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Functions — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question
If f(x)=3[2x]+1000∑r=12008x+r−[x+r]2008f(x) = 3[2x]+1000\displaystyle\sum_{r=1}^{2008}\dfrac{x+r-[x+r]}{2008}, then f(3)f(3) (where [⋅][\cdot] is GIF) is.
Solution
Answer: 18
Step 1: Simplify using fractional part
x+r−[x+r]1={x+r}\frac{x+r-[x+r]}{1} = \{x+r\}
f(x)=3[2x]+10002008∑r=12008{x+r}f(x) = 3[2x]+\frac{1000}{2008}\sum_{r=1}^{2008}\{x+r\}
 Step 2: Evaluate at x=3x = 3 [2×3]=6[2 \times 3] = 6. For each r∈{1,2,…,2008}r \in \{1,2,\ldots,2008\}: rr is a positive integer, so 3+r3+r is also an integer, hence {3+r}=0\{3+r\} = 0.
f(3)=3(6)+10002008×0=18f(3) = 3(6)+\frac{1000}{2008}\times 0 = 18
 Answer: 18
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