(4)

Consider the quadratic polynomials in the form of equation

$x^2+20x-2020=0\cdots \left(i\right)$

$x^2-20x+2020=0\cdots \left(ii\right)$

$\text{Since, }a\text{ and }b\text{ are roots of the equation }\left(i\right),\text{ then}$

$a+b=-20,ab=-2020$

$\because$ $c\text{ and }d\text{ are the roots of the equation }\left(ii\right),\text{ then}$

$c+d=20,cd=2020$

$\text{Now,}$

$ac(a-c)+ad(a-d)+bc(b-c)+bd(b-d)$

$=a^{2}c-a{c}^2+a^{2}d-a{d}^2+b^{2}c-b{c}^2+b^{2}d-b{d}^2$

$=a^2(c+d)+b^2(c+d)-c^2(a+b)-d^2(a+b)$

$=(c+d)(a^2+b^2)-(a+b)(c^2+d^2)$

$=(c+d)({(a+b)}^2-2ab)-(a+b)({(c+d)}^2-2cd)$

$=20[{\left(20\right)}^2+4040]+20[{\left(20\right)}^2-4040]$

$=20\times 800=16000$

(3)

$x^2-2x\sec \theta +1=0\Rightarrow x=\sec \theta \pm \tan \theta$

$\text{and }{x}^2+2x\tan \theta -1=0\Rightarrow x=-\tan \theta \pm \sec \theta$

$\because$ $-\frac{\pi }{6}<\theta <-\frac{\pi }{12}\Rightarrow \sec \frac{\pi }{6}>\sec \theta >\sec \frac{\pi }{12}$

$\text{and }-\tan \frac{\pi }{6}<\tan \theta <-\frac{\tan \pi }{12}$

$\text{Also }\tan \frac{\pi }{12}<-\tan \theta <\tan \frac{\pi }{6}$

$\text{Since, }{\alpha }_1,{\beta }_1\text{ are roots of }{x}^2-2x\sec \theta +1=0\text{ and }{\alpha }_1>{\beta }_1$

$\therefore$ ${\alpha }_1=\sec \theta -\tan \theta \text{ and }{\beta }_1=\sec \theta +\tan \theta$

$\text{Since, }{\alpha }_2,{\beta }_2\text{ are roots of }{x}^2+2x\tan \theta -1=0\text{ and }{\alpha }_2>{\beta }_2$

$\therefore$ ${\alpha }_2=-\tan \theta +\sec \theta ,{\beta }_2=-\tan \theta -\sec \theta$

$\therefore$ ${\alpha }_1+{\beta }_2=\sec \theta -\tan \theta -\tan \theta -\sec \theta =-2\tan \theta$

(4)

Quadratic equation with real coefficients and purely imaginary roots can be considered as

$p\left(x\right)=x^2+a=0\text{ where }a>0\text{ and }a\in \mathrm{ℝ}$

$\text{Then }p\left[p\right(x\left)\right]=0\Rightarrow {(x^2+a)}^2+a=0$

$\Rightarrow x^4+2a{x}^2+(a^2+a)=0$

$\Rightarrow x^2=\frac{-2a\pm \sqrt{4a^2-4(a^2+a)}}{2}$$=\frac{-2a\pm \sqrt{4a^2-4a^2-4a}}{2}$

$=\frac{-2a\pm \sqrt{-4a}}{2}$$\Rightarrow x^2=-a\pm \sqrt{a}i$

$\Rightarrow x=\sqrt{-a\pm \sqrt{a}i}=\alpha \pm i\beta ,\text{ where }\alpha ,\beta \ne 0$

$\therefore$ $p\left[p\right(x\left)\right]=0\text{ has complex roots which are neither purely real nor purely imaginary.}$

(3)

$\text{Consider }-3{(x-[x\left]\right)}^2+2[x-[x\left]\right]+a^2=0$

$\Rightarrow 3{\left\{x\right\}}^2-2\left\{x\right\}-a^2=0(\because x-[x]=\{x\left\}\right)$

$\Rightarrow 3({\left\{x\right\}}^2-\frac{2}{3}\{x\left\}\right)=a^2,a\ne 0$

$\Rightarrow a^2=3{\left(\right\{x\}-\frac{1}{3})}^2-\frac{1}{3}(\because 0\le \{x\}<1)$

$-\frac{1}{3}\le \left\{x\right\}-\frac{1}{3}<\frac{2}{3}\Rightarrow 0\le 3\left(\right\{x\left\}\frac{-1}{3}\right)<\frac{4}{3}$

$-\frac{1}{3}\le 3\left(\right\{x\left\}\frac{-1}{3}\right)-\frac{1}{3}<1$

$\text{For non-integral solution }0<a^2<1$

$\Rightarrow a\in (-1,0)\cup (0,1)$

(3)

$\because$ $\alpha ,\beta \text{ are the roots of }{x}^2-6x-2=0$

$\therefore$ ${\alpha }^2-6\alpha -2=0$

$\Rightarrow {\alpha }^{10}-6{\alpha }^9-2{\alpha }^8=0$

$\Rightarrow {\alpha }^{10}-2{\alpha }^8=6{\alpha }^9\cdots \left(i\right)$

$\text{Similarly }{\beta }^{10}-2{\beta }^8=6{\beta }^9\cdots \left(ii\right)$

$\text{On subtracting (ii) from (i),}$

${\alpha }^{10}-{\beta }^{10}-2({\alpha }^8-{\beta }^8)=6({\alpha }^9-{\beta }^9)$

