Let the function f : [ 0 , 1 ] 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 _______ . [2020]

Question

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 _______ . [2020]

Smart Achievers · Accepted Answer

(19)

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

$= \frac{4^{x}}{4^{x} + 2} + \frac{4 / 4^{x}}{4 / 4^{x} + 2} = \frac{4^{x}}{4^{x} + 2} + \frac{2}{2 + 4^{x}} = 1$

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

$= [f (\frac{1}{40}) + f (\frac{39}{40})] + [f (\frac{2}{40}) + f (\frac{38}{40})] + \dots + f (\frac{20}{40}) - f (\frac{1}{2})$

$= (1 + 1 + \dots 19 times) + f (\frac{1}{2}) - f (\frac{1}{2}) = 19$

Solution

Q.
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 _______ . [2020]

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Solution

Q. Let the function f:[0,1]→ℝ be defined by f(x)=4x4x+2. Then the value of f(140)+f(240)+f(340)+⋯+f(3940)-f(12) is _______ . [2020]

Ans. (19) Since, f(x)+f(1-x)=4x4x+2+41-x41-x+2 =4x4x+2+4/4x4/4x+2=4x4x+2+22+4x=1 ∴ f(140)+f(240)+⋯+f(3940)-f(12) =[f(140)+f(3940)]+[f(240)+f(3840)]+⋯+f(2040)-f(12) =(1+1+⋯19 times)+f(12)-f(12)=19

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Q.
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 _______ . [2020]