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…

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

 

Solution

fx=4x4x+2,  f1-x=41-x41-x+2=24x+2

fx+f1-x=1

f140+f240+..+f3940-f12

=f140+f3940+f240+f3840+.....f1940+f2140+f2040-f12

=19.

Asked in: JEE Advanced 2020 (Paper 2)

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