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For the curve sinx+siny=1 lying in first...

For the curve `sinx+siny=1` lying in first quadrant. If `lim_(xrarr0) x^(alpha)(d^(2)y)/(dx^(2))` exists and non-zero than `2alpha=`

A

3

B

4

C

5

D

1

Text Solution

Verified by Experts

The correct Answer is:
A
`sinx+siny=1`
`rArr" "cosx+cosy y'=0`
`rArr" "-sinx+cosy y''+(y')(-siny)=0`
`rArr" "y''=(y'siny+sinx)/(cosy)`
`=((-(cosx)/(cosy))^(2)siny+sinx)/(cosy)`
`=(sinxcos^(2)y+cos^(2)xsiny)/(cos^(3)y)`
Putting `sin x=t, sin y=1-t`, we get
`y''=(t^(-3//2)(1-t+t^(2)))/((2-t)^(3//2))`
`rArr" "alpha=(3)/(2)`
`rArr" "2alpha=3`
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