Let `y=f(x)` be definded parametrically as `y=t^(2)+t|t|,x=2t-|t|, t in R.` find f(x) and discuss its differentiability ,

Options

A. continuous and differentiable in [-1,1]
B. continuous but not differentiable in [-1,1]
C. continuous in [-1,1] and differentiable in [-1,1] only
D. none of these

Answer

A

Explanation

As `t in R`. So, the following cases arise:
CASE 1: When `t gt 0`
In this case, we have
`x=2t-t=t and y=t^(2)+t^(2)=2t^(2)`
`Rightarrow y=2x^(2), xge 0`
CASE II- When `t lt 0`
In this case, we have
`x=2t+t=3t and y=t^(2)-t^(2)=0`
`Rightarrow y=0"for all "x lt 0`
Thus, we have
`y=f(x)={{:(,2x^(2),x ge 0),(,0, x lt 0):}`
Clearly, f(x) is everywhere continuous and differentiable. Hence, f(x) is continuous and differentiable on [-1,1]