If fx-f1-x-1-x- - x -0 then maximum value of fx is

The correct option is C 1/2f(x) + f(1/x) = 1/x replace x with →1/x then f(1/x) + f(x) = x ⇒ x = 1/x x = ± 1 ⇒ f(x) is a constant for value at x = ± 1 f(1) + f( ...

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Byju's Answer

Standard XII

Mathematics

Global Maxima

If fx+f1/x=1/...

Question

If $f (x) + f (\frac{1}{x}) = \frac{1}{x}, \forall x \neq 0$ then maximum value of f(x) is

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-1

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\frac{1}{2}

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Solution

The correct option is C $\frac{1}{2}$
f(x) + f $(\frac{1}{x}) = \frac{1}{x}$
replace x with $\to \frac{1}{x} t h e n f (\frac{1}{x})$ + f(x) = x
$\Rightarrow x = \frac{1}{x}$
x = $\pm 1$
$\Rightarrow$ f(x) is a constant for value at x = $\pm 1$
f(1) + f(1) = 1
f(1) = $\frac{1}{2}$
f(-1) = $\frac{-}{1} 2$

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If f(x) + f(1x) = 1x, ∀ x ≠ 0 then maximum value of f(x) is

The correct option is C 12f(x) + f(1x)=1x replace x with →1xthenf(1x) + f(x) = x ⇒x=1x x = ±1 ⇒ f(x) is a constant for value at x = ±1 f(1) + f(1) = 1 f(1) = 12 f(-1) = −12

If $f (x) + f (\frac{1}{x}) = \frac{1}{x}, \forall x \neq 0$ then maximum value of f(x) is