JEE Main 2021 — Functions Question with Solution

$f\left(n+1\right)-f\left(n\right)=f\left(1\right)$

A.P. with common difference = $f\left(1\right)$

General term = $T_n = f\left(1\right) + (n-1)f\left(1\right) = nf\left(1\right)$

$\Rightarrow f\left(n\right)=nf\left(1\right)$$\Rightarrow f$ is one-one

Now, Let $f\left(g\left(x_2\right)\right)=f\left(g\left(x_1\right)\right)$

$\Rightarrow g\left(x_2\right)=g\left(x_1\right)$ (as $f$ is one-one)

$\Rightarrow x_1=x_2$ (as $fog$ is one-one)

$\Rightarrow g$ is one-one

For $f$ to be onto, $f\left(n\right)$ should take all the values of natural numbers.

As $f\left(x\right)$ is increasing, $f\left(1\right)=1$

$\Rightarrow f\left(n\right)=n$

Now, $f\left(g\left(n\right)\right)=g\left(n\right) f\left(1\right)$

may be many-one if

$g\left(n\right)$ is many-one

So “if $g$ is onto, then $fog$ is one-one” is incorrect.