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  • The sum of integral value of the elements in the domain of f(x)

Sir 3rd and 4th question 3 The sum of integral value of the elements in the domain of f(x) = sqrt(log_(1/3)|3-x|) is4 Number of integers in range of f(x)=x(x+2)(x+4)(x+6)+7, x forall [-4,2] is

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The sum of integral value of the elements in the domain of f(x) is

\mathrm{Domain\:of\:}\:\sqrt{\log _{0.5}\left(\left|3-x\right|\right)}\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:2\le \:x<3\quad \mathrm{or}\quad \:3<x\le \:4\:\\ \:\mathrm{Interval\:Notation:}&\:[2,\:3)\cup (3,\:4]\end{bmatrix}

So, there are two integer 4 and 2 

sum of integer = 4+2=6

 

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avinash.dongre

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So, there are two integer 4 and 2 

sum of integer = 4+2=6

\mathrm{Critical\:Points\:of}\:x\left(x+2\right)\left(x+4\right)\left(x+6\right)+7:\quad x=-3-\sqrt{5},\:x=-3,\:x=\sqrt{5}-3\\\mathrm{Range\:of\:}x\left(x+2\right)\left(x+4\right)\left(x+6\right)+7:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\ge \:-9\:\\ \:\mathrm{Interval\:Notation:}&\:[-9,\:\infty \:)\end{bmatrix}\\in \the\ domain\ [-4,2]\ minimum \ is\ at\sqrt{5}-3\\f(\sqrt{5}-3)=-9\\f(2)=391\\total\ number\ of\ intigers=391+9+1(of zero)=401

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Safeer PP

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