43.5, 13.6, 18.9,38.4,61.4,29.8
Largest value L= 61.4
Smallest value S = 13.6
R = L - S
= 61.4 - 13.6 = 4
Co-efficient of range = \(\frac { L-S }{ L+S } \)
= \(\frac { 47.8 }{ 75 } =0.64\)
Range = 47.8; co-efficient of range = 0.64.

f(x) = (x³ - 1) (x + 1)
= (x - 1) (x² + x + 1) (x + 1)
g(x) = x³ + 1 = (x + 1) (x² - x + 1)
GCD = x + 1
LCM = (x + 1) (x - 1) (x² + x + 1) (x² - x + 1)
f(x) x g(x) = (x³ - 1) (x + 1) (x³ + 1)
= ( x + 1) ((x³)² - (1)² )
= ( x + 1) (x⁶ - 1)
LCM x GCD = (x + 1) (x - 1)  (x² + x + 1) (x² - x + 1) (x + 1)
= (x + 1) (x² - x + 1) (x - 1) (x² + x + 1) (x + 1)
= (x³ + 1) (x - 1)(x + 1)
= (x⁶ - 1)(x + 1)
f(x) x g(x) = LCM x GCD
Hence verified.

Given, θ = 45⁰, y intercept, c = 11
Slope m = tan θ  = tan 45⁰ = 1
Therefore, equation of a straight line is of the form y = mx + c
Hence we get, y = x + 11 gives x − y + 11 = 0

Total number of cards = 52
Number of king cards = 4
Probability of drawing a king card = \(\frac{4}{52}\)
Number of queen cards = 4
Probability of drawing a queen card = \(\frac{4}{52}\)
Both the events of drawing a king and a queen are mutually exclusive
⇒ P(AUB) = P(A) + P(B)
Therefore, probability of drawing either a king or a queen = \(\frac { 4 }{ 52 } +\frac { 4 }{ 52 } =\frac { 2 }{ 13 } \).

When two coins are tossed together, the sample space is
S = {HH, HT, TH, TT} n(S) = 4
Let A be the event of getting different faces on the coins.
A = {HT, TH}; n(A) = 2
Probability of getting different faces on the coins is P(A) = \(\frac { n(A) }{ n(S) } =\frac { 2 }{ 4 } =\frac { 1 }{ 2 } \).

Standard deviation \(\sigma=6.5\)
Mean \(\bar{x}=12.5\)
Coefficient of variation C.V  \(=\frac{\sigma}{x} \times 100 \% \)
\(=\frac{6.5}{12.5} \times 100 \% \)
\(=\frac{65}{125} \times 100 \% \)
\(=\frac{13}{25} \times 100 \% \)
= 52 %
Co-efficient of variation is 52%

1 + 2 + 3 +...+ 60
1 + 2 + 3 +...+ n \(=\frac{n(n+1)}{2}\)
1 + 2 + 3 +...+ 60 \(=\frac{60 \times(60+1)}{2}=\frac{60 \times 61}{2}\)
1 + 2 + 3 +...+ 60 = 1830

Since the tangents at a point to a circle is
perpendicular to the radius through the point of contact
\(\therefore \angle O P O^{\prime}=90^{\circ}\)
OP² + OP² = (OO')²
[By Pythagoras theorem]
3² + 4² = (OO')²
9+16 = (OO')²
25 = (OO')²
OO' = 5cm
Since the line joining the centres of two intersecting circles is perpendicular bisector of their common chord.
\(\mathrm{OR} \perp \mathrm{PQ} \text { and } O^{\prime} \mathrm{R} \perp \mathrm{PQ}\)
AIso PR = QR
Let OR = x, then O'R = 5 - x
AIso at PR = QR = y cm
\(\text { In } \triangle O R P \text { and } \Delta O^{\prime} R P\)
Applying Pythagoras theorem
OP² = OR³ + RP² and O'P'² = O'R² + RP²
\(3^{2}=x^{2}+y^{2} and 4^{2}=(5-x)^{2}+y_{i}^{2} \)
\(Subtracting \Rightarrow 4^{2}-3^{2}=\left\{(5-x)^{2}+y^{2}\right\}-\left(x^{2}+y^{2}\right) \)
\(16-9=25-10 x+x^{2}+y^{2}-x^{2}-y^{2} \)
7 - 25 = 10x
10x = 25 - 7
10x = 18
x = 1.8 cm
3² = x² + y²
\(y=\sqrt{9-(1.8)^{2}}=\sqrt{5.76}\)
y = 2.4cm
Hence PR = QR = 2.4 cm
PQ = 2y = 4.8 cm

Given the n^th term of the A.P. is t_n = - 3 + 2n
Put n = 1 
t₁ = -3 + 2(1) = - 3 + 2
a = t₁ = -1
Put n = 2 
t₂ = -3 + 2(2) = - 3 + 4
t₂ = 1
Common difference d = t₂ - t₁ = 1 - (-1)
= 1 + 1 = 2
First term a = - 1; Common difference d = 2

Let r be the radius of the hemisphere.
Given that, base area = \(\pi\)r² = 1386 sq. m
T.S.A. = 3 \(\pi\)r² sq.m
= 3 x 1386 = 4158
Therefore, T.S.A. of the hemispherical solid is 4158 m².

