यदि `sqrt(x^(2)+y^(2))=a*e^(tan^(-1)(x//y)),agt0` हो तो `y''(0)` का मान होगा

Options

A. `(a)/(2)e^(-pi//2)`
B. `ae^(pi//2)`
C. `-(2)/(a)`
D. विद्यमान नहीं है

Answer

C

Explanation

`sqrt(x^(2)+y^(2))``=a*e^(tan^(-1)((y)/(x))),agt0` . . .(i)

दोनों पक्षों का अवकलन करने पर

`(1)/(2)*(1)/(sqrt(x^(2)+y^(2)))(2x+2yy')`

`=ae^(tan^(-1)((y)/(x)))*(1)/((1+(y^(2))/(x^(2))))((xy'-y)/(x^(2)))`

`(x+yy')/(sqrt(x^(2)+y^(2)))=a*e^(tan^(-1)((y)/(x)))xx(x^(2))/((x^(2)+y^(2)))xx(xy'-y)/(x^(2))`

[समी (i) से `a*e^(tan^(-1)((y)/(x)))` का मान रखने पर ]

`rArr(x+yy')/(sqrt(x^(2)+y^(2)))=sqrt(x^(2)+y^(2))xx(x^(2))/((x^(2)+y^(2)))xx(xy'-y)/(x^(2))`

`:.x+yy'=xy'-y`

`rArry'=(x+y)/(x-y)`

`:." "y''=(2(xy'-y))/((x-y)^(2))`

`:.y''(0)`

`=(2(0-y(0)))/({0-y(0)}^(2))`

`=(-2)/(y(0))`

`=(-2)/(a)e^((-pi)/(2))`

[ चूंकि `y(0)=axxe^(pi//2)` समी (i) से]