\(\sqrt{5+2\sqrt{6}}=\sqrt{3+2+2\sqrt{6}}\)
=\(\sqrt{(\sqrt{3}+\sqrt{2})^{2}}\)
=\(\sqrt{3}+\sqrt{2}\)
\(\sqrt{8-2\sqrt{15}}=\sqrt{5+3-2\sqrt{15}}\)
=\(\sqrt{(\sqrt{5}-\sqrt{3})^{2}}=\sqrt{5}-\sqrt{3}\)
\(\sqrt{5+2\sqrt{6}}+\sqrt{8-2\sqrt{15}}\)=\(\sqrt{3}+\sqrt{2}+\sqrt{5}-\sqrt{3}\)
=\(\sqrt{2}+\sqrt{5}\)

(0,0); (-p,0); (5,0); (0,-q)

We know that \(3^3+y^3=(x+y)(x^3-xy+y^2)\)
\(\Rightarrow\ x^3+y^3=(x+3)^3-3xy(x+y)\)
\(\Rightarrow\ x^3+y^3=(x+y){(x+y)^2-3xy}\)
\(\Rightarrow x^3+y^3=(x+y)(x^2+2xy+y^2-3xy)\) Using Identity
\(\Rightarrow\ x^3+y^3=(x+y)(x^2-xy+y^2)\)

\(8a^3-b^3-12a^2b+6ab^2\)
\(=(2a)^3-(b)^3-3(2a)(b)(2a-b)\)
\(=(2a-b)^3\) |  Using Identity VII
\(=(2a-b)(2a-b)(2a-b)\)

Let x = 0.99999...
Multiplying both sides by 10 we get,
10x = 0.99999...
10x = 9 + 0.99999...
10x = 9 + x
10x - x = 9
9x = 9
x = 9/9 = 1
Thus, 0.99999...= 1  =1/1
Here p = 1
q = 1
Since 0.99999...goes on for ever, so there is no gap between 1 and 0.99999...and hence they are equal.

(2x-y+z)²=4x²+y²+z²-4xy+2yz+4zx

2√50 x 3√32 x 4√18
= 10√2 x 12√2 x 12√2
= 2880√2

2x-1 = 0
\(\Rightarrow \ x=\frac { 1 }{ 2 } \)
By remainder theorem, if f(x) is divided by 2x-1, the remainder is \(f\left( \frac { 1 }{ 2 } \right) \)
\(\therefore \quad f\left( \frac { 1 }{ 2 } \right) =4{ \left( \frac { 1 }{ 2 } \right) }^{ 3 }-12{ \left( \frac { 1 }{ 2 } \right) }^{ 2 }+14\left( \frac { 1 }{ 2 } \right) -3\)
\(=4\times \frac { 1 }{ 8 } -12\times \frac { 1 }{ 4 } +14\times \frac { 1 }{ 2 } -3\)
\(=\frac { 1 }{ 2 } -3+7-3\)
\(=\frac { 1 }{ 2 } +1\)
\(=\frac { 3 }{ 2 } \)
Hence required remainder is \(\frac { 3 }{ 2 } \).

(2,3), (2,0), (0,3)

III

\(\quad a=-\frac { 179 }{ 171 } ,b=\frac { -20 }{ 171 } \)

The figure obtained is a rectangle
Area of the rectangle = 4 x 6
=24 square units.

The figure obtained is a right angled triangle.

\({ x }^{ 2 }-4={ x }^{ 2 }-{ 2 }^{ 2 }=(x+2)(x-2)\)
Let  \(p(x)=a{ x }^{ 4 }+2{ x }^{ 3 }-3{ x }^{ 2 }+bx-4\)
If  \(x^2-4\) is  a factor of p(x), then clearly each of x+2 and x-2 will be a factor of p(x).
If x+2 is a factor of p(x), then by factor theorem
\(p(-2)=0\ |x+2=0\Rightarrow x=-2\)
\(\Rightarrow a{ (-2) }^{ 4 }+2{ (-2) }^{ 3 }-3{ (-2) }^{ 2 }+b(-2)-4=0\)
\(\Rightarrow 16a-16-12-2b-4=0\)
\(\Rightarrow 16a-2b-32=0\)
\(\Rightarrow -b-16=0\ \ .......(1)\)
| Dividing throughout by 2
If x-2 is a factor of p(x), then by factor theorem
\(p(2)=0\ \ \ |x-2=0\Rightarrow x=2\)
\(\Rightarrow a{ (2) }^{ 4 }+2{ (2) }^{ 3 }-3{ (2) }^{ 2 }+b(-2)-4=0\)
\(\Rightarrow \ 16a+16-12+2b-4=0\)
\(\Rightarrow -2b=0\)
\(\Rightarrow 8a-b=0 \ \ .........(2)\)
Dividing throughout by 2
Solving (1) and (2), we get
\(a=1, b=-8\)
\(\therefore p(x)=a{ x }^{ 4 }+2{ x }^{ 3 }-3{ x }^{ 2 }+bx-4\)
\(={ x }^{ 2 }({ x }^{ 2 }-4)+1({ x }^{ 2 }-4)+2x({ x }^{ 2 }-4)\)
\(=({ x }^{ 2 }-4)({ x }^{ 2 }+2x+1)\)
\((x-2)(x+2)(x+1)(x+1)\)

