Tangent is drawn at the point `(x_i ,y_i)` on the curve `y=f(x),` which intersects the x-axis at `(x_(i+1),0)` . Now, again a tangent is drawn at `(x_(i+1,)y_(i+1))` on the curve which intersect the x-axis at `(x_(i+2,)0)` and the process is repeated `n` times, i.e. `i=1,2,3dot,ndot` If `x_1,x_2,x_3,ddot,x_n` from an arithmetic progression with common difference equal to `(log)_2e` and curve passes through `(0,2)dot` Now if curve passes through the point `(-2, k),` then the value of `k` is____

Answer

8

Explanation

Equation of tangent at `P(x_(1),y_(1))` of `y=f(x)` is
`y-y_(1) =(dy)/(dx)(x-x_(1))`
This tangent cuts the x-axis. So,
`x_(2)=x_(1)-y_(1)/((dy)/(dx))`
`x_(1),x_(2),x_(3),………..,x-(n)` are in A.P.
`x_(2)-x_(1)=-y_(1)/((dy)/(dx))=log_(2)e` (Given)
or `-y=log_(2)e(dy)/(dx)`
or `(dy)/ylog_(2)e=-dx`
Integrating both sides, we get
`log_(e)y=-xloge^(2)+c`
Since `y=f(x)` passes through (0,2),
`k=2`
`therefore y=2.e^(-xlog_(e)2)`
`therefore y=e^(1-x)`