(B) Consider the difference $g(n) - f(n) = 1 + (n+1)2^n - 2^{n+1}$.Since $2^{n+1} = 2 \cdot 2^n$,we have:$g(n) - f(n) = 1 + (n+1)2^n - 2 \cdot 2^n$$g(

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(B) Consider the difference $g(n) - f(n) = 1 + (n+1)2^n - 2^{n+1}$.Since $2^{n+1} = 2 \cdot 2^n$,we have:$g(n) - f(n) = 1 + (n+1)2^n - 2 \cdot 2^n$$g(n) - f(n) = 1 + (n+1-2)2^n$$g(n) - f(n) = 1 + (n-1)2^n$.For all $n \in N$,$n \geq 1$,which implies $(n-1) \geq 0$.Since $2^n > 0$,the term $(n-1)2^n \geq 0$.Therefore,$1 + (n-1)2^n > 0$,which means $g(n) - f(n) > 0$ or $g(n) > f(n)$.

Class 11 Mathematics Relations and Functions Different types of Function

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Let $f(n) = 2^{n+1}$ and $g(n) = 1 + (n+1)2^n$ for all $n \in N$. Then:

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$f(n) > g(n)$
B
$f(n) < g(n)$
C
$f(n)$ and $g(n)$ are not comparable
D
$f(n) > g(n)$ if $n$ is even and $f(n) < g(n)$ if $n$ is odd

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