JEE MAIN - Mathematics Hindi (2019 - 11th January Morning Slot - No. 2)
यदि xloge(logex) $$-$$ x2 + y2 = 4(y > 0), तो $${{dy} \over {dx}}$$ x = e पर के बराबर होता है :
$${{\left( {1 + 2e} \right)} \over {2\sqrt {4 + {e^2}} }}$$
$${{\left( {1 + 2e} \right)} \over {\sqrt {4 + {e^2}} }}$$
$${{\left( {2e - 1} \right)} \over {2\sqrt {4 + {e^2}} }}$$
$${e \over {\sqrt {4 + {e^2}} }}$$
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