CAT 2019

Slot 1

QA

Question & Solution

It is given that:
$$ f(x + y) = f(x) f(y) $$

Hence:
$$ f(2) = f(1+1) = f(1) \cdot f(1) = 2 \cdot 2 = 4 $$ $$ f(3) = f(2+1) = f(2) \cdot f(1) = 4 \cdot 2 = 8 $$ $$ f(4) = f(3+1) = f(3) \cdot f(1) = 8 \cdot 2 = 16 $$

$\dots \implies f(x) = 2^x$

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Now,
$$ f(a + 1) + f(a + 2) + \dots + f(a + n) = 16 (2^n - 1) $$

Putting $n = 1$, we get:
$$ f(a+1) = 16 $$ But from the functional equation $f(x+y) = f(x)f(y)$, we have:
$$ f(a+1) = f(a) \cdot f(1) $$ $$ 2^a \cdot 2 = 16 $$ $$ 2^{a+1} = 16 \implies a = 3 $$

Hence, $a = 3$.