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Question

The value of y′′(1) if x32x2y2+5x+y5=0 when y(1)>1, is equal to

A
227
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B
72128
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C
8
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D
82227
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Solution

The correct option is D 82227
Given expression is x32x2y2+5x+y5=0 ...(1)
Differentiating the given expression, we get
3x24xy24x2yy+5+y=0 ....(2)
Putting x=1 in the given expression, we get
32y2+5+y5=02y2y1=0
y=1 or y=12
Now y(1)>0y(1)=1
Differentiating again we get
6x4y28xyy8xyy4x2(y)24x2yy′′+y′′=0
Putting x=1,y=1 and y(1)=43, we get
648(43)8(43)4(169)3y′′(1)=0
y′′(1)=82227

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