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Functions question

2020 · 2 Sep · Shift 2 · Q24
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Functions question

2020 · 2 Sep · Shift 2 · Q24

JEE MainMathematicsFunctionsMCQ+4 / −1
Let f : R →\to→ R be a function which satisfies f(x + y) = f(x) + f(y) ∀\forall∀ x, y ∈\in∈ R. If f(1) = 2 and g(n) =∑k=1(n−1)f(k)\sum\limits_{k = 1}^{\left( {n - 1} \right)} {f\left( k \right)}k=1∑(n−1)​f(k), n ∈\in∈ N then the value of n, for which g(n) = 20, is :
  1. A
    20
  2. B
    9
  3. C
    5
  4. D
    4
View written solutionFree

Correct answer: C

  1. We are given that

    \,\forall x,y\in \mathbb R$$ and also $$f(1)=2.$$

  2. We need to compute g(n)=∑k=1n−1f(k),n∈N.g(n)=\sum_{k=1}^{n-1} f(k), \qquad n\in \mathbb N.g(n)=∑k=1n−1​f(k),n∈N.

  3. First find f(k)f(k)f(k) for natural numbers kkk.

    Using additivity repeatedly, f(2)=f(1+1)=f(1)+f(1)=2+2=4,f(2)=f(1+1)=f(1)+f(1)=2+2=4,f(2)=f(1+1)=f(1)+f(1)=2+2=4, f(3)=f(1+1+1)=3f(1)=6.f(3)=f(1+1+1)=3f(1)=6.f(3)=f(1+1+1)=3f(1)=6. In general, for any natural number kkk, f(k)=kf(1)=2k.f(k)=kf(1)=2k.f(k)=kf(1)=2k.

  4. Substitute into g(n)g(n)g(n): g(n)=∑k=1n−12k=2∑k=1n−1k.g(n)=\sum_{k=1}^{n-1} 2k = 2\sum_{k=1}^{n-1} k.g(n)=∑k=1n−1​2k=2∑k=1n−1​k.

    Now, ∑k=1n−1k=(n−1)n2.\sum_{k=1}^{n-1} k = \frac{(n-1)n}{2}.∑k=1n−1​k=2(n−1)n​.

    Therefore, g(n)=2⋅(n−1)n2=n(n−1).g(n)=2\cdot \frac{(n-1)n}{2}=n(n-1).g(n)=2⋅2(n−1)n​=n(n−1).

  5. We are given g(n)=20g(n)=20g(n)=20. So, n(n−1)=20.n(n-1)=20.n(n−1)=20.

    n2−n−20=0.n^2-n-20=0.n2−n−20=0.

    Factorizing, n2−n−20=(n−5)(n+4)=0.n^2-n-20=(n-5)(n+4)=0.n2−n−20=(n−5)(n+4)=0.

    Hence, n=5orn=−4.n=5 \quad \text{or} \quad n=-4.n=5orn=−4.

    Since n∈Nn\in \mathbb Nn∈N, we take n=5.n=5.n=5.

  6. Check options:

    • A: 202020 ❌
    • B: 999 ❌
    • C: 555 ✅
    • D: 444 ❌

Therefore, the correct option is C.

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