Mathematics - Functions Question with Solution | TestHub

Given, $f\left(1\right)+f\left(2\right)=f\left(3\right)$

Now possible values of $f\left(3\right)$ are $2,3,4,5$ and $6$ (it cannot be $1$ as minimum value $f\left(1\right) \& f\left(2\right)$ is $1$, so $1+1=2$ )

For each $k$, if $f\left(3\right)=k$ then there will be $\left(k-1\right)$ set of values for $f\left(1\right)$ and $f\left(2\right)$

For example if $f\left(3\right)=3$ then possible value of $f\left(1\right) \& f\left(2\right)$ will be $\left(1,2\right) \text{or} \left(2,1\right)\Rightarrow$ $2$ values 

Similarly for $f\left(3\right)=4$, possible values will be $\left(1,3\right),\left(3,1\right),\left(2,2\right)\Rightarrow$$3$ values.

And so on.

So number of possible combinations of $f\left(1\right), f\left(2\right)$ and $f\left(3\right)$ are $1+2+3+4+5=15$

And for each value of $f\left(3\right)$ there will be $6$ possible value of $f\left(4\right)$, so total cases will be $15\times 6=90$