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Question

The equation of the normal to the curve y(1+x2)=2−x where the tangent crosses x-axis is

A
5xy10=0
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B
x5y10=0
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C
5x+y+10=0
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D
x+5y+10=0
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Solution

The correct option is A 5xy10=0
Given curve can be written as, y=2x1+x2

To get the point where the above curve will cross the x-axis, put y=0

0=2x1+x2x=2

Thus the required point is (2,0)

Now differentiate the given curve using quotient rule,
dydx=(1+x2)(1)(2x)(2x)(1+x2)2

=x24x1(1+x2)2

Thus the slope of a tangent to this curve at point (2,0) is

m=dydx(2,0)=224(2)1(1+22)2=525=15

Therefore the slope of normal at this point will be 5

Hence required equation of normal is given by, (y0)=5(x2)
5xy10=0

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