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Question Number 103492 by abony1303 last updated on 15/Jul/20

$$\mathrm{Q}1:\mathrm{Evaluate}{\int }_{-1}^1\mid \mathrm{x}\mid \cdot ({\mathrm{x}}^3+1)\mathrm{dx}$$$$$$$$\mathrm{Q}2:\mathrm{Find}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{all}\mathrm{integers}k\mathrm{for}$$$$\mathrm{which}\mathrm{the}\mathrm{equation}2x^3-6x^2+k=0$$$$\mathrm{has}\mathrm{more}\mathrm{than}\mathrm{one}\mathrm{solution}.$$$$$$$$\mathrm{Q}3:\mathrm{Find}\mathrm{the}\mathrm{shortest}\mathrm{distance}\mathrm{from}\mathrm{a}$$$$\mathrm{point}\mathrm{on}\mathrm{the}\mathrm{curve}y=x^2-x\mathrm{to}\mathrm{the}\mathrm{line}$$$$y=x-3$$

Commented by abony1303 last updated on 15/Jul/20

$$\mathrm{pls}\mathrm{help}$$

Answered by mr W last updated on 15/Jul/20

$$Q2$$$$2x^2(x-3)+k=0$$$$k<0\Rightarrow oneroot:x>3$$$$k=0\Rightarrow tworoots:x=0or3$$$$k>0:$$$$\frac{1}{x^3}-\frac{6}{kx}+\frac{2}{k}=0$$$$suchthatthreeroots:$$$${\left(\frac{1}{k}\right)}^2+{(-\frac{2}{k})}^3<0$$$${\left(\frac{1}{k}\right)}^2(1-\frac{8}{k})<0$$$$1-\frac{8}{k}<0$$$$k<8$$$$\Rightarrow k=1,2,3,...,7$$$$\mathrm{\Sigma }k=\frac{7(1+7)}{2}=28$$

Commented by abony1303 last updated on 15/Jul/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}.\mathrm{but}\mathrm{I}\mathrm{drew}\mathrm{a}\mathrm{graph}\mathrm{and}\mathrm{k}=8$$$$\mathrm{also}\mathrm{correct}?\mathrm{And}\mathrm{pls}\mathrm{can}\mathrm{you}\mathrm{explain}\mathrm{the}$$$$\mathrm{raw}suchthatthreeroots...and\mathrm{below}it.$$

Commented by mr W last updated on 15/Jul/20

$$imisreadthequestionas:morethan$$$$tworoots.$$$$formorethanoneroot:theanswer$$$$is\mathrm{\Sigma }k=0+1+2+3+...+8=36$$$$$$$$wehave:$$$$\frac{1}{x^3}-\frac{6}{kx}+\frac{2}{k}=0$$$$lett=\frac{1}{x},then$$$$t^3-\frac{6}{k}t+\frac{2}{k}=0$$$$threerootsforx\Rightarrow threerootsfort.$$$$$$$$letf\left(t\right)=t^3+bt+c$$$$iff\left(t\right)=0hasthreeroots,itmeans$$$$f^{\prime }\left(t\right)=0\Rightarrow 3t^2+b=0\Rightarrow t_{1,2}=\pm \sqrt{-\frac{b}{3}}$$$$f\left(t_1\right)f\left(t_2\right)<0,seediagram.$$$$f\left(t_1\right)=t_1^3+{bt}_1+c=\frac{1}{3}{t}_1(3t_1^2+b)+\frac{2{bt}_1}{3}+c=\frac{2{bt}_1}{3}+c$$$$f\left(t_2\right)=\frac{2{bt}_2}{3}+c$$$$(\frac{2{bt}_1}{3}+c)(\frac{2{bt}_2}{3}+c)<0$$$$\frac{4b^2}{9}{t}_1{t}_2+\frac{2b}{3}(t_1+t_2)+c^2<0$$$$\frac{4b^2}{9}\times \frac{b}{3}+\frac{2b}{3}\times 0+c^2<0$$$$\Rightarrow {\left(\frac{b}{3}\right)}^3+{\left(\frac{c}{2}\right)}^2<0$$$$withb=-\frac{6}{k},c=\frac{2}{k}$$$$\Rightarrow {(-\frac{2}{k})}^3+{\left(\frac{1}{k}\right)}^2<0$$

Commented by mr W last updated on 15/Jul/20

Commented by mr W last updated on 15/Jul/20

Commented by mr W last updated on 15/Jul/20

Commented by mr W last updated on 15/Jul/20

Commented by mr W last updated on 15/Jul/20

Commented by mr W last updated on 15/Jul/20

Commented by mr W last updated on 15/Jul/20

$$threerootsonlyfor1⩽k⩽7.$$$$tworootsonlyfork=0or8.$$$$onerootonlyfork⩽-1ork⩾9.$$

Answered by mr W last updated on 15/Jul/20

$$Q3$$$$y=x^2-x$$$$y^{\prime }=2x-1=1\Rightarrow x=1,y=0$$$$d_{min}=\frac{\mid 1\times 1-1\times 0-3\mid }{\sqrt{1^2+{(-1)}^2}}=\sqrt{2}$$

Answered by mr W last updated on 15/Jul/20

$$Q1$$$${\int }_{-1}^1\mid \mathrm{x}\mid \cdot ({\mathrm{x}}^3+1)\mathrm{dx}$$$$={\int }_{-1}^1\mid \mathrm{x}\mid \cdot {\mathrm{x}}^3\mathrm{dx}+{\int }_{-1}^1\mid \mathrm{x}\mid \mathrm{dx}$$$$=0+2{\int }_0^1\mathrm{xdx}$$$$=1$$

