\(a\sqrt { 5 } -\frac { b }{ 11 } =\frac { \left( 3-\sqrt { 5 } \right) }{ \left( 3+2\sqrt { 5 } \right) } \times \frac { \left( 3-2\sqrt { 5 } \right) }{ \left( 3-2\sqrt { 5 } \right) } \)
\(=\frac { 9-6\sqrt { 5 } -3\sqrt { 5 } +2\times 5 }{ { \left( 3 \right) }^{ 2 }-{ \left( 2\sqrt { 5 } \right) }^{ 2 } } \)
\(=\frac { 9-9\sqrt { 5 } +10 }{ 9-20 } \)
\(=\frac { 19-9\sqrt { 5 } }{ -11 } \)
\(=\frac { 19 }{ -11 } -\frac { 9\sqrt { 5 } }{ -11 } =\frac { 9\sqrt { 5 } }{ 11 } -\frac { 19 }{ 11 } \)
\(a\sqrt { 5 } -\frac { b }{ 11 } =\frac { -19 }{ 11 } +\frac { 9\sqrt { 5 } }{ 11 } \)
\(\Rightarrow a\sqrt { 5 } -\frac { b }{ 11 } =\frac { 9 }{ 11 } \sqrt { 5 } -\frac { 19 }{ 11 } \)
On comparing both sides, we get
a=\(\frac{9}{11}\) , b = 19

Given equation is 3x+4y=6
(i) When it cuts x-axis then
put y=0, i.e 3x=6
x=2
Hence point on x-axis is (2,0)
(ii) when it cuts y-axis then
put x=0 i.e., 4y=6
y=3/2
hence point on y-axis is \((0,{3\over2})\)
3x+4y=6
\(\Rightarrow y=\frac{6-3x}{4}\)

Verification:
Divisor \(\times\) Quotient = (x²-3x+2)(3x+1)+0
= 3x³-8x²+3x+2
\(\therefore\) Dividend = Divisor \(\times\) Quotient + Remainder (Division algorithm)

\(\because\)  C is the midpoint of AB
\(\therefore \)  AC = CB
AC + AC = CB + AC
| If equals are added to equals, then   the wholes are equal (Euclid's Axiom (ii))]
\(\Rightarrow \)  2AC =  AB      I CB + AC coincides with AB 
\(\Rightarrow \)  \(1\over2\)(2AC) = \(1\over2\) AB
| Things which are halves of the same thing are equal (Euclid's Axiom (vii»]
\(\Rightarrow \)  AC =\(1\over2\)AB
\(\Rightarrow \) \(1\over2\)AC  = \(1\over2\)(\(1\over2\)AB)
| Things which are halves of the same thing are equal to one another (Euclid's Axiom (vii))]
\(1\over2\)AC = \(1\over2\)AB
AD = \(1\over4\)AB
\(\because\) D is the mid-point of AC
\(\therefore \)AD = DC =\(1\over2\)AC (as above)

x = 3 represents a line parallel to y-axis at a distance of 3 units to the right of the origin.
4x= 3y
\(\Rightarrow\ \ \ y={4x\over 3}\)
Table of solution

We plot the points (0,0) and (3, 4) on a graph paper and join the same by a ruler to get the line which is the graph of the equation 4x = 3y.

Area of the triangle GAB formed by the given two lines and the x-axis \(={3\times 4\over2}=6\) square units

\(A\rightarrow \left( -3,0 \right)\)
\(\\ B\rightarrow \left( 3,0 \right)\)
\(\\ C\rightarrow \left( 3,3 \right)\)
\(\\ D\rightarrow \left( -3,3 \right) \)

We know that
\((x+\frac{1}{x})^2=x^2+\frac{1}{x^2}+2(x)(\frac{1}{x})\) Using Identity I
\(\Rightarrow\ (x+\frac{1}{x})^2=x^2+\frac{1}{x^2}+2\)
\(\Rightarrow (x+\frac{1}{x})^2=23+2\)
\(\Rightarrow\ = (x+\frac{1}{x})^2=25\)
\(\Rightarrow \ x+\frac{1}{x}=5\)       .......(5)
Again, we know that 
\((x+\frac{1}{x})^3=x^3+\frac{1}{x^3}+3(x)(\frac{1}{x})(x+\frac{1}{x})\) Using Identity VI
\(\Rightarrow (x+\frac{1}{x})^3=x^3+\frac{1}{x^3}+3(x+\frac{1}{x})\)
\(\Rightarrow (5)^3=x^3+\frac{1}{x^3}+3(5)\)
\(\Rightarrow 125=x^3+\frac{1}{x^3}+15\)
\(\Rightarrow x^3+\frac{1}{x^3}=110\)

\(\frac{87^3+13^3}{87^2-87\times13+13^3}\)
\(=\frac{(87+13(87^2-87\times13+12^2)}{(87^2-87\times13+13^2)}\)
= 87+13=100

