Let f and g be functions satisfying f(x+y)=f(x)f(y), f(1)=7 and g(x+y)=g(xy), g(1)=1 for all x , y . If x = 1 n ( f ( x ) g ( x ) ) = 19607 , then n is equal to: [2026]

Question

Let $f$ and $g$ be functions satisfying $f (x + y) = f (x) f (y)$ , $f (1) = 7$ and $g (x + y) = g (x y)$ , $g (1) = 1$ for all $x, y \in ℕ$ . If $\sum_{x = 1}^{n} ((\frac{f (x)}{g (x)})) = 19607,$ then n is equal to: [2026]

Smart Achievers · Accepted Answer

(2)

$f (x + y) = f (x) f (y) \Rightarrow f (x) = a^{x}$

$(∵ f (1) = 7 \Rightarrow a^{1} = 7)$

$So f (x) = 7^{x}$

$Now$

$g (x + y) = g (x y) (put y = 1)$

$\Rightarrow g (x + 1) = g (x)$

$so g (1) = g (2) = g (3) = \dots = g (n) = 1$

$Given \sum_{x = 1}^{n} \frac{f (x)}{g (x)} = 19607$

$\sum_{x = 1}^{n} \frac{7^{x}}{1} = 19607$

$\Rightarrow 7 (\frac{7^{n} - 1}{7 - 1}) = 19607$

$7^{n} - 1 = \frac{6}{7} \times 19607$

$7^{n} = 16807$

$\Rightarrow n = 5$

Solution

Q.
Let $f$ and $g$ be functions satisfying $f (x + y) = f (x) f (y)$ , $f (1) = 7$ and $g (x + y) = g (x y)$ , $g (1) = 1$ for all $x, y \in ℕ$ . If $\sum_{x = 1}^{n} ((\frac{f (x)}{g (x)})) = 19607,$ then n is equal to: [2026]

1 7

2 5

3 6

4 4

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Solution

Q. Let f and g be functions satisfying f(x+y)=f(x)f(y), f(1)=7 and g(x+y)=g(xy), g(1)=1 for all x,y∈ℕ. If ∑x=1n(f(x)g(x))=19607, then n is equal to: [2026]

1 7

2 5

3 6

4 4

Ans. (2) f(x+y)=f(x) f(y)⇒f(x)=ax (∵f(1)=7⇒a1=7) So f(x)=7x Now g(x+y)=g(xy) (put y=1) ⇒g(x+1)=g(x) so g(1)=g(2)=g(3)=⋯=g(n)=1 Given ∑x=1nf(x)g(x)=19607 ∑x=1n7x1=19607 ⇒7(7n-17-1)=19607 7n-1=67×19607 7n=16807 ⇒n=5

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Q.
Let $f$ and $g$ be functions satisfying $f (x + y) = f (x) f (y)$ , $f (1) = 7$ and $g (x + y) = g (x y)$ , $g (1) = 1$ for all $x, y \in ℕ$ . If $\sum_{x = 1}^{n} ((\frac{f (x)}{g (x)})) = 19607,$ then n is equal to: [2026]