f(x)=x+22x−4+ x−22x−4 then f1(3)+f1(6)

f(x)=x+22x−4+x−22x−42≤x≤4⇒f(x)=x−2+2+|x−2−2|=22x>4⇒f(x)=x−2+2+x−2−2 f(x)=2x−2f1(x)=0 in(2,4) ,1x−2 in(4,∞)f1(3)+f1(6)=0+14=12

f(x)=x+22x−4+ x−22x−4 then f1(3)+f1(6)

f(x)=x+22x−4+x−22x−42≤x≤4⇒f(x)=x−2+2+|x−2−2|=22x>4⇒f(x)=x−2+2+x−2−2 f(x)=2x−2f1(x)=0 in(2,4) ,1x−2 in(4,∞)f1(3)+f1(6)=0+14=12

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$f (x) = \sqrt{x + 2 \sqrt{2 x - 4}} + \sqrt{x - 2 \sqrt{2 x - 4}}$ then $f^{1} (3) + f^{1} (6)$

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$\frac{1}{2}$

answer is D.

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Detailed Solution

$\begin{array}{l} f (x) = \sqrt{x + 2 \sqrt{2 x - 4}} + \sqrt{x - 2 \sqrt{2 x - 4}} \\ 2 \leq x \leq 4 \Rightarrow f (x) = \sqrt{x - 2} + \sqrt{2} + | \sqrt{x - 2} - \sqrt{2} | = 2 \sqrt{2} \\ x > 4 \Rightarrow f (x) = \sqrt{x - 2} + \sqrt{2} + \sqrt{x - 2} - \sqrt{2} \\ f (x) = 2 \sqrt{x - 2} \\ f^{1} (x) = 0 i n (2, 4), \frac{1}{\sqrt{x - 2}} i n (4, \infty) \\ f^{1} (3) + f^{1} (6) = 0 + \frac{1}{\sqrt{4}} = \frac{1}{2} \end{array}$