Let the function f:0,1→ℝ be defined by fx=4^x4^x+2. Then the value of f1/40+f2/40+f3/40+⋯+f39/40-f1/2 is_________

Correct answer: 19. fx=4^x4^x+2, f1-x=4^(1-x)4^(1-x)+2=2/(4^x+2) ⇒fx+f1-x=1 f1/40+f2/40+…..+f39/40-f1/2 =f1/40+f39/40+f2/40+f38/40+.....f19/40+f21/40+f20/40-f1/2 =19.

Q34 / 54 · JEE Advanced 2020 (Paper 2) · Mathematics

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_________

Correct answer: 19

Solution

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

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

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

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

$= 19$ .

Hence, the correct answer is 19.

Let the function f:0,1→ℝ be defined by fx=4x4x+2. Then the value of f140+f240+f340+⋯+f3940-f12 is_________

Solution

More Functions Questions

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_________