Question

Which of the following statements are incorrect?
I If $f\left(x\right)$ and $g\left(x\right)$ are one to one then $f\left(x\right)$ + $g\left(x\right)$
II If $f\left(x\right)$ and $g\left(x\right)$ are one to one then $f\left(x\right).g\left(x\right)$
III If $f\left(x\right)$ is odd then it is necessarily one to one.

• A. I and II only
• B. II and III only
• C. I, II and III
• D. None of the above

Answer 1

The correct option is C I, II and III

Let $f\left(x\right)=ax$ and $g\left(x\right)=-bx$
Both are injective, functions
$f\left(x\right)+g\left(x\right)=ax-bx$
$=x(a-b)$
Now for a special case where $a=b$ we get $f\left(x\right)+g\left(x\right)=0$, hence a constant function.
Therefore, A is incorrect.
$f\left(x\right)g\left(x\right)$
$=-ab{x}^2$
Now we know that functions of type $y=x^2$ or $y=-x^2$ are many-one functions.
That is $f\left(x_1\right)=f\left(x_2\right)$
$\to x_1=\pm x_2$ ...(hence many on one function).
Consider $f\left(x\right)=sin\left(x\right)$
$f(-x)=-f\left(x\right)$, since sin(x) is odd function.
However it is periodic with a period of $2\pi$
Hence it is not one on one function.
Hence all of the above are incorrect.