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

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

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

\(\text { Let } \log _{\sqrt{2}} 512=x\)
\(\text { Then }(\sqrt{2})^{x}=2^{9}\)
\(\Rightarrow 2^{\frac{x}{2}}=2^{9} \Rightarrow x / 2=9 \Rightarrow x=18\)

\(\text { In }(-\infty, 2), \frac{|x-2|}{x-2} \text { is negative }\)
\(\operatorname{In}[2, \infty), \frac{|x-2|}{x-2} \geq 0\)

\(\frac{\sin (A-B)}{\cos A \cos B} =\frac{\sin A \cos B-\cos A \sin B}{\cos A \cos B} \)
\(=\tan A-\tan B \)
\(\text { L.H.S } =\tan A-\tan B+\tan B-\tan C+\tan C-\tan A=0\)

\((A \times B) \cap(A \times C) =A \times(B \cap C) \)
\(n[(A \times B) \cap(A \times C)] =n(A) \times n(B \cap C) \)
\(\Rightarrow 8 =n(A) \times 2 \Rightarrow n(A)=4 \)

\(n(A \cap B)=1\)

\(\mathrm{f}: \mathbb{R} \rightarrow \mathbb{R} \text { is defined by }\)
\(\mathrm{f}(x)=1-|x|\)
\(\text { The range is }(-\infty, 1] \text { as } f(-\infty)=-\infty\)
\(f(0)=1\)
\(f(\infty) =-\infty\)