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JEE Mains · Maths · STD 12 - 1. relation and function

વિધેય \(\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}\) માટે \(\mathrm{f}(\mathrm{x}+\mathrm{y})=\mathrm{f}(\mathrm{x})+\mathrm{f}(\mathrm{y}) \forall \mathrm{x}, \mathrm{y} \in \mathrm{R}\) થાય જો \(\mathrm{f}(1)=2\) અને \(g(n)=\sum \limits_{k=1}^{(n-1)} f(k), n \in N\) હોય તો \(n\) કિમત મેળવો જ્યાં \(\mathrm{g}(\mathrm{n})=20\) થાય 

  1. A \(5\)
  2. B \(9\)
  3. C \(20\)
  4. D \(4\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(5\)

Step-by-step Solution

Detailed explanation

\(f(x+y)=f(x)+f(y)\) \(\Rightarrow f(n)=n f(1)\) \(f(n)=2 n\) \(g(n)=\sum_{k=1}^{n-1} 2 n=2\left(\frac{(n-1) n}{2}\right)=n(n-1)\) \(g(n)=20 \Rightarrow n(n-1)=20\) \(n=5\)
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