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Functions question

2023 · 29 Jan · Shift 2 · Q30
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  5. /2023 · 29 Jan · Shift 2 · Q30

Functions question

2023 · 29 Jan · Shift 2 · Q30

JEE MainMathematicsFunctionsMCQ+4 / −1
Consider a function f:N→Rf:\mathbb{N}\to\mathbb{R}f:N→R, satisfying f(1)+2f(2)+3f(3)+....+xf(x)=x(x+1)f(x);x≥2f(1)+2f(2)+3f(3)+....+xf(x)=x(x+1)f(x);x\ge2f(1)+2f(2)+3f(3)+....+xf(x)=x(x+1)f(x);x≥2 with f(1)=1f(1)=1f(1)=1. Then 1f(2022)+1f(2028)\frac{1}{f(2022)}+\frac{1}{f(2028)}f(2022)1​+f(2028)1​ is equal to
  1. A
    8000
  2. B
    8400
  3. C
    8100
  4. D
    8200
View written solutionFree

Correct answer: 4050

  1. Let
\cdots +xf(x).$$ Given, $$S_x=x(x+1)f(x), \qquad x\ge 2,$$ and also $f(1)=1$. 2. Write the relation for $x=n$ and $x=n-1$: $$S_n=n(n+1)f(n),$$ $$S_{n-1}=(n-1)n f(n-1).$$ But $$S_n=S_{n-1}+n f(n).$$ So, $$n(n+1)f(n)=(n-1)n f(n-1)+n f(n).$$ 3. Simplify: $$n[(n+1)f(n)-f(n)]=(n-1)n f(n-1)$$ $$n^2 f(n)=n(n-1)f(n-1).$$ For $n\ge 2$, $$n f(n)=(n-1)f(n-1),$$ so $$f(n)=\frac{n-1}{n}f(n-1).$$ 4. Use $f(1)=1$: $$f(2)=\frac{1}{2}f(1)=\frac12,$$ $$f(3)=\frac{2}{3}f(2)=\frac13,$$ and in general this telescopes to $$f(n)=\frac{1}{n}.$$ 5. Therefore, $$\frac{1}{f(2022)}+\frac{1}{f(2028)}=2022+2028=4050.$$ 6. Compare with options: - A: $8000$ - B: $8400$ - C: $8100$ - D: $8200$ None matches $4050$. So the stored correct answer $C$ is inconsistent with the given question. The correct value should be $$4050.$$
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