To represent \(-\frac { 2 }{ 11 } ,-\frac { 5 }{ 11 } and-\frac { 9 }{ 11 } \)on the number line we make 11 markings each being equal distance \(\frac { 1 }{ 11 } \) on the left of 0.
The point A represents \(\left( -\frac { 2 }{ 11 } \right) \) , the point B represents\(\left( -\frac { 2 }{ 11 } \right) \) and the point C represents \(\left( -\frac { 9 }{ 11 } \right) \)

Let the point A  (-1, -1), (1, 1) and (\(-\sqrt{13},\sqrt{13}\))
Distance = \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)
AB =\(\sqrt{(1+1)^2+(1+1)^2}\)
\(\sqrt{2^2+2^2}=\sqrt{4+4}=\sqrt{8}\)
BC =\(\sqrt{(-\sqrt{3}+1)^2+(\sqrt{3}-1)^2}\)
\(\sqrt{(\sqrt{3}+1)^2+(\sqrt{3}-1)^2}\)
\(\sqrt{3+1+2\sqrt{3}+3+1-2\sqrt{3}}\)
\(\sqrt{3+1+3+1}=\sqrt{8}\)
AC =\(\sqrt{(-\sqrt{3}+1)^2+(\sqrt{3}+1)^2}\)
\(=\sqrt{3+1-2\sqrt{3}+3+1+2\sqrt{3}}\)
\(=\sqrt{4+4}=\sqrt{8}\)

AB= BC=AC= \(\sqrt{8}\)
\(\therefore\)ABC is an equilateral triangle.

Let A(1, 1), B(5, 4) and C(-2, 5)
Distance =\(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)
AB =\(\sqrt{(5-1)^2+(4-1)^2}\)
\(=\sqrt{4^2+3^2}=\sqrt{16+9}=\sqrt{25}=5\)
BC =\(\sqrt{(-2-5)^2+(5-4)^2}\)
\(=\sqrt{(-7)^2+1^2}=\sqrt{49+1}=\sqrt{50}\)
AC =\(\sqrt{(-2-1)^2+(5-1)^2}\)
\(=\sqrt{(-3)^2+4^2}=\sqrt{9+16}=\sqrt{25}=5\)

AB = 5, AC = 5,
\(\therefore\) ABC is an isosceles triangle .................(1)
BC² = AB² + AC²
50 = 25 + 25\(\Rightarrow\) 50 = 50
\(\therefore\) \(\angle\)A = 90° ...................(2)
From (1) and (2) we get ABC is an isosceles right angle triangle.

Let A be the required number to be added.
(x⁴ - 3x² + 2x + 6 ) + A = x⁴ - 2x³ - x + 8
\(\therefore\) A = x⁴- 2x³ - x + 8 - (x⁴ - 3x² + 2x + 6)
= x⁴ - 2x³ - x + 8 - x⁴ + 3x² - 2x - 6
= -2x³ + 3x² - 3x + 2
Hence -2x³ + 3x² - 3x + 2 must be added.

Let the angles of the quadrilateral be 3x, 5x, 9x and 13x.
Sum of all the angles of quadrilateral = 360°.
3x + 5x + 9x + 13x = 360°
30x = 360°
x = \(\frac { { 360 }^{ 0 } }{ 30 } \)
=12⁰
3x =3 \(\times\) 12 = 36°
5x = 5 \(\times\) 12 = 60°
9x = 9 \(\times\) 12 = 108°
13x = 13 \(\times\) 12 = 156°
The required angles of quadrilateral are 36°, 60° 108° and 156°.

Let M be the set of students opting for Mathematics.
Let S be the set of students opting for Science.
n (M U S) = 40, n (M) = 26, n (S) = 24
n (M U S) = n (M) + n (S) - n (M ∩ S)
40 = 26 + 24 - n (M ∩ S)
n (M ∩ S) = 26 + 24 - 40
= 50 - 40 = 10
∴ Number of students opted for Mathematics and Science = 10

(i) A = {1,3,5,7} ; B = {1,2,4,6,8}
A-B = {1,3,5,7} - {1,2,4,6,8}
= {3,5,7}
B -A = {1,2,4,6, 8} - {1,3,5,7}
= {2,4,6,8}
AΔB = (A - B)U(B-A)
= {3,5,7} U {2,4,6,8}
= {2,3,4,5,6,7,8}.
(ii) BΔA = (B-A) U (A-B)
= {2,4,6,8} u {3,5,7}
= {2,3,4,5,6,7,8}

x² - 3x + 2.
The value is 0

(3x + 2) units

\({13\over 150} = {13\over 2\times 3\times 5^2}\)
Since it is not in the form of \({P\over 2^m\times 5^n}\)
\(\therefore {13\over 150}\) has non-terminating and recurring decimal expansion.

\({7\over 16}={7\over 2^4}={7\over 2^4\times 5^6}\)
\(\therefore {7\over 16}\) has a terminating decimal expansion.

(i) {-3,-2,-1,0,1,4} (ii) {-3,-2,-1,0,1,4} (iii) {-3,-2,0,1,3,4}

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