Question
Let f(x) = 24x+2 for x ϵ Real numbers. Find out the value of f(12001) + f(22001)+ ...... + f(20002001)
Answer: Option E
:
E
f(1 - x) = 241−x+2 =f(x) = 2∗4x4x+2 =f(x) = 4x4x+2
f(x) + f(1 - x) = 1
So, the desired sum is 1000.
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E
f(1 - x) = 241−x+2 =f(x) = 2∗4x4x+2 =f(x) = 4x4x+2
f(x) + f(1 - x) = 1
So, the desired sum is 1000.
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