Let \(f(x)={ x }^{ 3 }+2{ x }^{ 2 }-5ax-7\)
and  \(g(x)={ x }^{ 3 }-a{ x }^{ 2 }-12x+6\)
By remainder theorem,
\(f(-1)={ R }_{ 1 }\ \ \ \ ........(1)\)
\(x+1=0\Rightarrow x=-1\)
\(\Rightarrow { (-1) }^{ 3 }+2{ (-1) }^{ 2 }-5a(-1)-7={ R }_{ 1 }\) 
\(\Rightarrow \ -1+2+5a-7={ R }_{ 1 }\) 
\(\Rightarrow { R }_{ 1 }=a-6 \ \ ................(2)\)
and     \(g(2)={ R }_{ 2 }\ \ ...........(3)\)
\(x-2=0\Rightarrow x=2\)
\(\Rightarrow { (2) }^{ 3 }+a{ (2) }^{ 2 }-12(2)+6={ R }_{ 2 }\) 
\(\Rightarrow 8+4a-24+6={ R }_{ 2 }\) 
\(\Rightarrow { R }_{ 2 }=4a-10\ \ \ .....(4)\) 
According to the question,
\(2{ R }_{ 1 }+{ R }_{ 2 }=6\) 
\(\Rightarrow 2(5a-6)+(4a-10)=6\) 
\(\Rightarrow -12+4a-10=6\) 
\(\Rightarrow 14a=28\) 
\(\Rightarrow a=2\)

Mark the distance 4.5 units from a fixed point A on a given line to obtain a point B such that AB=4.5 units.From B, mark a distance of 1 unit and mark the new point as C.Find the mid-point of AC and mark that point as O.Draw a semi-circle with centre O and radius OC.Draw a semi-circle with Centre O and radius OC.Draw a line perpendicular to AC passing through B and intersecting the semi-circle at D.Then BD=\(\sqrt { 4.5 } \). By measurement,
BD = 2.2 units
\(\sqrt { 4.5 } \)= 2.2
Now, let us treat the line BC as the number line, with B as zero, C as 1 and so on.Draw an arc with centre B and radius BD, which intersects the number line in E.

\(\frac { 2 }{ \sqrt { 3 } +\sqrt { 5 } } +\frac { 5 }{ \sqrt { 3 } -\sqrt { 5 } } =a\sqrt { 3 } +b\sqrt { 5 } \)
\(\\ =\frac { 2\left( \sqrt { 3 } -\sqrt { 3 } \right) }{ \left( \sqrt { 3 } -\sqrt { 5 } \right) } +\frac { 5 }{ \sqrt { 3 } -\sqrt { 5 } } \frac { \sqrt { 3 } +\sqrt { 5 } }{ \sqrt { 3 } +\sqrt { 5 } } \)
\(\\ =a\sqrt { 3 } +b\sqrt { 5 } \)
\(\\ \frac { 2\left( \sqrt { 3 } -\sqrt { 5 } \right) }{ 3-5 } +\frac { 5\left( \sqrt { 3 } +\sqrt { 5 } \right) }{ 3-5 } =a\sqrt { 3 } +b\sqrt { 5 } \)
\(\\ -\left( \sqrt { 3 } -\sqrt { 5 } \right) -\frac { 5 }{ 2 } \left( \sqrt { 3 } +\sqrt { 5 } \right) =a\sqrt { 3 } +b\sqrt { 5 } \)
\(\\ -\frac { 7 }{ 2 } \sqrt { 3 } -\frac { 3 }{ 2 } \sqrt { 5 } =a\sqrt { 3 } +b\sqrt { 5 } \)
\(\\ a=-\frac { 7 }{ 2 } ,\ b=-\frac { 3 }{ 2 } \)