(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

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

(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

Medium

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

WBJEE 2022 All WBJEE

A
$f(n) > g(n)$
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

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Mathematics Questions

Chapter

Relations and Functions

Subtopic

Different types of Function

Previous Year

WBJEE 2022 Paper All WBJEE years →

Assertion $(A)$: $\coth x = \frac{1-k}{1+k}$ where $0 < k < 2$.
Reason $(R)$: The graph of $y = \tanh x$ always lies between the lines $y = -1$ and $y = 1$.
Choose the correct option:

MediumTS EAMCET 2022

View Solution

Let $R$ be the set of real numbers. Define the real function $f: R \rightarrow R$ by $f(x) = x + 10$ and sketch the graph of this function.

Easy

View Solution

Let $g: (-\infty, \infty) \to (-\frac{\pi}{2}, \frac{\pi}{2})$ be defined by $g(x) = 2 \tan^{-1}(e^x) - \frac{\pi}{2}$. Then $g(x)$ is...

Difficult

View Solution

The equation $6^{x}+8^{x}=10^{x}$ has

MediumWBJEE 2021

View Solution

Let the function $g : (-\infty, \infty) \to \left( - \frac{\pi}{2}, \frac{\pi}{2} \right)$ be given by $g(u) = 2 \tan^{-1}(e^u) - \frac{\pi}{2}$. Then,$g$ is -

Medium

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

Similar Questions

Assertion $(A)$: $\coth x = \frac{1-k}{1+k}$ where $0 < k < 2$.Reason $(R)$: The graph of $y = \tanh x$ always lies between the lines $y = -1$ and $y = 1$.Choose the correct option:

Let $R$ be the set of real numbers. Define the real function $f: R \rightarrow R$ by $f(x) = x + 10$ and sketch the graph of this function.

Let $g: (-\infty, \infty) \to (-\frac{\pi}{2}, \frac{\pi}{2})$ be defined by $g(x) = 2 \tan^{-1}(e^x) - \frac{\pi}{2}$. Then $g(x)$ is...

The equation $6^{x}+8^{x}=10^{x}$ has

Let the function $g : (-\infty, \infty) \to \left( - \frac{\pi}{2}, \frac{\pi}{2} \right)$ be given by $g(u) = 2 \tan^{-1}(e^u) - \frac{\pi}{2}$. Then,$g$ is -

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Assertion $(A)$: $\coth x = \frac{1-k}{1+k}$ where $0 < k < 2$.
Reason $(R)$: The graph of $y = \tanh x$ always lies between the lines $y = -1$ and $y = 1$.
Choose the correct option: