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If the primitive of $\frac{1}{f (x)}$ is equal to $\log {f (x)}^{2} + C$ then f(x) is

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x + d

x2+d

x22+d

x2 + d

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detailed solution

Correct option is B

Given that,∫1f(x)dx=log⁡{f(x)}2+C⇒1f(x)=1{f(x)}22f(x)f′(x)⇒f′(x)=12⇒f(x)=x2+d[where, d is a integration constant]

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If the primitive of 1f(x) is equal to log⁡{f(x)}2+C then f(x) is

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If the primitive of $\frac{1}{f (x)}$ is equal to $\log {f (x)}^{2} + C$ then f(x) is