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QJEEAdv19Maae48: JEE Advanced 2019 Mathematics - Class 12 JEEAdv19Maae48 Solution: Let denote a curve y = y(x) which is in the first quadrant and…

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

Multiple Choice Options

  • A xy′+1−x2=0xy' + \sqrt {1 - {x^2}} = 0
  • B xy′−1−x2=0xy' - \sqrt {1 - {x^2}} = 0
  • C y=log⁡e(1+1−x2x)−1−x2y = {\log _e}\left( {{{1 + \sqrt {1 - {x^2}} } \over x}} \right) - \sqrt {1 - {x^2}}
  • D y=−log⁡e(1+1−x2x)+1−x2y = - {\log _e}\left( {{{1 + \sqrt {1 - {x^2}} } \over x}} \right) + \sqrt {1 - {x^2}}

Step-by-Step Verified Solution

Let a point P(h, k) on the curve y = y(x), so equation of tangent to the curve at point P is

y−k=(dydx)h,k(x−h)y - k = {\left( {{{dy} \over {dx}}} \right)_{h,k}}(x - h) ....(i)

Now, the tangent (i) intersect the Y-axis at Yp, so coordinates Yp is (0,k−hdydx)\left( {0,k - h{{dy} \over {dx}}} \right) ,

where dydx{{{dy} \over {dx}}} = (dydx){\left( {{{dy} \over {dx}}} \right)} (h,k)

So, PYp = 1 (given)

⇒h2+h2(dydx)2=1\Rightarrow \sqrt {{h^2} + {h^2}{{\left( {{{dy} \over {dx}}} \right)}^2}} = 1

⇒dydx=±1−x2x\Rightarrow {{dy} \over {dx}} = \pm {{\sqrt {1 - {x^2}} } \over x}

[on replacing h by x]

⇒dy=±1−x2xdx\Rightarrow dy = \pm {{\sqrt {1 - {x^2}} } \over x}dx

On puting x = sin θ\theta , dx = cos θ\theta d θ\theta , we get

dy=±1−sin⁡2θsin⁡θcos⁡θdθdy = \pm {{\sqrt {1 - {{\sin }^2}\theta } } \over {\sin \theta }}\cos \theta d\theta

=±cos⁡2θsin⁡θdθ= \pm {{{{\cos }^2}\theta } \over {\sin \theta }}d\theta

=±(cos⁡ecθ−sin⁡θ)dθ= \pm (\cos ec\theta - \sin \theta )d\theta

⇒y=±[1n(cos⁡ecθ−cot⁡θ)+cos⁡θ]+C\Rightarrow y = \pm [1n(\cos ec\theta - \cot \theta ) + \cos \theta ] + C

⇒y=±[1n(1−cos⁡θsin⁡θ)+cos⁡θ]+C\Rightarrow y = \pm \left[ {1n\left( {{{1 - \cos \theta } \over {\sin \theta }}} \right) + \cos \theta } \right] + C

⇒y=±[1n(1−1−sin⁡2θsin⁡θ)+1−sin⁡2θ]+C\Rightarrow y = \pm \left[ {1n\left( {{{1 - \sqrt {1 - {{\sin }^2}\theta } } \over {\sin \theta }}} \right) + \sqrt {1 - {{\sin }^2}\theta } } \right] + C

⇒y=±[1n(1−1−x2x)+1−x2]+C\Rightarrow y = \pm \left[ {1n\left( {{{1 - \sqrt {1 - {x^2}} } \over x}} \right) + \sqrt {1 - {x^2}} } \right] + C

[ ∵\because x = sin θ\theta ]

=±[−1n1+1−x2x+1−x2]+C= \pm \left[ { - 1n{{1 + \sqrt {1 - {x^2}} } \over x} + \sqrt {1 - {x^2}} } \right] + C

[on rationalization]

∵\because The curve is in the first quadrant so y must be positive, so

y=1n(1+1−x2x)−1−x2+Cy = 1n\left( {{{1 + \sqrt {1 - {x^2}} } \over x}} \right) - \sqrt {1 - {x^2}} + C

As curve passes through (1, 0), so

0=0−0+c⇒c=00 = 0 - 0 + c \Rightarrow c = 0 , so required curve is

y=1n(1+1−x2x)−1−x2y = 1n\left( {{{1 + \sqrt {1 - {x^2}} } \over x}} \right) - \sqrt {1 - {x^2}}

and required differential eaquation is

dydx=−1−x2x{{dy} \over {dx}} = - {{\sqrt {1 - {x^2}} } \over x}

⇒xy′+1−x2=0\Rightarrow xy' + \sqrt {1 - {x^2}} = 0

Hence, options (a) and (c) are correct.

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