The value of
\(\log _{3} 11 \log _{11} 13 \log _{13} 15 \log _{15} 27 \log _{27} 81=\log _{3} 81 \)
\(=\log _{3} 3^{4}=4 \log _{3} 3=4 \)

\(\frac{1-2 x}{3+2 x-x^{2}}= \frac{\mathrm{A}}{3-x}+\frac{\mathrm{B}}{x+1} \)
\( 1-2 x=\mathrm{A}(x+1)+\mathrm{B}(3-x) \)
\( \text { Put } x=3,-5=\mathrm{A}(4) \Rightarrow \mathrm{A}=\frac{-5}{4} \)
\( \text { Put } x=-1,3=\mathrm{B}(4) \Rightarrow \mathrm{B}=\frac{3}{4} \)
\(\mathrm{~A}+\mathrm{B}= \frac{-5}{4}+\frac{3}{4}=\frac{-2}{4}=\frac{-1}{2} \)

\(\frac{2}{x+2}+\frac{1}{x-1}=\frac{2 x-2+x+2}{(x+2)(x-1)}=\frac{3 x}{(x+2)(x-1)}= k = 3\)

\(x^{2}-\mathrm{k} x+\mathrm{c}=0\)
a and b are the roots
\(\therefore a+b=k, a b=c\)
To find
\(\sqrt{(a-b)^{2}+0^{2}}=a-b=\sqrt{(a+b)^{2}-4 a b}=\sqrt{k^{2}-4 c}\)
 

\(x^{2}+a x+c=0 \)
\(x^{2}+d x+b=0 \)
\(8 \& 2 \text { are the roots }\)\(\text { 3. } 3 \text { are the roots }\)
\(\therefore a=-10 ; c=16 \quad d=-6, \quad b=9\)
\(x^{2}+a x+b =0 \)
\(x^{2}-10 x+9 =0 \)
\(\Rightarrow(x-1)(x-9) =0 \)
\(\therefore x =1 \text { (or) } 9 \)

\(3 x^{2}-5 x-7=0\)
\(\text { Let the roots be } \alpha, \beta\)
\(\Rightarrow \operatorname{Sum}: \alpha+\beta=\frac{5}{3}\)
\(\text { Product : } \alpha \beta=\frac{-7}{3}\)
\(\text { Now take the roots are }-\alpha,-\beta\)
\(\Rightarrow \text { Sum : }-\alpha-\beta \quad-(\alpha+\beta)=-\frac{5}{3}\)
\(\text { Product : }(-\alpha)(-\beta)=\alpha \beta=\frac{-7}{3}\)
\(\text { The reouired equation }\)
\(x^{2}+\frac{5}{3} x-\frac{7}{3}=0 \)
\(3 x^{2}+5 x-7=0 \)

\(x^{2}+|x-1| =1\)
\(|x-1| =1-x^{2} \)
\(1-x^{2} =x-1 \text { (or) } 1-x^{2} =-x+1 \)
\(x^{2}+x-2 =0 \ x^{2}-x =0 \)
\((x+2)(x-1) =0 \ x(x-1) =0 \)
\(x =-2 \text { (or) } 1 \ x=0 ; x =1 \)
The roots are -2, 0, 1. Since -2 does not satisfy. The roots are 0 and 1.
.'. Number of solution is 2.

\(x^{2}-k x+16=0, a^{} \& b \text { are roots }\)
\(a+b =k ; a b=16 \)
\(a^{2}+b^{2} =(a+b)^{2}-2 a b \)
\(32 =k^{2}-32 \)
\(k^{2} =64 \)
\(k =\pm 8 \Rightarrow k=-8,8 \)

\(2 x^{2}+(a-3) x+3 a-5=0\)
\(\text {Sum }=\frac{-(a-3)}{2} ; \quad \text { Product }=\frac{3 a-5}{2}\)
\(\text {Given they are equal, } \frac{-(a-3)}{2}=\frac{3 a-5}{2}\)
\(-4 a =-8 \)
\(a =2 \)