$\Rightarrow a_{10}-2a_8=6a_9$

$\Rightarrow \frac{a_{10}-2a_8}{2a_9}=3$

(3)

$\text{Given : }{\left(2x\right)}^{\ln 2}={\left(3y\right)}^{\ln 3}$

$\Rightarrow \ln 2·\ln \left(2x\right)=\ln 3·\ln \left(3y\right)$

$\Rightarrow \ln 2·\ln \left(2x\right)=\ln 3·(\ln 3+\ln y)\cdots \left(i\right)$

$\text{Also given : }{3}^{\ln x}=2^{\ln y}$

$\Rightarrow \ln x·\ln 3=\ln y·\ln 2\Rightarrow \ln y=\frac{\ln x·\ln 3}{\ln 2}\cdots \left(ii\right)$

$\text{From equation (i) and (ii), we get}$

$\ln 2·\ln \left(2x\right)=\ln 3[\ln 3+\frac{\ln x·\ln 3}{\ln 2}]$

$\Rightarrow {\left(\ln 2\right)}^2\ln \left(2x\right)={\left(\ln 3\right)}^2\ln 2+{\left(\ln 3\right)}^2\ln x$

$\Rightarrow {\left(\ln 2\right)}^2\ln \left(2x\right)={\left(\ln 3\right)}^2(\ln 2+\ln x)$

$\Rightarrow {\left(\ln 2\right)}^2\ln \left(2x\right)-{\left(\ln 3\right)}^2\ln \left(2x\right)=0$

$\Rightarrow [{\left(\ln 2\right)}^2-{\left(\ln 3\right)}^2]\ln \left(2x\right)=0$$\Rightarrow \ln \left(2x\right)=0$

$\Rightarrow 2x=1\Rightarrow x=\frac{1}{2}$

(2)

$\text{Given : }\alpha +\beta =-p\text{ and }{\alpha }^3+{\beta }^3=q$

$\Rightarrow {(\alpha +\beta )}^3-3\alpha \beta (\alpha +\beta )=q$

$\Rightarrow -p^3-3\alpha \beta (-p)=q\Rightarrow \alpha \beta =\frac{p^3+q}{3p}$

$\text{Now for required quadratic equation,}$

$\text{Sum of roots }=\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{{\alpha }^2+{\beta }^2}{\alpha \beta }$

$=\frac{{(\alpha +\beta )}^2-2\alpha \beta }{\alpha \beta }$$=\frac{p^2-2\left(\frac{p^3+q}{3p}\right)}{\frac{p^3+q}{3p}}=\frac{p^3-2q}{p^3+q}$

$\text{and Product of roots }=\frac{\alpha }{\beta }·\frac{\beta }{\alpha }=1$

$\therefore$ $\text{Required equation is }$$x^2-\left(\frac{p^3-2q}{p^3+q}\right)x+1=0$

$\Rightarrow (p^3+q)x^2-(p^3-2q)x+(p^3+q)=0$

(4)

$\text{Since }\alpha \text{ and }\beta \text{ are the roots of }{x}^2-px+r=0$

$\therefore$ $\alpha +\beta =p\cdots \left(i\right)$

      $\text{and }\alpha \beta =r\cdots \left(ii\right)$

$\text{Also }\frac{\alpha }{2}\text{ and }2\beta \text{ are the roots of }{x}^2-qx+r=0$

$\therefore$ $\frac{\alpha }{2}+2\beta =q\Rightarrow \alpha +4\beta =2q\cdots \left(iii\right)$

$\text{Solving (i) and (iii) for }\alpha \text{ and }\beta , \text{we get}$

$\beta =\frac{1}{3}(2q-p)\text{ and }\alpha =\frac{2}{3}(2p-q)$

$\text{On substituting the values of }\alpha \text{ and }\beta \text{ in equation (ii),}$

$\text{we get }\frac{2}{9}(2p-q)(2q-p)=r.$

(1)

$\because$ $a,b,c\text{ are sides of a triangle and }a\ne b\ne c$

$\therefore$ $|a-b|$$<$$\left|c\right|$$\Rightarrow a^2+b^2-2ab<c^2\cdots \left(i\right)$

$\text{Similarly,}$

$b^2+c^2-2bc<a^2\cdots \left(ii\right), c^2+a^2-2ca<b^2\cdots \left(iii\right)$

$\text{On adding (i), (ii) and (iii) we get}$

$a^2+b^2+c^2<2(ab+bc+ca)$

$\Rightarrow \frac{a^2+b^2+c^2}{ab+bc+ca}<2\cdots \left(iv\right)$

$\because$ $\text{Roots of the given equation are real}$

$\therefore$ ${(a+b+c)}^2-3\lambda (ab+bc+ca)\ge 0$

$\Rightarrow \frac{a^2+b^2+c^2}{ab+bc+ca}\ge 3\lambda -2\cdots \left(v\right)$

$\text{From (iv) and (v), }3\lambda -2<2\Rightarrow \lambda <\frac{4}{3}$

(1)

$x^2+px+q=0$

$\text{Let roots be }\alpha \text{ and }{\alpha }^2, \text{then}$

$\alpha +{\alpha }^2=-p,\alpha ·{\alpha }^2=q\Rightarrow \alpha =q^{1/3}$

$\therefore$ ${\left(q\right)}^{1/3}+{\left(q^{1/3}\right)}^2=-p$

$\text{On taking cube on both sides, we get}$

$q+q^2+3q(q^{1/3}+q^{2/3})=-p^3$

$\Rightarrow q+q^2-3pq=-p^3$

$\Rightarrow p^3+q^2-q(3p-1)=0$