JEE MAIN - Mathematics Hindi (2019 - 12th January Morning Slot - No. 19)

यदि x > 1 के लिए, (2x)2y = 4e2x$$-$$2y,

तब (1 + loge 2x)2 $${{dy} \over {dx}}$$ के बराबर है :
$${{x\,{{\log }_e}2x - {{\log }_e}2} \over x}$$
loge 2x
x loge 2x
$${{x\,{{\log }_e}2x + {{\log }_e}2} \over x}$$

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