The maximum value of log 10 1001-x2-x-1 is k- Then the value of log1-k-log 3 is

The correct option is B Negative log_10100^(1/x^2+x+1) = 2/(x^2+x+1) = 2/((x+(1/2))^2+ ( 3/4 ) = ( 8/3 ) k = ( 8/3 ) log ( 1/k ) log3 = log ( 3/8 ) ...

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Question

The maximum value of $l o g_{10} 100^{(\frac{1}{x^{2} + x + 1})}$ is k. Then the value of $l o g (\frac{1}{k})$ - log3 is

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Solution

The correct option is B
Negative

$l o g_{10} 100^{(\frac{1}{x^{2} + x + 1})} = \frac{2}{(x^{2} + x + 1)} = \frac{2}{((x + (\frac{1}{2}))^{2} + (\frac{3}{4})} = (\frac{8}{3}) k = (\frac{8}{3}) l o g (\frac{1}{k}) - l o g 3 = l o g (\frac{3}{8}) - l o g 3 = - l o g 8$

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The maximum value of log10100(1x2+x+1) is k. Then the value of log(1k) - log3 is

The correct option is B Negative log10100(1x2+x+1)=2(x2+x+1)=2((x+(12))2+(34)=(83)k=(83)log(1k)−log3=log(38)−log3=−log 8

The maximum value of $l o g_{10} 100^{(\frac{1}{x^{2} + x + 1})}$ is k. Then the value of $l o g (\frac{1}{k})$ - log3 is

The correct option is B
Negative

$l o g_{10} 100^{(\frac{1}{x^{2} + x + 1})} = \frac{2}{(x^{2} + x + 1)} = \frac{2}{((x + (\frac{1}{2}))^{2} + (\frac{3}{4})} = (\frac{8}{3}) k = (\frac{8}{3}) l o g (\frac{1}{k}) - l o g 3 = l o g (\frac{3}{8}) - l o g 3 = - l o g 8$