In \(\triangle\)ABC, AD is the bisector of \(\angle\)A
Therefore by Angle Bisector of \(\angle\)A
\(\frac { AB }{ AC } =\frac { BD }{ DC } \)
\(\frac{4}{3}=\frac{6}{A C}\) gives 4AC = 18. Hence, AC \(=\frac{9}{2}=4.5 \mathrm{~cm}\)

Given A = {-2, -1, 0, 1, 2} and  f(x) = x² + x + 1
f(-2) = (-2)² + (-2) + 1 = 3;
f(-1) = (-1)² + (-1) + 1 = 1
f(0) = 0² + 0 + 1 = 1; 
f(1) = 1 ² + 1 + 1 = 3
f(2) = 2² + 2 + 1 = 7
Therefore, B = {1,3,7}

8, 24, 72, .......

Each term is obtained by multiplying the previous term by 3.
Next three terms are 216, 648, 1944.

Let the distance be 'd'
Speed = 500 km/hr
Time = 't' hours
Distance = Time x Speed
d(t) = 500 t

f(x) = 2x + 5, x ≠ 0.
\(\frac{f(x+2)-f(2)}{x} =\frac{[2(x+2)+5]-[2(2)+5]}{x} \) 
\(=\frac{2 x+4+5-9}{x}=\frac{2 x+9-9}{x} \)
\(=\frac{2 x}{x}=2\)

f = {(x, y) / x, y \(\in \) N and y = 2x}
Given that y = 2x
x = {1,2,3,..}

f = {(1,2), (2, 4), (3, 6), (4, 8)..}
Domain of f = {1, 2, 3, 4...........}
Co domain = {1,2,3, 4.......}
Range of f = {2,4,6, 8......}
Here, the first elements (x) are having unique images. So, this relation is a function.

f(x) = x² - 2 where x \(\in \){ -2, -1, 0, 3}
(i) f( -2) = ( -2)² - 2 = 2; f( -1) = ( -1)² - 2 = -1
f(0) = (0)² - 2 = - 2 ; f(3) = (3)² - 2 = 7
Therefore, f = {(-2, 2), (-1, -1), (0, -2), (3, 7)}
(ii) We note that each element in the domain of f has a unique image. Therefore f is a function.

15 \(\equiv \) 3 (mod d) means 15 - 3 = kd, for some integer k,
12 = kd
gives d divides 12.
The divisors of 12 are 1,2,3,4,6,12. But d should be larger than 3 and so the possible values for d are 4, 6, 12.

for some natural number n,
4ⁿ = (2)ⁿ 
So 2 is a factor of 4ⁿ
By fundamental theorem of arithmetic, we know the factorization of 4n is unique.
Only factor of 4ⁿ is 2, even number of times.
But 4ⁿ always end with 4 or 6.
If n is odd then 4ⁿ end with 4.
If n is even, then 4ⁿ end with the digit 6.

\(\Delta AOQ\) and \(\Delta BOP,\angle OAQ=\angle OBP=90^{ 0 } \)
\(\angle AOQ=\angle BOP\) (Vertically opposite angles)
Therefore, by AA Criterion of similarity,
\(\Delta AOQ\sim \Delta BOP\)
\(\frac { AO }{ BO } =\frac { OQ }{ OP } =\frac { AQ }{ BP } \)
\(\frac { 10 }{ 6 } =\frac { AQ }{ 9 } \) gives \(AQ=\frac { 10\times 9 }{ 6 } =15cm\)

Given Set =  {(x, y) / y = x + 3, x \(\in \) {0, 1, 2, 3, 4, 5}}
When x = 0, y = 0 + 3 = 3
When x = 1, y = 1 + 3 = 4
When x = 2,y = 2 + 3 = 5
When x = 3, y = 3 + 3 = 6
When x = 4, y = 4 + 3 = 7
When x = 5, y = 5 + 3 = 8
Relation R = {(0, 3), (1,4), (2,5), (3,6), (4,7), (5,8)}
Domain of R = {0, 1, 2, 3, 4, 5}
Range of R = {3, 4, 5, 6, 7, 8}

A x B = {(3,1), (3,7), (3,10), (4,1), (4,7), (4,10), (7,1), (7,7), (7,10), (8,1), (8,7), (8,10)}
We note that, R₁ ⊆  A x B. Thus, R₁ is a relation from A to B.

A = {1,2,3} B = {2,3,5,7} 
A x B = {(1,2),(1,3),(1,5),(1,7),(2,2),(2,3),(2,5),(2,7),(3,2),(3,3),(3,5),(3,7)}
B x A = {(2,1),(2,2),(2,3),(3,1),(3,2),(3,3),(5,1),(5,2),(5,3),(7,1),(7,2),(7,3)}

A x B = {(3,2), (3,4), (5,2), (5,4)}
We have A = {set of all first coordinates of elements of A x B}. Therefore, A = {3,5}
B = {set of all second coordinates of elements of A x B}. Therefore, B = {2,4}
Thus A = {3,5} and B = {2,4}.

Plot the points in a rough diagram and take them in counter-clockwise order.
Let the vertices be


The area of Δ ABC is 
= \(\frac{1}{2}\) { (x₁y₂ + x₂y₃ + x₃y₁) - (x₂y₁ + x₃y₂ + x₁y₃) }
= \(\frac{1}{2}\) { (6 + 30 + 25) - (25 - 10 18) }
= \(\frac{1}{2}\) { 61 + 3 }
= \(\frac{1}{2}\) (64) = 32 sq. units