If 2≤x≤4 then the max value of f(x)=(x−2)6(4−x)5 is

f1(x)=(x−2)5(4−x)4(34−11x)=0⇒x=3411⇒f(3411)=(1211)0(1011)5

If 2≤x≤4 then the max value of f(x)=(x−2)6(4−x)5 is

f1(x)=(x−2)5(4−x)4(34−11x)=0⇒x=3411⇒f(3411)=(1211)0(1011)5

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If $2 \leq x \leq 4$ then the max value of $f (x) = (x - 2)^{6} (4 - x)^{5}$ is

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${(\frac{2}{11})}^{6} {(\frac{10}{11})}^{5}$

${(\frac{11}{12})}^{5} {(\frac{11}{10})}^{5}$

${(\frac{12}{11})}^{6} {(\frac{10}{11})}^{5}$

${(\frac{2}{11})}^{6} {(\frac{10}{9})}^{5}$

answer is C.

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Detailed Solution

$f^{1} (x) = (x - 2)^{5} (4 - x)^{4} (34 - 11 x) = 0$

$\Rightarrow x = \frac{34}{11} \Rightarrow f (\frac{34}{11}) = {(\frac{12}{11})}^{0} {(\frac{10}{11})}^{5}$

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If 2≤x≤4 then the max value of f(x)=(x−2)6(4−x)5 is

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If $2 \leq x \leq 4$ then the max value of $f (x) = (x - 2)^{6} (4 - x)^{5}$ is