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Functions question

2023 · 29 Jan · Shift 1 · Q42
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Functions question

2023 · 29 Jan · Shift 1 · Q42

JEE MainMathematicsFunctionsNumerical+4 / −1
Suppose fff is a function satisfying f(x+y)=f(x)+f(y)f(x + y) = f(x) + f(y)f(x+y)=f(x)+f(y) for all x,y∈Nx,y \in Nx,y∈N and f(1)=15f(1) = {1 \over 5}f(1)=51​. If ∑n=1mf(n)n(n+1)(n+2)=112\sum\limits_{n = 1}^m {{{f(n)} \over {n(n + 1)(n + 2)}} = {1 \over {12}}}n=1∑m​n(n+1)(n+2)f(n)​=121​, then mmm is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 10

  1. Since f(x+y)=f(x)+f(y)f(x+y)=f(x)+f(y)f(x+y)=f(x)+f(y) for all x,y∈Nx,y\in Nx,y∈N, the function is additive on natural numbers.

  2. Using repeated addition, f(n)=f(1+1+⋯+1)=nf(1)f(n)=f(1+1+\cdots+1)=nf(1)f(n)=f(1+1+⋯+1)=nf(1) for every n∈Nn\in Nn∈N.

  3. Given f(1)=15f(1)=\frac15f(1)=51​, we get f(n)=n5.f(n)=\frac{n}{5}.f(n)=5n​.

  4. Substitute into the sum:

    =\sum_{n=1}^m \frac{n/5}{n(n+1)(n+2)} =\frac15\sum_{n=1}^m \frac{1}{(n+1)(n+2)}.$$

  5. Now decompose: 1(n+1)(n+2)=1n+1−1n+2.\frac{1}{(n+1)(n+2)}=\frac{1}{n+1}-\frac{1}{n+2}.(n+1)(n+2)1​=n+11​−n+21​.

  6. Therefore the sum telescopes:

    =\frac15\left(\frac12-\frac{1}{m+2}\right).$$

  7. This is given equal to 112\frac{1}{12}121​, so 15(12−1m+2)=112.\frac15\left(\frac12-\frac{1}{m+2}\right)=\frac{1}{12}.51​(21​−m+21​)=121​.

  8. Multiply by 555: 12−1m+2=512.\frac12-\frac{1}{m+2}=\frac{5}{12}.21​−m+21​=125​.

  9. Hence, 1m+2=12−512=112.\frac{1}{m+2}=\frac12-\frac{5}{12}=\frac{1}{12}.m+21​=21​−125​=121​.

  10. So, m+2=12  ⟹  m=10.m+2=12 \implies m=10.m+2=12⟹m=10.

Therefore, the required integer is 101010.

Comparison with stored answer: the derived answer matches the stored correct answer.

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