For the functions defined parametrically by the equations
`f(t)=x={{:(2t+t^(2)sin.(1)/(t),,,tne0),(0,,,t=0):}` and
`g(t)=y={{:((1)/(t)"sint"^(2),t ne0),(0,t =0):}`

Options

A. equation of tangent at t = 0 is `x-2y=0`
B. equation of normal at t = 0 is `2x+y=0`
C. tangent does not exist at t = 0
D. normal does not exist at `t = 0`

Answer

A, B

Explanation

It can be easily checked that function is continuous and differentiable at x = 0.
Now `(dy)/(dx)=(g'(t))/(f'(t))`
`g'(0)=underset(trarr0)(lim)((g(t)-g(0))/(t-0))=underset(trarr0)(lim)(((1)/(t)sint^(2)-0)/(t))=1`
`f'(0)=underset(t rarr0)(lim)((f(t)-f(0))/(t-0))=underset(t rarr0)(lim)(2t+t^(2)sin.(1)/(t))/(t)=2+0=2`
`rArr" "(dy)/(dx)=(1)/(2)`
So the equation of tangent `y-0=(1)/(2)(x-0)`
and equation of normal `y-0=-2(x-0)`