JEE Mains · Maths · STD 12 - 1. relation and function
Let \(f : N \rightarrow R\) be a function such that \(f(x+y)=2 f(x) f(y)\) for natural numbers \(x\) and \(y\). If \(f(1)=2\), then the value of \(\alpha\) for which \(\sum \limits_{k=1}^{10} f(\alpha+k)=\frac{512}{3}\left(2^{20}-1\right)\) holds, is
- A \(2\)
- B \(3\)
- C \(4\)
- D \(6\)
Answer & Solution
Correct Answer
(C) \(4\)
Step-by-step Solution
Detailed explanation
\(f : N \rightarrow R , f ( x + y )=2 f ( x ) f ( y )\) \(f (1)=2\), \(\sum_{ k =1}^{10} f (\alpha+ k )=2 f (\alpha) \sum_{ k =1}^{10} f ( k )\) \(=2 f (\alpha)( f (1)+ f (2)+\ldots .+ f (10))\) From \((1)\) \(f (2)=2 f ^{2}(1)=2^{3}\) \(\left.f (3)=2 f (2) f (1)=2^{5}\right)\)…
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