(A) Using the identity $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$,let $a = (n+1)^{1/3}$ and $b = n^{1/3}$.Then $f_n = a - b = \frac{a^3 - b^3}{a^2 + ab + b^2

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(A) Using the identity $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$,let $a = (n+1)^{1/3}$ and $b = n^{1/3}$.Then $f_n = a - b = \frac{a^3 - b^3}{a^2 + ab + b^2} = \frac{1}{(n+1)^{2/3} + (n+1)^{1/3}n^{1/3} + n^{2/3}}$.Since $n Thus,$3n^{2/3} Taking the reciprocal,we get $\frac{1}{3(n+1)^{2/3}} Replacing $n$ with $n+1$,we get $\frac{1}{3(n+2)^{2/3}} Combining these inequalities,we have $f_{n+1} Therefore,$A = N$.

Class 12 Mathematics Relation and Function Mix Examples of Relation and Function

Let $N$ be the set of positive integers. For all $n \in N$,let $f_n = (n+1)^{1/3} - n^{1/3}$ and $A = \{n \in N : f_{n+1} < \frac{1}{3(n+1)^{2/3}} < f_n\}$. Then,

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A
$A = N$
B
$A$ is a finite set
C
the complement of $A$ in $N$ is nonempty,but finite
D
$A$ and its complement in $N$ are both infinite

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Mix Examples of Relation and Function

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Which of the following statements is correct?

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Let $N$ be the set of positive integers. For all $n \in N$,let $f_n = (n+1)^{1/3} - n^{1/3}$ and $A = \{n \in N : f_{n+1} < \frac{1}{3(n+1)^{2/3}} < f_n\}$. Then,

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Let $g(x) = 1 + x - [x]$ and $f(x) = \begin{cases} -1, & x < 0 \\ 0, & x = 0 \\ 1, & x > 0 \end{cases}$,where $[x]$ denotes the greatest integer less than or equal to $x$. Then for all $x$,$f(g(x)) = $

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Suppose $f:[-2, 2] \to R$ is defined by $f(x) = \begin{cases} -1 & \text{for } -2 \le x \le 0 \\ x - 1 & \text{for } 0 < x \le 2 \end{cases}$,then find the set $\{ x \in (-2, 2) : x \le 0 \text{ and } f(|x|) = x \}$.

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