The least value of the function $$F(x)$$ $$=$$∫$$x2$$ $$log1/3$$⁡tdt,x∈[$$1/10$$,$$4$$] is at x $$=$$

F′(x)=$$−$$log1/3$$⁡$$x=log$$⁡xlog⁡$$3$$. So F′$$(x) $$1$$. Hence f has least value at x $$=$$ $$1$$.

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The least value of the function $F (x) =$
$\int_{x}^{2} \log_{1 / 3} t d t, x \in [1 / 10, 4]$ is at $x =$

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Correct option is C

F′(x)=−log1/3⁡x=log⁡xlog⁡3. So F′(x)<0 for 1100 for x > 1. Hence f has least value at x = 1.

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The least value of the function F(x) =∫x2 log1/3⁡tdt,x∈[1/10,4] is at x =

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The least value of the function $F (x) =$
$\int_{x}^{2} \log_{1 / 3} t d t, x \in [1 / 10, 4]$ is at $x =$