JEE MAIN - Mathematics (2019 - 11th January Morning Slot - No. 2)
If xloge(logex) $$-$$ x2 + y2 = 4(y > 0), then $${{dy} \over {dx}}$$ at x = e is equal to :
$${{\left( {1 + 2e} \right)} \over {2\sqrt {4 + {e^2}} }}$$
$${{\left( {1 + 2e} \right)} \over {\sqrt {4 + {e^2}} }}$$
$${{\left( {2e - 1} \right)} \over {2\sqrt {4 + {e^2}} }}$$
$${e \over {\sqrt {4 + {e^2}} }}$$
Explanation
Differentiating with respect to x,
$$x.{1 \over {\ell nx}}.{1 \over x} + \ell n(\ell nx) - 2x + 2y.{{dy} \over {dx}} = 0$$
at $$x = e$$ we get
$$1 - 2e + 2y{{dy} \over {dx}} = 0 \Rightarrow {{dy} \over {dx}} = {{2e - 1} \over {2y}}$$
$$ \Rightarrow {{dy} \over {dx}} = {{2e - 1} \over {2\sqrt {4 + {e^2}} }}\,\,$$
as $$y(e) = \sqrt {4 + {e^2}} $$
$$x.{1 \over {\ell nx}}.{1 \over x} + \ell n(\ell nx) - 2x + 2y.{{dy} \over {dx}} = 0$$
at $$x = e$$ we get
$$1 - 2e + 2y{{dy} \over {dx}} = 0 \Rightarrow {{dy} \over {dx}} = {{2e - 1} \over {2y}}$$
$$ \Rightarrow {{dy} \over {dx}} = {{2e - 1} \over {2\sqrt {4 + {e^2}} }}\,\,$$
as $$y(e) = \sqrt {4 + {e^2}} $$
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