Let \(p(x)=x^{ 4 }+ax^{ 3 }-3x^{ 2 }+2x+b\)
If (x+1) and (x-1) are factors of p(x), then by factor theorem,
\(p(-1)=0 \ \ .......(1)\ x+1=0\ \ \ \Rightarrow \ x=-1\)
and   \(p(1)=0\ \ \ .......(2)|\ -1=0\Rightarrow x=1\)
Now, \(p(-1)=0\)
\(\Rightarrow (-1)^{ 4 }+a(-1)^{ 3 }-3(-1)^{ 2 }+2(-1)+b=0\)
\(\Rightarrow 1-a-3-2+b=0\)
\(\Rightarrow -a+b=4\ \ .......(3)\)
and   \(p(1)=0\)
\(\Rightarrow (1)^{ 4 }+a(1)^{ 3 }-3(1)^{ 2 }+2(1)+b=0\)
\(\Rightarrow 1+a-3+2+b=0\)
\(\Rightarrow a+b=0\ .......(4)\)
Solving (3) from (2), we get
a = -2,b = 2

1, -2, 3

\({ 5 }^{ p-3 }\times { 3 }^{ 2p-8 }=225\)
\(\\ { 5 }^{ p-3 }\times { 3 }^{ 2p-8 }={ 5 }^{ 2 }\times { 3 }^{ 2 }\)
\(\\ p-3=2\)
\(\\ 2p-8=2\)
\(\\ p=5\)

(i) Let x =2.4 \(\overline { 178 } \) 
Then x = 2.4178178178...
⇒  10x = 24.178178178.....       . ...(1)
⇒  10000x = 24178.178178178  ....(2)
Subracting (1) from (2), we get
99990x = 24154
⇒ x = \(24154\over9990\)
⇒ x = \(12077\over4995\)
(ii) p = 12077
(iii) q = 4995
(iv) p is a natural number or a whole number or a positive integer.
(v) q is a natural number or a whole number or a positive integer.
(vi) Let us find out the HCF of p and q
p = 12077 = 13 x 929
q = 4995 = 3 x 3 x 3 x 5 x 37
\(\therefore\)  HCF (p,q) = 1
\(\therefore\) p and q are co-prime.
\(\therefore\)  Apala is correct.
So, the value 'deductive' is depicted by her declaration.
(vii) The mathematical concept 'Number system' has been converted in this problem.
(viii) The formulae used in the solution are as follows:
1. Method of expressing a recurring and non-terminating decimal in the form of a rational number.
2. Concept of natural numbers or whole numbers or positive integer.
3. reduction of a rational in its lowest terms.
4. Concept of co-prime numbers.

Let x = \(0.15\overline { 9 } \)
x = 0.159999...
100x = 15.9999...      ...(1)
1000x = 159.9999...    ...(2)
Subtracting (1) from (2), we get
900x = 144
\(x=\frac { 144 }{ 900 } \)
\(\\ x=\frac { 4 }{ 25 } \)
Here, p = 4, q = 25(\(\neq 0\))

\(\frac { 1 }{ 5 } =\frac { 1\times 6 }{ 5\times 6 } =\frac { 6 }{ 30 } =\frac { 6\times 10 }{ 30\times 10 } =\frac { 60 }{ 300 } \) 
\(\\ \frac { 1 }{ 6 } =\frac { 1\times 5 }{ 6\times 5 } =\frac { 5 }{ 30 } =\frac { 5\times 30 }{ 30\times 10 } =\frac { 50 }{ 300 }\)
\( \\ \because 50<51<52<53<54<60\)
\(\\ \therefore \frac { 50 }{ 300 } <\frac { 51 }{ 300 } <\frac { 52 }{ 300 } <\frac { 53 }{ 300 } <\frac { 54 }{ 300 } <\frac { 60 }{ 300 } \)
Hence, four rational numbers between \(\frac { 1 }{ 5 } \) and \(\frac { 1 }{ 6 } \) can be taken as
\(\frac { 51 }{ 300 } ,\frac { 52 }{ 300 } ,\frac { 53 }{ 300 } \)and \(\frac { 54 }{ 300 } \)
or \(\frac { 17 }{ 100 } ,\frac { 13 }{ 75 } ,\frac { 53 }{ 300 } \)and \(\frac { 9 }{ 50 } \)

\(\frac { 3 }{ 5 } =\frac { 3\times 8 }{ 5\times 8 } =\frac { 24 }{ 40 } \)
\(\\ \frac { 7 }{ 8 } =\frac { 7\times 5 }{ 8\times 5 } =\frac { 35 }{ 40 } \)
\(\\ \because 24<25<26<27<35\)
\(\\ \therefore \frac { 24 }{ 40 } <\frac { 25 }{ 40 } <\frac { 26 }{ 40 } <\frac { 27 }{ 40 } <\frac { 35 }{ 40 } \)
Hence, three rational numbers between  \(\frac { 3 }{ 5 } \) and \(\frac { 7 }{ 8 } \) can be taken as
\(\frac { 25 }{ 40 } ,\frac { 26 }{ 40 } \) and \(\frac { 27 }{ 40 } \)
\(\frac { 5 }{ 8 } ,\frac { 13 }{ 20 } \) and \(\frac { 27 }{ 40 } \)