Question

Let $y=f\left(x\right)$ be defined parametrically as $y=t^2+t|t|,x=2t-|t|,t\in R$ and $f\left(x\right)=k$ has at least one real solution, then

• A. $k\in R$
• B. $k\in R-\left\{0\right\}$
• C. $k\in R^{+}\cup \left\{0\right\}$
• D. $k\in R^{-}$

Answer 1

The correct option is C $k\in R^{+}\cup \left\{0\right\}$

$y=t^2+t|t|,x=2t-|t|$
Case $1:$ when $t<0$
$\Rightarrow x=3t$ and $y=0,\forall x<0$

Case $2:$ when $t\ge 0$
$\Rightarrow x=t$ and $y=2t^2,\forall x\ge 0$
$\therefore f\left(x\right)=\{\begin{array}{cc}2x^2& ;x\ge 0\\ 0& ;x<0\end{array}$

Since, $f\left(x\right)$ is polynomial and continuous at $x=0$, so continuous in $R.$
$\therefore$ Range of $f\left(x\right)$ is $[0,\infty )$
From $I.V.T.$ : we can say $f\left(x\right)=k$ has at least one solution if $k\in [0,\infty )$