sample paper

Question

If $f(x)=\frac{4^{x}}{4^{1-x}+ 4^{x}}$, then $f(\frac{1}{65}) + f(\frac{2}{65}) +f(\frac{3}{65}) +..............f(\frac{64}{65}) +$ is equal to

Answers:

a. 16

b. 0

c. 32

d. 64

Answer 1

eq. 1:     f(x) = 4^x / 4(^1-x) + 4^x 

replace x with 1-x, we get

eq 2:     f(1- x) = 4^(1-x) / 4(^1-x) + 4^x

add eq 1 and eq 2

f(x) + f(1-x) =   [ 4^x + 4^(1-x) ] / [ 4^(1-x) + 4^x  ]

f(x) + f(1-x) = 1

x = 1/65

f(1/65) + f(64/65) = 1

f(2/65) + f(63/65) = 1

f(3/65) + f(62/65) = 1

f(4/65) + f(61/65) = 1

f(5/65) + f(60/65) = 1 

.

.

.

f(30/65) + f(35/65) = 1 

f(31/65) + f(34/65) = 1  

f(32/65) + f(33/65) = 1 

 on summing all we get

f(1/65) + f(2/65) + f(3/65) + ...................................................................... + f(64/65)  = 32

so answer is 32