Question

Consider $f\left(x\right)=\frac{x^2}{1+x^2}$ and $g\left(x\right)=px$ where $p\in R.$ Then which of the following statements is (are) CORRECT?

• A. If $f\left(x\right)=g\left(|x|\right)$ has exactly three distinct solutions, then number of all possible real values of $p$ is $1.$
• B. If $f\left(x\right)=g\left(|x|\right)$ has exactly five distinct solutions, then range of all possible real values of $p$ is $(0,\frac{1}{2}).$
• C. Number of distinct tangents that can be drawn to the curve $y=f\left(x\right)$ from the origin is $3.$
• D. Number of distinct tangents that can be drawn to the curve $y=f\left(x\right)$ from the origin is $1.$

Answer 1

Answer: (A), (B), (C)

The correct options are
A If $f\left(x\right)=g\left(|x|\right)$ has exactly three distinct solutions, then number of all possible real values of $p$ is $1.$
B If $f\left(x\right)=g\left(|x|\right)$ has exactly five distinct solutions, then range of all possible real values of $p$ is $(0,\frac{1}{2}).$
C Number of distinct tangents that can be drawn to the curve $y=f\left(x\right)$ from the origin is $3.$

$f\left(x\right)\ge 0$
$\underset{x\to \infty }{lim}f\left(x\right)=1$
So, range of $f$ is $[0,1)$

Clearly, $f\left(x\right)=g\left(|x|\right)$ has $x=0$ as one solution.
$f\left(x\right)=g\left(|x|\right)$ has three solutions iff each branch of $g\left(|x|\right)=p|x|,p>0$ is tangent to $y=f\left(x\right)$
$\Rightarrow f^{\prime }\left(x\right)=p$
$\Rightarrow \frac{2x}{(1+x^2{)}^2}=p$

Intersection point of $f\left(x\right)=g\left(|x|\right):$
$\frac{x^2}{1+x^2}=\frac{2x^2}{(1+x^2{)}^2}$
$\Rightarrow 1+x^2=2$
$\Rightarrow x=\pm 1$
$\Rightarrow p=\frac{1}{2}$ as $p>0$

If $p>\frac{1}{2},$ $y=f\left(x\right)$ will not intersect $y=g\left(|x|\right)$ and hence, there will be no solution.
For five distinct solutions, $p\in (0,\frac{1}{2})$