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Functions question

2003 · Shift 0 · Q68
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Functions question

2003 · Shift 0 · Q68

JEE MainMathematicsFunctionsMCQ+4 / −1
If f:R→Rf:R \to Rf:R→R satisfies fff(x + y) = fff(x) + fff(y), for all x, y ∈\in∈ R and fff(1) = 7, then ∑r=1nf(r)\sum\limits_{r = 1}^n {f\left( r \right)}r=1∑n​f(r) is
  1. A
    7n(n+1)2{{7n\left( {n + 1} \right)} \over 2}27n(n+1)​
  2. B
    7n2{{7n} \over 2}27n​
  3. C
    7(n+1)2{{7\left( {n + 1} \right)} \over 2}27(n+1)​
  4. D
    7n+(n+1)7n + \left( {n + 1} \right)7n+(n+1)
View written solutionFree

Correct answer: A

  1. We are given that

    \quad \forall x,y\in \mathbb R$$ and $$f(1)=7.$$

  2. We need to find ∑r=1nf(r).\sum_{r=1}^{n} f(r).∑r=1n​f(r).

  3. Using the additive property repeatedly for natural numbers: f(2)=f(1+1)=f(1)+f(1)=14,f(2)=f(1+1)=f(1)+f(1)=14,f(2)=f(1+1)=f(1)+f(1)=14, f(3)=f(2+1)=f(2)+f(1)=14+7=21.f(3)=f(2+1)=f(2)+f(1)=14+7=21.f(3)=f(2+1)=f(2)+f(1)=14+7=21. In general, for any positive integer rrr, f(r)=rf(1)=7r.f(r)=rf(1)=7r.f(r)=rf(1)=7r.

  4. Therefore, ∑r=1nf(r)=∑r=1n7r=7∑r=1nr.\sum_{r=1}^{n} f(r)=\sum_{r=1}^{n} 7r=7\sum_{r=1}^{n} r.∑r=1n​f(r)=∑r=1n​7r=7∑r=1n​r.

  5. Now use the formula ∑r=1nr=n(n+1)2.\sum_{r=1}^{n} r=\frac{n(n+1)}{2}.∑r=1n​r=2n(n+1)​. Hence, \sum_{r=1}^{n} f(r)=7\cdot \frac{n(n+1)}{2}= rac{7n(n+1)}{2}.

  6. Comparing with the options:

    • A: 7n(n+1)2\frac{7n(n+1)}{2}27n(n+1)​ ✅
    • B: 7n2\frac{7n}{2}27n​ ❌
    • C: 7(n+1)2\frac{7(n+1)}{2}27(n+1)​ ❌
    • D: 7n+(n+1)7n+(n+1)7n+(n+1) ❌

So the correct answer is A.

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