1 markMathematicsSets, relations and functionsFunctional equations and special functionsModerateabout 2 min

IIT JAM 2022 Question 44: Sets, relations and functions

Let $a, b \in \mathbb{R}$ . If $f(x) = ax + b$ satisfies $a + b = 4$ and $f(x+y) = f(x) + f(y) - 2$ for all $x, y \in \mathbb{R}$ , then $\displaystyle\sum_{n=1}^{50} f(n) =$ _______ (in integer).

Answer and worked solution

Answer: 2650 to 2650

With $f(x) = ax + b$ , $f(x+y) = a(x+y) + b$ while $f(x) + f(y) - 2 = ax + ay + 2b - 2$ . Matching gives $b = 2b - 2$ , so $b = 2$ , and $a + b = 4$ gives $a = 2$ . Hence $f(x) = 2x + 2$ .

Then $\sum_{n=1}^{50}(2n + 2) = 2\cdot\dfrac{50\cdot51}{2} + 2\cdot50 = 2550 + 100 = 2650$ . The answer is $2650$ .

Plain-text version

Let a, b ∈ ℝ. If f(x) = ax + b satisfies a + b = 4 and f(x+y) = f(x) + f(y) - 2 for all x, y ∈ ℝ, then Σₙ₌₁⁵⁰ f(n) = _ (in integer).

Also written as: Let a, b in R. If f(x) = ax + b satisfies a + b = 4 and f(x+y) = f(x) + f(y) - 2 for all x, y in R, then sum_(n=1)^(50) f(n) = _ (in integer).

Sets, relations and functions has come up 65 times in 6 exams since 2004.

What this question tests

This question is on Functional equations and special functions, part of Sets, relations and functions in Mathematics.

Functions defined by a rule they must satisfy, or by recursion, or with unusual behaviour.

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