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Question

Let denote a curve y=yx which is in the first quadrant and let the point 1,0 lie on it. Let the tangent to at a point P intersect the y-axis at YP. If PYP has length 1 for each point P on , then which of the following is options is/are correct?


A

y=loge1+1-x2x-1-x2

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B

xy'-1-x2=0

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C

y=loge1+1-x2x+1-x2

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D

xy'+1-x2=0

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Solution

The correct option is D

xy'+1-x2=0


Explanation for correct options:

Determining the correct option:

Draw curve y=yx at point 1,0

Let, Px,y,y=fx

The Equation of the tangent is

Y-y=dydxX-x

Y-y=-xdydxY=y-xdydx

So, 0,y-xdydx=0,YP

From given data

PYP=1x-02+y-xdydx-y2=1x-02+xdydx2=1

Squaring on both the sides

x2+xdydx2=1xdydx2=1-x2xdydx=±1-x2dydx=±1-x2xdy=±1-x2xdx.............(1)

Let,x=sinθ

On differentiating with respect to x,

dx=cosθdθ

Then,

dy=±1-sin2θsinθcosθdθdy=±cosθsinθcosθdθdy=±cos2θsinθdθdy=±1-sin2θsinθdθdy=±1sinθ-sin2θsinθdθdy=±cosecθ-sinθdθ

Integrating on both sides

y=±logcosecθ-cotθ+cosθ+cy=±logcosecθ+cotθ+cosθ+c

We know that,

sinθ=xsin2θ=x21-cos2θ=x2cos2θ=x2-1cosθ=x2-1cotθ=cosθsinθ=1-x2x[cosθ=x2-1andsinθ=x]

Then,

y=±-log1x+1-x2x+1-x2+cy=±-log1+1-x2x+1-x2+c

As the curve lies in the first quadrant so y must be positive

y=log1+1-x2x-1-x2+c

When,

y=fx,1,0atx=1,y=0y1=0

Then,

0=log1+1-121-1-12+c0=0+0+cc=0

Since, y=log1+1-x2x-1-x2

Hence, the correct answer is option (A).

Again using the negative value of equation 1.

dy=-1-x2xdxdydx=-1-x2xxy'+1-x2x=0

Hence, the correct answer is option (D).

Therefore, the correct answers are options (A) and (D).


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