Question

Assertion :Let function $f:R\to R$ is such that $f\left(x\right)f\left(y\right)-f\left(xy\right)=x+y$ for all $x,y\in R$
$f\left(x\right)$ is a Bijective function. Reason: $f\left(x\right)$ is a linear function.

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

Consider $f\left(x\right)f\left(y\right)-f\left(xy\right)=x+y$
Let $f\left(x\right)=\lambda +x$ where $\lambda$ is a constant.
Then $f\left(x\right).f\left(y\right)-f\left(xy\right)$
$=(\lambda +x)(\lambda +y)-(\lambda +xy)$
$={\lambda }^2+\lambda (x+y)+xy-(\lambda +xy)$
$=({\lambda }^2-\lambda )+\lambda (x+y)$
$=x+y$
Hence, $({\lambda }^2-\lambda )=(x+y)[1-\lambda ]$
$({\lambda }^2-\lambda )-(x+y)[1-\lambda ]=0$
$({\lambda }^2-\lambda )-(x+y)[\lambda -1]=0$
$(\lambda -1)(\lambda -(x+y\left)\right)=0$
Since $\lambda$ is a constant, hence $\lambda =1$
Thus $f\left(x\right)=1+x$.
Since $f\left(x\right)$ is linear, it is bijective.