A = { 1, 2, 3, 7}, B = { 3, 0, -1, 7}
A x B = {(1, 3), (1, 0), (1, - 1), (1, 7),(2,3), (2, 0), (2, -1), (2, 7), (3,3), (3, 0), (3, - 1), (3, 7), (7, 3)., (7, 0), (7, -1), (7 ,7)}
R₃ = {( 2, -1), (7, 7) , (1, 3)}
It is clear that  \(R_{3} \subseteq A \times B\)
R₃ is a relation from A to B.

It is a function. Since a Vertical line intersects the curve at only one point.

\( \frac{\cos \theta}{1+\sin \theta} =\sec \theta-\tan \theta \)
\(\text { LHS } =\frac{\cos \theta}{1+\sin \theta} \)
\(=\frac{\cos \theta}{1+\sin \theta} \times \frac{1-\sin \theta}{1-\sin \theta} \)
[Multiplying the Numerator and Denominator by 1 - sin\(\theta\)]
\( =\frac{\cos \theta(1-\sin \theta)}{1^{2}-\sin ^{2} \theta} \)
\(\left[\because(a+b)(a-b)=a^{2}-b^{2}\right] \)
\(=\frac{\cos \theta(1-\sin \theta)}{\cos ^{2} \theta} \)
\(\left[\because 1-\sin ^{2} \theta=\cos ^{2} \theta\right] \)
\(=\frac{(1-\sin \theta)}{\cos \theta} \)
\(=\frac{1}{\cos \theta}-\frac{\sin \theta}{\cos \theta} \)
\(=\sec \theta-\tan \theta \)
= RHS

tan⁴\(\theta \) + tan²\(\theta \) = sec⁴\(\theta \) - sec²\(\theta \) 
L.H.S = tan²θ (tan²θ + 1)
= tan²θ.sec²θ 
= sec⁴θ - sec²θ
= R.H.S

 9a²b²x² - 24abcdx + 16c²d² = 0
     a               b              c
∆ = b²-4ac
= (-24abcd)² - 4 x 9a²b² x 16c²d²
= 576a²b²c²d² - 576a²b²c²d²
= 0
∴ The roots are real and equal.

x²-x-1 = 0, Here a = 1, b = -1, c = -1
∆ = b²-4ac = (-1)²-4 x 1 x -1
= 1 + 4
= 5 > 0
∴ The roots are real and unequal 

\(\sqrt { 2 } { f }^{ 2 }-6f+3\sqrt { 2 } \) = 0
       a          b        c
\(x=\frac { -b\sqrt { { b }^{ 2 }-4ac } }{ 2a } \)
Here \(f=\frac { -(-6)\pm \sqrt { { (-6) }^{ 2 }-4\times \sqrt { 2 } \times 3\sqrt { 2 } } }{ 2\times \sqrt { 2 } } \)
\(=\frac { 6\pm \sqrt { 36-24 } }{ 2\sqrt { 2 } } \)
\(=\frac { 6\pm \sqrt { 12 } }{ 2\sqrt { 2 } } \Rightarrow \frac { 3+\sqrt { 3 } }{ \sqrt { 2 } } ,\frac { 3-\sqrt { 3 } }{ 2 } \)

x² + 3x = 0
Comparing with ax² + bx + c = 0
a = 1,b = 3, c = 0
Sum of the roots \(-\frac{b}{a}=-\frac{3}{1}=-3\)
Products of the roots \(\frac{c}{a}=\frac{0}{1}=0\)

Given sum of the roots = -(2 - a)²
product of the roots = (a + 5)²
General form of the Quadratic equation is x² - (∝β) x - ∝β = 0
⇒ x² - (-(2 - a)²)x + (a + 5)² = 0  

\(\frac { 1 }{ 2 } \), 1, 2, 4,...
\(\frac { { t }_{ 2 } }{ { t }_{ 1 } } =\frac { 1 }{ \frac { 1 }{ 2 } } =2;\frac { { t }_{ 3 } }{ { t }_{ 2 } } =\frac { 2 }{ 1 } =2;\frac { { t }_{ 4 } }{ { t }_{ 3 } } =\frac { 4 }{ 2 } =2\)
Here the ratios between successive terms are equal. Therefore the sequence \(\frac { 1 }{ 2 } \),1,2,4.... is a Geometric Progression with common ratio r = 2.