Answered by OlafThorendsen last updated on 15/Jul/20

$$\mathrm{Q}1.$$$$\mathrm{I}={\int }_{-1}^1\mid x\mid (x^3+1)dx$$$$\mathrm{I}={\int }_{-1}^0-x(x^3+1)dx+{\int }_0^{1}x(x^3+1)dx$$$$\mathrm{I}={[-\frac{x^5}{5}-\frac{x^2}{2}]}_{-1}^0+{[\frac{x^5}{5}+\frac{x^2}{2}]}_0^1$$$$\mathrm{I}=(-\frac{1}{5}+\frac{1}{2})+(\frac{1}{5}+\frac{1}{2})$$$$\mathrm{I}=1$$$$\mathrm{Q}2.$$$${\mathrm{\Delta }}^{\prime }={(-3)}^2-\left(2\right)\left(k\right)=9-2k$$$$\mathrm{More}\mathrm{than}\mathrm{one}\mathrm{solution}\iff {\mathrm{\Delta }}^{\prime }>0$$$$\iff k<\frac{9}{2}$$$$\iff k\in \{0;1;2;3;4\}$$$$\Rightarrow \mathrm{sum}=0+1+2+3+4=10$$$$\mathrm{Q}3.$$$$\mathrm{A}\mathrm{is}\mathrm{a}\mathrm{point}\mathrm{of}\mathrm{the}\mathrm{curve}$$$$\mathrm{B}\mathrm{is}\mathrm{a}\mathrm{point}\mathrm{of}\mathrm{the}\mathrm{line}$$$$\mathrm{A}\left(\begin{array}{c}a\\ a^2-a\end{array}\right)\mathrm{and}\mathrm{B}\left(\begin{array}{c}b\\ b-3\end{array}\right)$$$${\mathrm{AB}}^2={(b-a)}^2+{(b-3-a^2+a)}^2$$$$\mathrm{for}\mathrm{a}\mathrm{given}a,{\mathrm{AB}}^2=f\left(b\right)$$$$f\left(b\right)={(b-a)}^2+{(b-3-a^2+a)}^2$$$$f^{\prime }\left(b\right)=2(b-a)+2(b-3-a^2+a)$$$$f^{\prime }\left(b\right)=4b-2a^2-6$$$$f^{\prime }\left(b\right)=0\iff b=\frac{a^2+3}{2}$$$$\mathrm{Then}:$$$${\mathrm{AB}}^2={(\frac{a^2+3}{2}-a)}^2+{(\frac{a^2+3}{2}-3-a^2+a)}^2$$$${\mathrm{AB}}^2={(\frac{a^2}{2}-a+\frac{3}{2})}^2+{(-\frac{a^2}{2}+a-\frac{3}{2})}^2$$$${\mathrm{AB}}^2=2{(\frac{a^2}{2}-a+\frac{3}{2})}^2$$$$\mathrm{AB}=\sqrt{2}\mid \frac{a^2}{2}-a+\frac{3}{2}\mid \mathrm{and}\mathrm{then}:$$$$\mathrm{A}\left(\begin{array}{c}a\\ a^2-a\end{array}\right)\mathrm{and}\mathrm{B}\left(\begin{array}{c}\frac{a^2+3}{2}\\ \frac{a^2-3}{2}\end{array}\right)$$

Commented by abony1303 last updated on 15/Jul/20

$$\mathrm{no},\mathrm{no}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{ok}.:)$$

Commented by OlafThorendsen last updated on 15/Jul/20

$$\mathrm{sorry}!\mathrm{I}\mathrm{need}\mathrm{to}\mathrm{change}\mathrm{my}$$$$\mathrm{glasses}!:-)$$

Commented by abony1303 last updated on 15/Jul/20

$$\mathrm{Thank}\mathrm{you}\mathrm{ser}.\mathrm{Everything}\mathrm{is}\mathrm{clear},\mathrm{but}$$$$\mathrm{can}\mathrm{you}\mathrm{pls}\mathrm{explain}\mathrm{what}\mathrm{is}{△}^{\prime }\mathrm{in}\mathrm{Q}2?$$

Commented by bemath last updated on 15/Jul/20

$$\mathrm{\Delta }=b^2-4ac$$$$discriminant$$

Commented by abony1303 last updated on 15/Jul/20

$$\mathrm{but}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{not}\mathrm{the}\mathrm{quadratic}\mathrm{equation}?$$

Commented by OlafThorendsen last updated on 15/Jul/20

$${\mathrm{\Delta }}^{\prime }:\mathrm{reduced}\mathrm{discriminant}$$$$\left(\mathrm{sorry}\mathrm{my}\mathrm{english}\mathrm{is}\mathrm{not}\mathrm{fluent}\right)$$$$x=\frac{-b\pm \sqrt{\mathrm{\Delta }}}{2a}\mathrm{with}\mathrm{\Delta }=b^2-4ac$$$$\mathrm{but}\mathrm{when}b\mathrm{is}\mathrm{even}\mathrm{you}\mathrm{can}\mathrm{use}:$$$$x=\frac{-b^{\prime }\pm \sqrt{{\mathrm{\Delta }}^{\prime }}}{a}$$$$\mathrm{with}{b}^{\prime }=\frac{b}{2}\mathrm{and}{\mathrm{\Delta }}^{\prime }=b^{\prime 2}-ac$$$$\left(\mathrm{the}\mathrm{calculation}\mathrm{is}\mathrm{faster}\right)$$

Commented by abony1303 last updated on 15/Jul/20

$$\mathrm{ser}\mathrm{but}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{not}\mathrm{the}\mathrm{quadratic}\mathrm{equation}?$$

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