(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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(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 Relations and Functions 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,

KVPY 2019 All KVPY

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

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Mathematics Questions

Chapter

Relations and Functions

Subtopic

Mix Examples of Relation and Function

Previous Year

KVPY 2019 Paper All KVPY years →

Let the function $f(x) = x^2 + x + \sin x - \cos x + \log(1 + |x|)$ be defined over the interval $[0, 1]$. The odd extension of $f(x)$ to the interval $[-1, 1]$ is:

Difficult

View Solution

$f: R \rightarrow R$ is a function defined by $f(x) = \frac{1}{e^x + 2e^{-x}}$. Assertion $(A): f(c) = \frac{1}{3}$ for some values of $c \in R$. Reason $(R): 0 < f(x) \leq \frac{1}{2\sqrt{2}}$ for all $x \in R$. Then which of the following options is correct?

MediumTS EAMCET 2023

View Solution

If $a+\alpha=1, b+\beta=2$ and $af(x)+\alpha f\left(\frac{1}{x}\right)=bx+\frac{\beta}{x}$ for $x \neq 0$,then the value of the expression $\frac{f(x)+f\left(\frac{1}{x}\right)}{x+\frac{1}{x}}$ is ..... .

DifficultJEE MAIN 2021

View Solution

The function $f(x) = x^{\frac{1}{\ln x}}$ is:

View Solution

If the function $f:(-\infty,-1] \rightarrow(a, b]$ defined by $f(x)=e^{x^3-3 x+1}$ is one-one and onto,then the distance of the point $P(2 b+4, a+2)$ from the line $x+e^{-3} y=4$ is:

DifficultJEE MAIN 2024

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

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,

Similar Questions

Let the function $f(x) = x^2 + x + \sin x - \cos x + \log(1 + |x|)$ be defined over the interval $[0, 1]$. The odd extension of $f(x)$ to the interval $[-1, 1]$ is:

$f: R \rightarrow R$ is a function defined by $f(x) = \frac{1}{e^x + 2e^{-x}}$. Assertion $(A): f(c) = \frac{1}{3}$ for some values of $c \in R$. Reason $(R): 0 < f(x) \leq \frac{1}{2\sqrt{2}}$ for all $x \in R$. Then which of the following options is correct?

If $a+\alpha=1, b+\beta=2$ and $af(x)+\alpha f\left(\frac{1}{x}\right)=bx+\frac{\beta}{x}$ for $x \neq 0$,then the value of the expression $\frac{f(x)+f\left(\frac{1}{x}\right)}{x+\frac{1}{x}}$ is ..... .

The function $f(x) = x^{\frac{1}{\ln x}}$ is:

If the function $f:(-\infty,-1] \rightarrow(a, b]$ defined by $f(x)=e^{x^3-3 x+1}$ is one-one and onto,then the distance of the point $P(2 b+4, a+2)$ from the line $x+e^{-3} y=4$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module