Basic Maths

We know if \( |\mathrm{a}|+|\mathrm{b}|=|\mathrm{a}+\mathrm{b}|\)
Then \( \mathrm{a} \cdot \mathrm{~b} \geq 0 \)
\( \Rightarrow \mathrm{x}\left(\mathrm{x}^2-9\right) \geq 0\)
\( \mathrm{x} \in[-3,0] \cup[3, \infty)\)
\( \alpha^2+\beta^2+\gamma^2=18\)

$|x-2{|}^2+|x-2|-2=0$

$\left(\right|x-2|+2)\left(\right|x-2|-1)=0$

$|x-2|+2=0,|x-2|-1=0$

(not possible), $|x-2|=1$

$x=2\pm 1$

$x=3,1$

and $x^2-2(x-3)-5=0$

$\text{Case I:-\hspace{0.17em}\hspace{0.17em}}x⩾3$

$x^2-2x+6-5=0$

${(x-1)}^2=0$

$x=1$ (not possible)

$\text{Case II:-\hspace{0.17em}\hspace{0.17em}}x<3$

$x^2+2x-6-5=0$

$x^2+2x-11=0$

Let $\alpha , \beta$ be the roots of this equation then

${\alpha }^2+{\beta }^2={(\alpha +\beta )}^2-2\alpha \beta$

$={(-2)}^2-2(-11)=4+22=26$

Sum of the squares of the roots $=3^2+1^2+26=36$

$x|x-2|+3|x-3|+1=0$

$\mathrm{Case} \mathrm{I}:‐ \mathrm{If}x<2$

$-x^2+2x+3x+9+1=0$

$x^2+x-10=0x=\frac{-1-\sqrt{41}}{2},$

$x=\frac{-1+\sqrt{41}}{2}\left(x\right)$

$\text{CaseII:‐ If\hspace{0.17em} }2⩽x<3$

$x^2-2x-3x+9+1=0$

$x^2-5x+10=0$

$D=25-40=-15<0$

$\text{Case III:‐ If }x⩾3$

$x^2-2x+3x-9+1=0$

$x^2+x-8=0$

$x=\frac{-1-\sqrt{33}}{2}, x=\frac{-1+\sqrt{33}}{2}\left(x\right)$

$\Rightarrow$Number of real roots = 1

Given: \(\log _{(x+3)}\left(6 x^2+28 x+30\right)=5-2 \log _{(6 x+10)}\left(x^2+6 x+9\right)\)

$\Rightarrow {\log }_{x+3}\left[\left(x+3\right)\left(6x+10\right)\right]=5-2{\log }_{6x+10}{\left(x+3\right)}^2$

$\Rightarrow 1+{\log }_{x+3}(6x+10)=5-4{\log }_{6x+10}(x+3)$

Let ${\log }_{(x+3)}(6x+10)=A$

$1+A=5-\frac{4}{A}$

$\begin{array}{l}\Rightarrow A+\frac{4}{A}=4\Rightarrow A=2\\ \Rightarrow {\log }_{(x+3)}\left(6x+10\right)=2\\ \Rightarrow 6x+10={(x+3)}^{\mathbf{2}}\\ \Rightarrow 6x+10=x^2+9+6x\\ \Rightarrow x^2=1\Rightarrow x=\pm 1\end{array}$

So sum of roots \(= 0\)

$|x-2{|}^2+|x-2|-2=0$

$\left(\right|x-2|+2)\left(\right|x-2|-1)=0$

$|x-2|+2=0,|x-2|-1=0$

(not possible), $|x-2|=1$

$x=2\pm 1$

$x=3,1$

and $x^2-2(x-3)-5=0$

$\text{Case I:-\hspace{0.17em}\hspace{0.17em}}x⩾3$

$x^2-2x+6-5=0$

${(x-1)}^2=0$

$x=1$ (not possible)

$\text{Case II:-\hspace{0.17em}\hspace{0.17em}}x<3$

$x^2+2x-6-5=0$

$x^2+2x-11=0$

Let $\alpha , \beta$ be the roots of this equation then

${\alpha }^2+{\beta }^2={(\alpha +\beta )}^2-2\alpha \beta$

$={(-2)}^2-2(-11)=4+22=26$

Sum of the squares of the roots $=3^2+1^2+26=36$

Given: \(e^{5(\ln x)^2+3}=x^8\)

Taking \(\log\) on both sides:

\(\begin{aligned} & \Rightarrow \ln e^{5(\ln x)^2+3}=\ln x^8 \\ & \Rightarrow 5(\ln x)^2+3=8 \ln x\end{aligned}\)

Let \(\ln x=t\)

\(\begin{aligned} & \Rightarrow 5 t^2-8 t+3=0 \\ & t_1+t_2=\frac{8}{5} \\ & \Rightarrow \ln x_1+\ln x_2=\frac{8}{5} \\ & \Rightarrow \ln x_1 x_2=\frac{8}{5} \\ & \Rightarrow x_1 x_2=e^{8 / 5}\end{aligned}\)

\(\begin{aligned}
& e^{5(l n x)^2+3}=x^8 \\
& 5(\ell n x)^2+3=8 \ell n x \\
& \Rightarrow \quad \ell \mathrm{nx}=\mathrm{t} \\
& \Rightarrow \quad 5 t^2-8 t+3=0 \\
& \Rightarrow \quad(t-1)(5 t-3)=0 \\
& t=\frac{3}{5}, 1 \\
& \ell \mathrm{nx}=\frac{3}{5} \quad ; \quad \ell \mathrm{nx}=1 \\
& \mathrm{x}=\mathrm{e}^{3 / 5} \quad ; \quad \mathrm{x}=\mathrm{e}^1 \\
& \text { Product }=\mathrm{e}^{3 / 5} \cdot \mathrm{e}^1=\mathrm{e}^{8/5} \text { Ans. }
\end{aligned}\)

\(\begin{aligned}
& e^{5(l n x)^2+3}=x^8 \\
& 5(\ell n x)^2+3=8 \ell n x \\
& \Rightarrow \quad \ell \mathrm{nx}=\mathrm{t} \\
& \Rightarrow \quad 5 t^2-8 t+3=0 \\
& \Rightarrow \quad(t-1)(5 t-3)=0 \\
& t=\frac{3}{5}, 1 \\
& \ell \mathrm{nx}=\frac{3}{5} \quad ; \quad \ell \mathrm{nx}=1 \\
& \mathrm{x}=\mathrm{e}^{3 / 5} \quad ; \quad \mathrm{x}=\mathrm{e}^1 \\
& \text { Product }=\mathrm{e}^{3 / 5} \cdot \mathrm{e}^1=\mathrm{e}^{8/5} \text { Ans. }
\end{aligned}\)

$18x^2-11x+1>0$

$\Rightarrow (2x-1)(9x-1)>0$

$\Rightarrow x<\frac{1}{9}\text{ or }\frac{1}{2}<x$

Also, $10x^2-17x+7>0$

$\Rightarrow (x-1)(10x-7)>0$

$\Rightarrow x<\frac{7}{10}\text{ or }1<x$

Also, $10x^2-17x+7\ne 1$

\(\Rightarrow 10 x^2-17 x+6 \neq 0 \)
\(\Rightarrow (5 x-6)(2 x-1) \neq 0 \)
\(\Rightarrow x \neq \frac{6}{5}, \frac{1}{2} \)

Hence, $x\in \left(-\infty ,\frac{1}{9}\right)\cup \left(\frac{1}{2},\frac{7}{10}\right)\cup (1,\infty )-\left\{\frac{6}{5}\right\}$

$90(a+b+c+d+e)=90\left(\frac{1}{9}+\frac{1}{2}+\frac{7}{10}+1+\frac{6}{5}\right)$

