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Functions question

2020 · 6 Sep · Shift 2 · Q29
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Functions question

2020 · 6 Sep · Shift 2 · Q29

JEE MainMathematicsFunctionsNumerical+4 / −1
Suppose that a function f : R →\to→ R satisfies f(x + y) = f(x)f(y) for all x, y ∈\in∈ R and f(1) = 3. If ∑i=1nf(i)=363\sum\limits_{i = 1}^n {f(i)} = 363i=1∑n​f(i)=363 then n is equal to ‾\underline{\hspace{2cm}}​ .
Numerical answer
View written solutionFree

Correct answer: 5

  1. We are given

    \quad \text{for all } x,y\in \mathbb{R},$$ and $$f(1)=3.$$

  2. We need to find f(i)f(i)f(i) for positive integers iii.

    Using the functional equation repeatedly: f(2)=f(1+1)=f(1)f(1)=3⋅3=9,f(2)=f(1+1)=f(1)f(1)=3\cdot 3=9,f(2)=f(1+1)=f(1)f(1)=3⋅3=9, f(3)=f(2+1)=f(2)f(1)=9⋅3=27.f(3)=f(2+1)=f(2)f(1)=9\cdot 3=27.f(3)=f(2+1)=f(2)f(1)=9⋅3=27.

    In general, for any positive integer iii, f(i)=3i.f(i)=3^i.f(i)=3i.

  3. Therefore, ∑i=1nf(i)=∑i=1n3i=363.\sum_{i=1}^n f(i)=\sum_{i=1}^n 3^i=363.∑i=1n​f(i)=∑i=1n​3i=363.

  4. This is a geometric series: ∑i=1n3i=3(3n−1)3−1=3n+1−32.\sum_{i=1}^n 3^i=\frac{3(3^n-1)}{3-1}=\frac{3^{n+1}-3}{2}.∑i=1n​3i=3−13(3n−1)​=23n+1−3​.

    So, 3n+1−32=363.\frac{3^{n+1}-3}{2}=363.23n+1−3​=363.

  5. Solve for nnn: 3n+1−3=726,3^{n+1}-3=726,3n+1−3=726, 3n+1=729.3^{n+1}=729.3n+1=729.

    Since 729=36,729=3^6,729=36, we get n+1=6  ⟹  n=5.n+1=6 \implies n=5.n+1=6⟹n=5.

  6. Hence, the required integer is 5.\boxed{5}.5​.

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