Question

The function $y=f\left(x\right)$, defined parametrically as $x=2t-|t-1|$ and $y=2t^2+t\left|t\right|$, is

• A. Continuous and differentiable for $x\in R$
• B. Continuous for $x\in R$ and differentiable for $x\in R-\text{{2}}$
• C. Continuous for $x\in R$ and differentiable for $x\in R-\text{{-1,2}}$
• D. None of these

Answer 1

Answer: (B)

Continuous for $x\in R$ and differentiable for $x\in R-\text{{2}}$

Here, $x=2t-|t-1|$ and $y=2t^2+t\left|t\right|$

Now, when $t<0;$

$x=2t-\{-(t-1)\}=3t-1$

and $y=2t^2-t^2=t^2\Rightarrow y=\frac{1}{9}{(x+1)}^2$

when $0\le t<1;$

$x=2t-(-(t-1))=3t-1$

and $y=2t^2+t^2=3t^2\Rightarrow y=\frac{1}{3}{(x-1)}^2$

when $t>1;$

$x=2t-(t-1)-t+1$

and $y-2t^2+t^2=3t^2\Rightarrow y=3{(x-1)}^2$

Thus, $y=f\left(x\right)=\{\begin{array}{c}\begin{array}{cc}\frac{1}{9}{(x+1)}^2,& x<-1\end{array}\\ \begin{array}{cc}\frac{1}{3}{(x+1)}^2,& -1\le x<2\end{array}\\ \begin{array}{cc}3{(x+1)}^2,& x\ge 1\end{array}\end{array}$

We have to check continuity and differentiability at $x=-1$ and $2.$

Differentiability at $x=-1;$

$L.H.D.=L{f}^{\prime }(-1)=\underset{h\to 0}{lim}\frac{f(-1-h)-f(-1)}{-h}=\underset{h\to 0}{lim}\frac{\frac{1}{9}{(-1-h+1)}^2-0}{-h}=0$

$R.H.D.=R{f}^{\prime }(-1)=\underset{h\to 0}{lim}\frac{f(-1+h)-f(-1)}{h}=\underset{h\to 0}{lim}\frac{\frac{1}{3}{(-1+h+1)}^2-0}{h}=0$

Therefore, $f\left(x\right)$ is differentiable and hence continuous at $x=-1$

Differentiability at $x=2;$

$L.H.D.L{f}^{\prime }\left(2\right)=\underset{h\to 0}{lim}\frac{f(2-h)-f\left(2\right)}{-h}=\underset{h\to 0}{lim}\frac{\frac{1}{3}{(2-h+1)}^2-3}{-h}=2$

$R.H.D.R{f}^{\prime }\left(2\right)=\underset{h\to 0}{lim}\frac{f(2+f)-f\left(2\right)}{h}=\underset{h\to 0}{lim}\frac{3{(2+h-1)}^2-3}{h}=6$

Hence, $f\left(x\right)$ is not differentiable at $x=2.$

Since $L{f}^{\prime }\left(2\right)$ and $R{f}^{\prime }\left(2\right)$ are finite, therefore $f\left(x\right)$ is continuous at $x=2.$

Hence, $f\left(x\right)$is continuous for all $x$ and differentiable for $x$ except $x=2.$