JEE 2023 — Mathematics PYQ

Step 1: Function $f(n)$ ki value nikalna

Di gayi condition $f(x+y)=f(x)+f(y)$ ek linear function ko darshati hai.

Kyuki $f(1)=\frac{1}{5}$, toh:

$f(2)=f(1+1)=f(1)+f(1)=2\cdot f(1)$

$f(3)=f(2+1)=f(2)+f(1)=3\cdot f(1)$

Isi tarah, kisi bhi $n\in \mathbb{N}$ ke liye:

Step 2: Summation mein value rakhna

Ab hum $f(n)=\frac{n}{5}$ ko di gayi equation mein rakhte hain:

Yahan $n$ se $n$ cancel ho jayega aur constant $\frac{1}{5}$ ko sum ke bahar nikal lenge:

Dono side 5 se multiply karne par:

Step 3: Telescopic Series ka upyog (Method of Differences)

Hum $\frac{1}{(n+1)(n+2)}$ ko aise likh sakte hain:

Ab values put karte hain ($n=1$ se $m$ tak):

For $n=1:\left(\frac{1}{2}-\frac{1}{3}\right)$

For $n=2:\left(\frac{1}{3}-\frac{1}{4}\right)$

...

For $n=m:\left(\frac{1}{m+1}-\frac{1}{m+2}\right)$

In sabko jodne par beech ki saari terms cancel ho jayengi:

Step 4: $m$ ke liye solve karna

Iska matlab hai:

Answer: $m$ ki value 10 hai.