QJEEM22Ma1f40: JEE Main 2022 Mathematics - Class 11 JEEM22Ma1f40 Solution: The total number of functions, such that , is equal to :

Given, $f(1) + f(2) = f(3)$

It means $f(1),f(2)$ and $f(3)$ are dependent on each other. But there is no condition on $f(4)$ , so $f(4)$ can be $f(4) = 1,2,3,4,5,6$ .

For $f(1),f(2)$ and we have to find how many functions possible which will satisfy the condition $f(1) + f(2) = f(3)$

Case 1 :

When $f(3) = 2$ then possible values of $f(1)$ and $f(2)$ which satisfy $f(1) + f(2) = f(3)$ is $f(1) = 1$ and $f(2) = 1$ .

And $f(4)$ can be = 1, 2, 3, 4, 5, 6

$\therefore$ Total possible functions $=1\times6=6$

Case 2 :

When $f(3) = 3$ then possible values

(1) $f(1) = 1$ and $f(2) = 2$

(2) $f(1) = 2$ and $f(2) = 1$

And $f(4)$ can be = 1, 2, 3, 4, 5, 6.

$\therefore$ Total functions $= 2 \times 6 = 12$

Case 3 :

When $f(3) = 4$ then

(1) $f(1) = 1$ and $f(2) = 3$

(2) $f(1) = 2$ and $f(2) = 2$

(3) $f(1) = 3$ and $f(2) = 1$

And $f(4)$ can be = 1, 2, 3, 4, 5, 6

$\therefore$ Total functions $= 3 \times 6 = 18$

Case 4 :

When $f(3) = 5$ then

(1) $f(1) = 1$ and $f(4) = 4$

(2) $f(1) = 2$ and $f(4) = 3$

(3) $f(1) = 3$ and $f(4) = 2$

(4) $f(1) = 4$ and $f(4) = 1$

And $f(4)$ can be = 1, 2, 3, 4, 5 and 6

$\therefore$ Total functions $= 4 \times 6 = 24$

Case 5 :

When $f(3) = 6$ then

(1) $f(1) = 1$ and $f(2) = 5$

(2) $f(1) = 2$ and $f(2) = 4$

(3) $f(1) = 3$ and $f(2) = 3$

(4) $f(1) = 4$ and $f(2) = 2$

(5) $f(1) = 5$ and $f(2) = 1$

And $f(4)$ can be = 1, 2, 3, 4, 5 and 6

$\therefore$ Total possible functions $= 5 \times 6 = 30$

$\therefore$ Total functions from those 5 cases we get

$= 6 + 12 + 18 + 24 + 30$

$= 90$