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JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy
1 of 1
Application +4 | -1
If f(x)=log⁑e{u(x)Ο…(x)},u(1)=Ο…(1)f\left( x \right) = {\log _e}\left\{ {\frac{{u\left( x \right)}}{{\upsilon \left( x \right)}}} \right\},u\left( 1 \right) = \upsilon \left( 1 \right) and uβ€²(1)=Ο…β€²(1)=2u'\left( 1 \right) = \upsilon '\left( 1 \right) = 2, then fβ€²(1)f'\left( 1 \right) is equal to
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Correct option: A
0
Option A (0) is correct because when u(1)=v(1)u(1) = v(1) and uβ€²(1)=vβ€²(1)=2u'(1) = v'(1) = 2, the derivative fβ€²(1)f'(1) becomes zero.

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