f : R → R is given by f ( x ) = a x + a a x ∀ x ∈ R , then f ( 11 1 ) + f ( 11 2 ) + … . + f ( 11 9 ) + f ( 11 10 ) =

f ( 1 − x ) = 1 − f ( x ) ⇒ f ( x ) + f ( 1 − x ) = 1 ⇒ f ( 11 1 ) + f ( 11 10 ) = 1

Mathematics - Functions Question with Solution | TestHub

MathematicsFunctionsGeneral (Definition,type and special type)Easy2 minQB

MathematicsEasyinteger

$f : R \to R$ is given by $f (x) = \frac{a ^{x}}{a ^{x} + a} \forall x \in R$ , then $f (\frac{1}{11}) + f (\frac{2}{11}) + \dots . + f (\frac{9}{11}) + f (\frac{10}{11}) =$

Answer:

Solution:

$f (1 - x) = 1 - f (x) \Rightarrow f (x) + f (1 - x) = 1$

$\Rightarrow f (\frac{1}{11}) + f (\frac{10}{11}) = 1$

Stream:JEESubject:MathematicsTopic:FunctionsSubtopic:General (Definition,type and special type)

⏱ 2mℹ️ Source: QB

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