JEE MainMathematicsFunctions
Let f(x) = _e(x+2) _e(x+2) + _e(103-x) . Then the value of _ k=1 ¹⁰⁰ f(k) is equal to
Options
- A50
- B100
- C51
- D49
Correct answer
A. 50
Step-by-step solution
Given f(x) = _e(x+2) _e(x+2) + _e(103-x) Evaluate f(101-x) : f(101-x) = _e(101-x+2) _e(101-x+2) + _e(103-(101-x)) f(101-x) = _e(103-x) _e(103-x) + _e(x+2) Adding f(x) and f(101-x) : f(x) + f(101-x) = _e(x+2) + _e(103-x) _e(x+2) + _e(103-x) = 1 The required sum is S = f(1) + f(2) + + f(100) We can pair the terms as follows: S = (f(1) + f(100)) + (f(2) + f(99)) + + (f(50) + f(51)) Since f(k) + f(101-k) = 1 , each pair sums to 1 . There are 50 such pairs. Therefore, S = 50 1 = 50 . Answer: 50