JEE Advanced 2025 — Paper 1 — MATH-III solution

JEE Advanced2025Paper 1MATH-III

Let $R$ denote the set of all real numbers. Let $f : R \to R$ be a function such that $f (x) > 0$ for all $x \in R$ , and $f (x + y) = f (x) f (y)$ for all $x, y \in R$ .

Let the real numbers $a_{1}, a_{2}, \dots, a_{50}$ be in an arithmetic progression. If $f (a_{31}) = 64 f (a_{25})$ , and

$i = 1 \sum 50 f (a_{i}) = 3 (2^{25} + 1),$

then the value of

$i = 6 \sum 30 f (a_{i})$

is _________.

Your answer

Solution

The Cauchy-type equation $f (x + y) = f (x) f (y)$ with $f > 0$ has the form $f (x) = k^{x}$ for some $k > 0$ .

Let $a_{i} = a + (i - 1) d$ . Then $f (a_{i}) = k^{a + (i - 1) d}$ , a geometric progression with first term $k^{a}$ and common ratio $r = k^{d}$ .

From $f (a_{31}) = 64 f (a_{25})$ : $k^{a + 30 d} = 64 \cdot k^{a + 24 d} \Rightarrow k^{6 d} = 64 \Rightarrow r = k^{d} = 2.$

Sum to $50$ terms: $i = 1 \sum 50 f (a_{i}) = k^{a} \cdot \frac{r ^{50} - 1}{r - 1} = k^{a} (2^{50} - 1) = k^{a} (2^{25} - 1) (2^{25} + 1) .$

Setting this equal to $3 (2^{25} + 1)$ gives $k^{a} = \frac{3}{2 ^{25} - 1} .$

Now the partial sum from $i = 6$ to $i = 30$ has $25$ terms with first term $f (a_{6}) = k^{a} \cdot r^{5}$ :

$i = 6 \sum 30 f (a_{i}) = k^{a} \cdot 2^{5} \cdot \frac{2 ^{25} - 1}{2 - 1} = \frac{3}{2 ^{25} - 1} \cdot 32 \cdot (2^{25} - 1) = 96.$

$∴$ The answer is $96$ .

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