8 + f(d) = f(b)$

$\therefore$ Possible values of $f(d) = 0$

$\therefore$ Total possible function in this case = 1

$\therefore$ Total possible functions when $f(c) = 2$ and $f(a) = 0,1$ are

$= 3 + 1 = 4$

Case IV:

(1) When $f(c) = 3$ and $f(a) = 0$ then

$3 \times 3 + 2 \times 0 + f(d) = f(b)$

$\Rightarrow 9 + f(d) = f(b)$

$\therefore$ Possible values of $f(d) = 1$

$\therefore$ Total one-one functions from four cases

$= 17 + 9 + 4 + 1 = 31$