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Question

# Suppose f and g are functions having second derivative f'' and g'' respectively at all points. If f(x)⋅g(x)=1 for all x and f' and g' are never zero, then f''(x)f'(x)−g''(x)g'(x)equals

A
2f(x)f(x)
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B
2g(x)g(x)
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C
2f(x)f(x)
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D
2f(x)f(x)
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Solution

## The correct option is D 2f′(x)f(x) We know that, f(x)⋅g(x)=1⇒g=1f⇒g′=−1f2f′⇒g′′=−[−2f3f′2+1f2f′′] So, f''(x)f'(x)−g''(x)g'(x)=f''f'−−[−2f3f′2+1f2f′′]−1f2f′=f''f'−(−2f′f+f′′f)=2f′f Also, g⋅f=1⇒g′f+gf′=0⇒f′f=−g′g Hence, f''(x)f'(x)−g''(x)g'(x)=2f′(x)f(x)=−2g′(x)g(x)

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