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JEE Mathematics 2023 Question with Solution

Suppose ff is a function satisfying f(x+y)=f(x)+f(y)f(x + y) = f(x) + f(y) for all x,y∈Nx, y \in \mathbb{N} and f(1)=15f(1) = \frac{1}{5}. If the sum from n=1n = 1 to mm of [f(n)n(n+1)(n+2)]\left[\frac{f(n)}{n(n + 1)(n + 2)}\right] equals 112\frac{1}{12}, then mm is equal to:

  • A

    55

  • B

    1010

  • C

    1515

  • D

    2020

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: f(x+y)=f(x)+f(y)f(x + y) = f(x) + f(y) for all x,y∈Nx, y \in \mathbb{N}, f(1)=15f(1) = \frac{1}{5}, and

∑n=1mf(n)n(n+1)(n+2)=112\sum_{n=1}^{m} \frac{f(n)}{n(n+1)(n+2)} = \frac{1}{12}

Find: mm

From the functional equation on natural numbers, f(n)=nf(1)f(n) = nf(1). Hence,

f(n)=n5f(n) = \frac{n}{5}

Substitute this into the given sum:

∑n=1mn5n(n+1)(n+2)=112\sum_{n=1}^{m} \frac{\frac{n}{5}}{n(n+1)(n+2)} = \frac{1}{12}

So,

15∑n=1m1(n+1)(n+2)=112\frac{1}{5}\sum_{n=1}^{m} \frac{1}{(n+1)(n+2)} = \frac{1}{12}

Therefore,

∑n=1m1(n+1)(n+2)=512\sum_{n=1}^{m} \frac{1}{(n+1)(n+2)} = \frac{5}{12}

Now use partial fractions:

1(n+1)(n+2)=1n+1−1n+2\frac{1}{(n+1)(n+2)} = \frac{1}{n+1} - \frac{1}{n+2}

Thus the series telescopes:

∑n=1m(1n+1−1n+2)=12−1m+2\sum_{n=1}^{m} \left(\frac{1}{n+1} - \frac{1}{n+2}\right) = \frac{1}{2} - \frac{1}{m+2}

So,

12−1m+2=512\frac{1}{2} - \frac{1}{m+2} = \frac{5}{12}

Hence,

1m+2=12−512=112\frac{1}{m+2} = \frac{1}{2} - \frac{5}{12} = \frac{1}{12}

Therefore,

m+2=12m+2 = 12

and so,

m=10m = 10

Therefore, the correct option is B.

Telescoping Observation

Given: f(x+y)=f(x)+f(y)f(x+y)=f(x)+f(y) on N\mathbb{N} and f(1)=15f(1)=\frac{1}{5}

Find: mm

Since additivity on natural numbers gives f(n)=nf(1)f(n)=nf(1), we get

f(n)=n5f(n)=\frac{n}{5}

Then the summand becomes

f(n)n(n+1)(n+2)=15(n+1)(n+2)\frac{f(n)}{n(n+1)(n+2)}=\frac{1}{5(n+1)(n+2)}

Use directly

1(n+1)(n+2)=1n+1−1n+2\frac{1}{(n+1)(n+2)}=\frac{1}{n+1}-\frac{1}{n+2}

So the sum is

15(12−1m+2)=112\frac{1}{5}\left(\frac{1}{2}-\frac{1}{m+2}\right)=\frac{1}{12}

This gives

12−1m+2=512\frac{1}{2}-\frac{1}{m+2}=\frac{5}{12}

and hence

1m+2=112\frac{1}{m+2}=\frac{1}{12}

Therefore,

m=10m=10

The correct option is B.

Common mistakes

  • Assuming f(x+y)=f(x)+f(y)f(x+y)=f(x)+f(y) means f(x)=kxf(x)=kx for all real numbers without using the given domain. Here the domain is N\mathbb{N}, and on natural numbers additivity is enough to conclude f(n)=nf(1)f(n)=nf(1). Use repeated addition, not an unjustified real-variable theorem.

  • Failing to cancel the factor nn correctly in f(n)n(n+1)(n+2)\frac{f(n)}{n(n+1)(n+2)} after substituting f(n)=n5f(n)=\frac{n}{5}. The correct simplification is 15(n+1)(n+2)\frac{1}{5(n+1)(n+2)}, not 15n(n+1)(n+2)\frac{1}{5n(n+1)(n+2)}.

  • Using the wrong partial fraction decomposition for 1(n+1)(n+2)\frac{1}{(n+1)(n+2)}. The correct identity is 1n+1−1n+2\frac{1}{n+1}-\frac{1}{n+2}. Any sign error will spoil the telescoping cancellation.

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