JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

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

Choose the correct answer:

Explanation

Solution

Maaniye $S_x = \sum_{i=1}^x i f(i)$ . Humein diya gaya hai:

S_x = x(x+1)f(x)

$x$ ki jagah $x-1$ rakhne par:

S_{x-1} = (x-1)xf(x-1)

Humein pata hai ki $S_x - S_{x-1} = xf(x)$ , isliye:

x(x+1)f(x) - (x-1)xf(x-1) = xf(x)

(x+1)f(x) - f(x) = (x-1)f(x-1)

xf(x) = (x-1)f(x-1)

Is pattern se:

xf(x) = (x-1)f(x-1) = (x-2)f(x-2) = ... = 1 \cdot f(1)

Chunki $f(1) = 1$ , toh $xf(x) = 1$ , yani $f(x) = \frac{1}{x}$ .

Lekin yeh formula $x \ge 2$ ke liye check karna hoga. $x=2$ ke liye:

f(1) + 2f(2) = 2(3)f(2) \implies 1 + 2f(2) = 6f(2) \implies 4f(2) = 1 \implies f(2) = \frac{1}{4}

General form $f(x) = \frac{1}{x(x+1)}$ nahi, balki steps follow karke $f(x) = \frac{1}{2x}$ jaisa nikalta hai. Asal mein iska pattern $\frac{1}{f(x)} = 2x$ banta hai for $x \ge 2$ .

Toh:

\frac{1}{f(2022)} + \frac{1}{f(2028)} = 2(2022) + 2(2028) = 4044 + 4056 = 8100

Correct Option: (1)