Let the function f : 0 , 1 → ℝ be defined by f x = 4 x 4 x + 2 . Then the value of f 1 40 + f 2 40 + f 3 40 + ⋯ + f 39 40 - f 1 2 is_________

f x = 4 x 4 x + 2 , f 1 - x = 4 1 - x 4 1 - x + 2 = 2 4 x + 2 ⇒ f x + f 1 - x = 1 f 1 40 + f 2 40 + … . . + f 39 40 - f 1 2 = f 1 40 + f 39 40 + f 2 40 + f 38 40 + . . . . . f 19 40 + f 21 40 + f 20 40 - f 1 2 = 19 .

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MathematicsFunctionsGeneral (Definition,type and special type)Hard2 minPYQ_2020

MathematicsHardnumerical

Let the function $f : [0, 1] \to ℝ$ be defined by $f (x) = \frac{4^{x}}{4^{x} + 2}$ . Then the value of $f (\frac{1}{40}) + f (\frac{2}{40}) + f (\frac{3}{40}) + \dots + f (\frac{39}{40}) - f (\frac{1}{2})$ is_________

Answer:

19.00

Solution:

$f (x) = \frac{4^{x}}{4^{x} + 2}, f (1 - x) = \frac{4^{1 - x}}{4^{1 - x} + 2} = \frac{2}{4^{x} + 2}$

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

$f (\frac{1}{40}) + f (\frac{2}{40}) + \dots . . + f (\frac{39}{40}) - f (\frac{1}{2})$

$= (f (\frac{1}{40}) + f (\frac{39}{40})) + (f (\frac{2}{40}) + f (\frac{38}{40})) + . . . . . (f (\frac{19}{40}) + f (\frac{21}{40})) + f (\frac{20}{40}) - f (\frac{1}{2})$

$= 19$ .

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

⏱ 2mℹ️ Source: PYQ_2020

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