The value of parameter t so that the line `(4-t)x+ty+(a^(3)-1)=0` is normal to the curve xy = 1 may lie in the interval

Options

A. `(1,4)`
B. `(-oo,0)uu(4,oo)`
C. `(-4,4)`
D. `[3,4]`

Answer

B

Explanation

Slope of line `(4-t)x+ty+(a^(3)-1)=0" is "(t-4)/(t)`
For `xy=1, (dy)/(dx)=(-y)/(x)=(-1)/(x^(2))`
`therefore" Slope of normal "=-x^(2)=(t-4)/(t)`
As `x^(2)gt0,(t-4)/(t)gt0`
`therefore" "t in (-oo,0)uu(4,oo)`