Help request(Spams are reported.) How would you solve this question? The answer is 415, but there's no explanation in the book. Topic: Floor Function What is the value of [tex]\left \lfloor{(\sqrt{3} +1})^6}\right \rfloor[/tex]? ([tex]\left \lfloor{n}\right \rfloor[/tex] means the greatest integer which isn't greater than [tex]n[/tex].). Created by TakenName. Math

Topic :-FunctionsGiven :-where[n] represents Greatest Integer Function.To Find :-The value of given expression.Solution :-Technique 1Solve the brackets using algebraic addition.So, greatest integer will be,Technique 2Solve the brackets using Substitution.Replacing values,So, greatest integer will be,415Answer :-So, the greatest integer function / floor function value of given expression will be 415.Note :There are many ways to solve this problem. The main point/technique to solve this question is to solve the value of given exponent.

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assingh

Verified answer

Topic :-

Functions

Given :-

$[ (\sqrt{3}+1)^6]$

where

[n] represents Greatest Integer Function.

To Find :-

The value of given expression.

Solution :-

Technique 1

Solve the brackets using algebraic addition.

$[ (\sqrt{3}+1)^6]$

$[ (1.7320+1)^6]$

$[ (2.7320)^6]$

$[415.80]$

So, greatest integer will be,

$415$

Technique 2

Solve the brackets using Substitution.

$[ (\sqrt{3}+1)^6]$

$\because \cos 15^{\circ}=\dfrac{\sqrt{3}+1}{2\sqrt{2}}$

$\therefore \sqrt{3}+1=2\sqrt{2}\cos 15^{\circ}$

Replacing values,

$[ (2\sqrt{2}\cos 15^{\circ})^6]$

$[ (2\sqrt{2})^6(\cos 15^{\circ})^6]$

$[ 512\cos^6 15^{\circ}]$

$\because \cos^615^{\circ}=0.8121$

$[ 512\times 0.8121]$

$[415.80]$

So, greatest integer will be,

415

Answer :-

So, the greatest integer function / floor function value of given expression will be 415.

Note :

There are many ways to solve this problem. The main point/technique to solve this question is to solve the value of given exponent.

mathdude500

$\large\underline{\bold{Given \:Question - }}$

$\bf \:What \: is \: the \: value \: of \: \lfloor \sf \: {( \sqrt{3} + 1) }^{6} \rfloor$

$\large\underline{\bold{Solution-}}$

We know,

By expanding using Binomial Theorem,

$\sf \:{(x + 1)}^{6}=^6C_0 +^6 C_1x+ ^6 C_2 {x}^{2}+^6 C_3 {x}^{3}+^6 C_4 {x}^{4}+^6 C_5 {x}^{5}+^6 C_6 {x}^{6}$

$\sf \: {(x + 1)}^{6} = 1 + 6x + 15 {x}^{2} + 20 {x}^{3} + 15 {x}^{4} + 6 {x}^{5} + {x}^{6}$

$\bf \: Put \: x \: = \: \sqrt{3} \: \: we \: get$

$\sf \: {( \sqrt{3}+1)}^{6}=1+6\sqrt{3} + 15 {(\sqrt{3})}^{2}+20{(\sqrt{3})}^{3}+ 15 {(\sqrt{3})}^{4}+6 {(\sqrt{3})}^{5}+ {(\sqrt{3})}^{6}$

$\sf \: {(\sqrt{3} + 1)}^{6} = 1 + 6 \sqrt{3} + 45 + 60 \sqrt{3} + 135 + 54 \sqrt{3} + 27$

$\sf \: {( \sqrt{3} + 1) }^{6} = 208 + 120 \sqrt{3}$

$\sf \: {( \sqrt{3} + 1) }^{6} = 208 + 120 \times (1.732)$

$\sf \: {( \sqrt{3} + 1) }^{6} = 208 + 120 \times (1 + 0.732)$

$\sf \: {( \sqrt{3} + 1) }^{6} = 208 + 120 + 120 \times 0.732$

$\sf \: {( \sqrt{3} + 1) }^{6} = 328 + 120(0.7 + 0.032)$

$\sf \: {( \sqrt{3} + 1) }^{6} = 412 + 120 \times (0.025 + 0.007)$

$\sf \: {( \sqrt{3} + 1) }^{6} = 412 + 3 + 0.84$

$\sf \: {( \sqrt{3} + 1) }^{6} = 415 + 0.84$

Hence,

$\lfloor \sf \: {( \sqrt{3} + 1) }^{6} \rfloor = \lfloor(415 + 0.84) \rfloor$

$\therefore \bf \: \lfloor \sf \: { \bf( \sqrt{3} + 1) }^{6} \rfloor \: = \bf \: 415$

${\bf{Hence, \: Proved}}$

Note :-

The floor function (also known as the greatest integer function) ⌊⋅⌋ of a real number x denotes the greatest integer less than or equal to x.

Mathematically,

$\sf \: \lfloor x \rfloor = \max \{ n \in \mathbb{Z} \mid n \leq x \}$

and

Ceiling Function, mathematically given by

$\lceil x \rceil = \min \{ n \in \mathbb{Z} \mid n \geq x \}$

Important inequalities:

$x-1 \;$

At the same time, we have the right-going implications of the following statements (x is a real number and n an integer):

$\begin{aligned} n=\lfloor x \rfloor \quad &\Longleftrightarrow \quad n \leq x < n+1, \\ n=\lfloor x \rfloor \quad &\Longleftrightarrow \quad x-1 < n \leq x, \\ n=\lceil x \rceil \quad &\Longleftrightarrow \quad n-1 < x \leq n, \\ n=\lceil x \rceil \quad &\Longleftrightarrow \quad x \leq n < x+1. \\ \end{aligned}$

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Verified answer

Topic :-

Given :-

To Find :-

Solution :-

Technique 1

Technique 2

Answer :-

Note :-

Important inequalities:

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