$=10+45+63+90+108=316$

Given: \(x|x+4|+3|x+2|+10=0\)

Case 1: \(x \geq-2\)

\(|x+4|=x+4, |x+2|=x+2\)

\(\therefore x(x+4)+3(x+2)+10=0\)

\(\Rightarrow x^2+7 x+16=0\)

\(D=49-64=-15<0 \Rightarrow\) no real solution

Case 2: \(-4 \leq x<-2\)

\(|x+4|=x+4, |x+2|=-(x+2)\)

\(\therefore x(x+4)+3(-(x+2))+10=0\)

\(\Rightarrow x^2+x+4=0\)

\(D=1-16=-15<0 \Rightarrow\) no real solution

Case 3: \(x<-4\)

\(|x+4|=-(x+4), |x+2|=-(x+2)\)

\(\therefore x(-(x+4))+3(-(x+2))+10=0\)

\(\Rightarrow x^2+7 x-4=0\)

\(\Rightarrow x=\frac{-7 \pm \sqrt{65}}{2}\)

Only \(\frac{-7-\sqrt{65}}{2}<-4\) is valid.

Number of distinct real solutions \(=1\)

$|x-2{|}^2+|x-2|-2=0$

$\left(\right|x-2|+2)\left(\right|x-2|-1)=0$

$|x-2|+2=0,|x-2|-1=0$

(not possible), $|x-2|=1$

$x=2\pm 1$

$x=3,1$

and $x^2-2(x-3)-5=0$

$\text{Case I:-\hspace{0.17em}\hspace{0.17em}}x⩾3$

$x^2-2x+6-5=0$

${(x-1)}^2=0$

$x=1$ (not possible)

$\text{Case II:-\hspace{0.17em}\hspace{0.17em}}x<3$

$x^2+2x-6-5=0$

$x^2+2x-11=0$

Let $\alpha , \beta$ be the roots of this equation then

${\alpha }^2+{\beta }^2={(\alpha +\beta )}^2-2\alpha \beta$

$={(-2)}^2-2(-11)=4+22=26$

Sum of the squares of the roots $=3^2+1^2+26=36$

$|x-2{|}^2+|x-2|-2=0$

$\left(\right|x-2|+2)\left(\right|x-2|-1)=0$

$|x-2|+2=0$ (not possible)

$|x-2|=1$

$x=2\pm 1$

$x=3,1$

Now, $x^2-2\left|x-3\right|-5=0$

Case l:- \(x \geqslant 3\)

$x^2-2x+6-5=0$

${(x-1)}^2=0$

$x=1$ (not possible)

Case II:- \(x<3\)

$x^2+2x-6-5=0$

$x^2+2x-11=0$

Let $\alpha , \beta$ be the roots of this equation, then

${\alpha }^2+{\beta }^2={(\alpha +\beta )}^2-2\alpha \beta$

\(=(-2)^2-2(-11)=4+22=26\)

Sum of the squares of the roots $=3^2+1^2+26=36$

$x|x-2|+3|x-3|+1=0$

Case I:- If \(x<2\)

$-x^2+2x-3x+9+1=0$

$$\begin{gathered} x^2+x-10=0 \\ x=\frac{-1-\sqrt{41}}{2} \end{gathered}$$

$x=\frac{-1+\sqrt{41}}{2}$

\(\frac{-1+\sqrt{41}}{2}>2\) must be rejected.

Case ll:- If \(2 \leqslant x<3\)

$x^2-2x-3x+9+1=0$

$x^2-5x+10=0$

$D=25-40=-15<0$

Case III:- If \(x \geqslant 3\)

$x^2-2x+3x-9+1=0$

$x^2+x-8=0$

$x=\frac{-1-\sqrt{33}}{2}, x=\frac{-1+\sqrt{33}}{2}$

Both roots \(<3\), so no solution here

$\Rightarrow$Number of real roots \(= 1\)