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Tangent is drawn at the point (xi ,yi) o...

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____

A

(1, 4)

B

(5, 1/16)

C

(2, 1/2)

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
B

Equation of Tangent at `P(x_(1), y_(1))"of y" = f(x)`
`y - y_(1) = (dy)/(dx)(x-x_(1))`
This tangent cuts the x-axis, where
`x_(2) = x_(1)-(y_(1))/(((dy)/(dx)))`
`because" "x_(1), x_(2), x_(3),.....x_(n)` are in AP
`" "x_(2) - x_(1) = = - (y_(1))/((dy)/(dx))=log_(2)e`
`" "-y = (log_(2)e)(dy)/(dx)`
`therefore" "(dy)/(y) log_(2) e = - dx`
Integrating both sides
`log y = -x log_(e) 2 + log c`
`therefore" "y = c 2^(-x)`
`because" "y = f(x)` passes through (0, 2).
`2 = ce^(0) rArr c = 2`
`therefore" "y = 2^(1-x)`
`therefore" "y(5) = (1)/(16)`
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