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

Consider a function \(f : N \rightarrow R\), satisfying \(f(1)+2 f(2)+3 f(3)+\ldots+x f(x)=x(x+1) f(x) ; x \geq 2\) with \(f(1)=1\). Then \(\frac{1}{f(2022)}+\frac{1}{f(2028)}\) is equal to

  1. A \(8200\)
  2. B \(8000\)
  3. C \(8400\)
  4. D \(8100\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(8100\)

Step-by-step Solution

Detailed explanation

Given for \(x \geq 2\) \(f(1)+2 f(2)+\ldots \ldots+x f(x)=x(x+1) f(x)\) \(\text { replace } x \text { by } x +1\) \(\Rightarrow \quad x(x+1) f(x)+(x+1) f(x+1)\) \(=(x+1)(x+2) f(x+1)\) \(\Rightarrow \quad \frac{x}{f(x+1)}+\frac{1}{f(x)}=\frac{(x+2)}{f(x)}\)…
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