JEE Main 2026 22 January Evening Shift
Exam JEE Main
Year 2026
Date 22 January
Shift Evening Shift
Subject Mathematics
Chapter Functions JEE Main 2026 · 22 January · Evening Shift · Mathematics · Functions
Let f and g be functions satisfying f(x+y)=f(x) f(y), f(1)=7 and...
Let
f f f and
g g g be functions satisfying
f ( x + y ) = f ( x ) f ( y ) , f ( 1 ) = 7 f(x+y)=f(x) f(y), f(1)=7 f ( x + y ) = f ( x ) f ( y ) , f ( 1 ) = 7 and
g ( x + y ) = g ( x y ) , g ( 1 ) = 1 g(x+y)=g(x y), g(1)=1 g ( x + y ) = g ( x y ) , g ( 1 ) = 1 , for all
x , y ∈ N x, y \in \mathbf{N} x , y ∈ N . If
∑ x = 1 n ( f ( x ) g ( x ) ) = 19607 \sum_{x=1}^{\mathrm{n}}\left(\frac{f(x)}{\mathrm{g}(x)}\right)=19607 ∑ x = 1 n ( g ( x ) f ( x ) ) = 19607 , then n is equal to:
Step-by-step solution
f ( x + y ) = f ( x ) f ( y ) f(x+y)=f(x)f(y) f ( x + y ) = f ( x ) f ( y ) with
f ( 1 ) = 7 f(1)=7 f ( 1 ) = 7 gives
f ( x ) = 7 x f(x)=7^x f ( x ) = 7 x .
For
g g g : put
y = 1 y=1 y = 1 in
g ( x + y ) = g ( x y ) g(x+y)=g(xy) g ( x + y ) = g ( x y ) to get
g ( x + 1 ) = g ( x ) g(x+1)=g(x) g ( x + 1 ) = g ( x ) .
So
g ( x ) = g ( 1 ) = 1 g(x)=g(1)=1 g ( x ) = g ( 1 ) = 1 for all
x ∈ N x \in \mathbb{N} x ∈ N .
∑ x = 1 n f ( x ) g ( x ) = ∑ x = 1 n 7 x = 7 ( 7 n − 1 ) 6 = 19607 \sum_{x=1}^{n}\frac{f(x)}{g(x)}=\sum_{x=1}^{n}7^x=\frac{7(7^n-1)}{6}=19607 ∑ x = 1 n g ( x ) f ( x ) = ∑ x = 1 n 7 x = 6 7 ( 7 n − 1 ) = 19607 7 n + 1 − 7 = 117642 7^{n+1}-7=117642 7 n + 1 − 7 = 117642 ⇒ 7 n + 1 = 117649 = 7 6 \Rightarrow 7^{n+1}=117649=7^6 ⇒ 7 n + 1 = 117649 = 7 6 n + 1 = 6 ⇒ n = 5 n+1=6 \Rightarrow n=5 n + 1 = 6 ⇒ n = 5
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