A curve is represented parametrically by the equations `x=f(t)=a^(In(b^t))and y=g(t)=b^(-In(a^(t)))a,bgt0 and a ne 1, b ne 1" Where "t in R.`
The value of `(d^(2)y)/(dx^(2))` at the point where f(t)=g(t) is

Options

A. 0
B. `(1)/(2)`
C. 1
D. 2

Explanation

`x=f(t)=a^(In(b^t))=a^(t" In "b)" (1)"`
`y=g(t)=b^(-In(a^(t)))=(b^(In a))^(-t)=(a^(In b))^(-t)=a^(-t" In b")`
`therefore" "y=g(t)=a^(In(b^(-t)))=f(-t)" (2)"`
From equations (1), and (2),
`xy=1`
`f(t)=g(t)rArrf(t)=f(-t)rArrt=0" "[because f(t)" is one-one function"]`
`"At "t=0, x=y=1`
`because" "xy=1, (dy)/(dx)=(-1)/(x^(2))and (d^(2)y)/(dx^(2))=(2)/(x^(3))" At "x=1, (d^(2)y)/(dx^(2))=2`