JEE Advanced 2024 — Paper 2 — MATH-III solution

JEE Advanced2024Paper 2MATH-III

Let $f : R \to R$ be a function such that $f (x + y) = f (x) + f (y)$ for all $x, y \in R$ , and $g : R \to (0, \infty)$ be a function such that $g (x + y) = g (x) g (y)$ for all $x, y \in R$ . If $f (\frac{- 3}{5}) = 12$ and $g (\frac{- 1}{3}) = 2$ , then the value of $(f (\frac{1}{4}) + g (- 2) - 8) g (0)$ is ______.

Your answer

Solution

Cauchy's additive equation with continuity (implicit) gives $f (x) = k x$ . From $f (- 3/5) = 12$ : $- \frac{3}{5} k = 12 \Rightarrow k = - 20$ , so $f (x) = - 20 x$ .

The multiplicative form $g (x + y) = g (x) g (y)$ with positive range gives $g (x) = a^{x}$ . From $g (- 1/3) = 2$ : $a^{- 1/3} = 2 \Rightarrow a = \frac{1}{8}$ , so $g (x) = (\frac{1}{8})^{x}$ .

Note $g (0) = 1$ . Compute the bracket:

$f (\frac{1}{4}) = - 20 \cdot \frac{1}{4} = - 5, g (- 2) = 8^{2} = 64.$

$(- 5 + 64 - 8) \cdot 1 = 51.$

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