Question

A function y=f(x) is defined parametrically as $y=t^2+t|t|,x=2t–|t|,tϵR$. Then at x= 0, f(x) is:

• A. Continuous but not differentiable
• B. Differentiable
• C. Discontinuous
• D. None of these

Answer 1

The correct option is B Differentiable



$y=t^2+t|t|,$
$X=2t-|t|,tϵR.$
$\Rightarrow y=\{\begin{array}{cc}0:& t<0\\ 2t^2:& t\ge 0\end{array}$ and $x=\{\begin{array}{cc}3t:& t<0\\ t:& t\ge 0\end{array}$
$\Rightarrow f\left(x\right)=y=\{\begin{array}{cc}0:& x<0\\ 2x^2:& x\ge 0\end{array}$
$\Rightarrow f^{\prime }\left(x\right)=\{\begin{array}{cc}0:& x<0\\ 4x:& x>0\end{array}$
And Lf'(0) = 0, Rf'(0) = 0 $\Rightarrow$ f '(0) = 0.