\(\lim _{x \rightarrow \infty} \frac{\sqrt{x^{2}-1}}{2 x+1}=\lim _{x \rightarrow \infty} \frac{\sqrt{1-\frac{1}{x^{2}}}}{2+\frac{1}{x}}=\frac{\sqrt{1-0}}{2+0}=\frac{1}{2}\)

\(\left|\begin{array}{lll} k a & k b & k c \\ k x & k y & k z \\ k p & k q & k r \end{array}\right|=k^{3}\left|\begin{array}{lll} a & b & c \\ x & y & z \\ p & q & r \end{array}\right|=k^{3} \Delta\)
\(\text { Take } k \text { from } R_{1}, R_{2}, R_{3}\)

\(\mathrm{AM} \geq \mathrm{GM} \)
\(\Rightarrow a \geq g \)

\(\tan \theta=\frac{2 \sqrt{b^{2}-a b}}{a+b}=\frac{2 \sqrt{\left(\frac{-1}{2}\right)^{2}+6}}{-5}\)
\(=\left|\frac{2 \sqrt{\frac{1}{4}+6}}{-5}\right|\)
\(=\frac{2\left(\frac{5}{2}\right)}{5}=1 \Rightarrow \theta=45^{\circ}\)
\(\frac{2 \cos \theta+3 \sin \theta}{4 \sin \theta+5 \cos \theta}=\frac{\frac{2}{\sqrt{2}}+\frac{3}{\sqrt{2}}}{\frac{4}{\sqrt{2}}+\frac{5}{\sqrt{2}}}=\frac{5}{9}\)

Let y = 2x + c be the required line.
Given perpendicular distance from origin'to this line is \(\sqrt5\)
\(\frac{c}{\sqrt{1+4}}=\sqrt{5}\)
c  = 5
This required line is y - 2x + 5
2x - y + 5 = 0

\(\frac{1}{\sqrt{1}+\sqrt{3}} =\frac{1}{\sqrt{3}+\sqrt{1}} \times \frac{\sqrt{3}-1}{\sqrt{3}-1}=\frac{\sqrt{3}-1}{2} \)
\(\frac{1}{\sqrt{3}+\sqrt{5}} =\frac{1}{\sqrt{5}+\sqrt{3}} \times \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}} \)
\(=\frac{\sqrt{5}-\sqrt{3}}{2} \)
\(\text { Sum to } \mathrm{n} \text { terms }=\frac{(\sqrt{3}-1)}{2}+\frac{(\sqrt{5}-\sqrt{3})}{2}+\ldots . .\left(\frac{\sqrt{2 n+1}-\sqrt{2 n-1}}{2}\right)\)
\(=\frac{\sqrt{2 n+1}-1}{2}\)

Product of r consecutive positive integers is divisible by r! (by theorem).

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

\(\text { i) Let } y=x \text { then } x=y \Rightarrow f^{-1}(x)=x\)
\(\text { ii) Let } y=x^{2} \text { then } y=\sqrt{x} \Rightarrow f^{-1}(x)=\sqrt{x}\)
\(\text { iii) Let } y=8 \sqrt{x} \text { then } \frac{y^{2}}{64}=x \Rightarrow \mathrm{f}^{-1}(x)=\frac{x^{2}}{64}\)
\(\therefore \mathrm{f}^{-1}(x)= \begin{cases}x & x<1 \\ \sqrt{x} & 1 \leq x \leq 16 \\ \frac{x^{2}}{64} & x>16\end{cases}\)