JEE Main 2022 — Functions Question with Solution

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 $$=1\times 6=6$$

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

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

$$=90$$