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Let $f: R \rightarrow R$ and $g: R \rightarrow R$ be the functions defined by $f(x)=\frac{x}{1+x^2}$, $x \in R, g(x)=\frac{x^2}{1+x^2}, x \in R$
Then, the correct statement (s) among the following is/are :
(a) both $f . g$ are one-one
(b) both $f . g$ are onto
(c) both $f . g$ are not one-one as well as not onto
(d) $f$ and $g$ are onto but not one-one
Options:
Then, the correct statement (s) among the following is/are :
(a) both $f . g$ are one-one
(b) both $f . g$ are onto
(c) both $f . g$ are not one-one as well as not onto
(d) $f$ and $g$ are onto but not one-one
Solution:
2278 Upvotes
Verified Answer
The correct answer is:
C
We have,
$$
\begin{aligned}
& f(x)=\frac{x}{1+x^2}, x \in R \\
& g(x)=\frac{x^2}{1+x^2}, x \in R \\
& f^{\prime}(x)=\frac{1+x^2-2 x^2}{\left(1+x^2\right)^2} \\
& f^{\prime}(x)=\frac{1-x^2}{\left(1+x^2\right)^2}
\end{aligned}
$$
Clearly $f(x)$ is not monotonic.
$\therefore f(x)$ is not one-one function range of $f(x)$, is
$$
\left[-\frac{1}{2}, \frac{1}{2}\right]
$$
$\therefore f(x)$ is not onto
Clearly $g(x)$ is even function
$\therefore g(x)$ is not one-one function
Range of $g(x)$ is $[0,1]$
$\therefore g(x)$ is also not onto. Here, $f(x)$ and $g(x)$ both are neither one-one not onto.
$$
\begin{aligned}
& f(x)=\frac{x}{1+x^2}, x \in R \\
& g(x)=\frac{x^2}{1+x^2}, x \in R \\
& f^{\prime}(x)=\frac{1+x^2-2 x^2}{\left(1+x^2\right)^2} \\
& f^{\prime}(x)=\frac{1-x^2}{\left(1+x^2\right)^2}
\end{aligned}
$$
Clearly $f(x)$ is not monotonic.
$\therefore f(x)$ is not one-one function range of $f(x)$, is
$$
\left[-\frac{1}{2}, \frac{1}{2}\right]
$$
$\therefore f(x)$ is not onto
Clearly $g(x)$ is even function
$\therefore g(x)$ is not one-one function
Range of $g(x)$ is $[0,1]$
$\therefore g(x)$ is also not onto. Here, $f(x)$ and $g(x)$ both are neither one-one not onto.
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