Given points (2, 3) and (- 7, - 1)
Equation of the line passing through (x₁ , y₁) and (x₂, y₂) is
\( \frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}} \)
\(\frac{y-3}{-1-3}=\frac{x-2}{-7-2} \)
9 (y - 3) - 4 (x - 2)
9y - 27 = 4 x - 8
4 x - 9y +19 = 0

\(\frac { x+2 }{ x+3 } +\frac { x-1 }{ x-2 } =\frac { (x-2)(x+2)+(x+3)(x-1) }{ (x+3)(x-2) } \)
\(=\frac { { x }^{ 2 }-4+{ x }^{ 2 }+2x-3 }{ (x+3)(x-2) } \)
\(=\frac { 2{ x }^{ 2 }+2x-7 }{ \left( x+3 \right) \left( x-2 \right) } \)

From (i) and (ii) we get
\(\begin{array}{r} A C^{2}+A B^{2}=A D^{2}+B C \cdot D E+\frac{1}{4} B C^{2}+A D^{2}- B C \cdot D E+\frac{B C^{2}}{4} \end{array}\)
\(=2 A D^{2}+2\left(\frac{B C^{2}}{4}\right)\)
\(A C^{2}+A B^{2}=2 A D^{2}+\frac{B C^{2}}{2}\)
\(b^{2}+c^{2}=2 p^{2}+\frac{a^{2}}{2}\)

Given AB = 5.6 cm; AD = 1.4 cm;
AC = 7.2 cm; AE = 1.8 cm
DB = AB - AD
= 5.6 - L.4 = 4.2 cm
EC = AC - AE
= 7.2 - 1.8 = 5.4 cm
Now \(\frac{A D}{D B}=\frac{1.4}{4.2} \)
\(\frac{A D}{D B}=\frac{1}{3} \)
\(\frac{A E}{E C}=\frac{1.8}{5.4} \)
\(\frac{A E}{E C}=\frac{1}{3} \)
From (1) and (2)
\(\frac{A D}{D B}=\frac{A E}{E C}\)
DE || BC

(–10, –4), (–8, –1) and (–3, –5)

Area of triangle \(=\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+\right. \left.x_{3}\left(y_{1}-y_{2}\right)\right] \text { sq. units } \)
\(=\frac{1}{2}[-10(-1+5)-8(-5+4)-3(-4+1)] \)
\(=\frac{1}{2}[-10(4)-8(-1)-3(-3)] \)
\(=\frac{1}{2}[-40+8+9] \)
\(=\frac{1}{2}[-40+17]=-\frac{23}{2}=-11.5 \)
[ Area cannot be negative ]
Area of triangle = 11.5 sq. units.

We have
\(\Delta \)QMO ~ \(\Delta \)RPN 
\(\frac { MQ }{ RP } =\frac { QO }{ RN } \) 
\(\frac{M Q}{Q R}=\frac{Q R}{R N}\)
[ \(\because\)OQRP is a square PR = QR and QO = QR]
QR x QR = MQ x RN
QR² = MQ x RN 

\(\frac { { x }^{ 4 }{ b }^{ 2 } }{ x-1 } \times \frac { { x }^{ 2 }-1 }{ { a }^{ 4 }{ b }^{ 3 } } =\frac { { x }^{ 4 }\times { b }^{ 2 } }{ x-1 } \times \frac { \left( x+1 \right) \left( x-1 \right) }{ { a }^{ 4 }\times { b }^{ 3 } } =\frac { { x }^{ 4 }\left( x+1 \right) }{ { a }^{ 4 }b } \)

a_n =  2n² - 6
n^th term is given by a_n =  2n² - 6
First term a₁ = 2(1)² - 6 = - 4
Second term a₂ = 2(2)² - 6 = 2
Third term a₃ = 2(3)² - 6 = 12
Fourth term a₄ = 2(4)² - 6 = 26
∴ The first four terms are -4, 2, 12, 26, ...

Here, the expression is defined for all real values of 'x'
i.e., x \(\epsilon\) R

\(\frac { 1 }{ 2 } ,\frac { 2 }{ 3 } ,\frac { 3 }{ 4 } \)
a₁ = \(\frac { 1 }{ 2 } \); a₂ = \(\frac { 2 }{ 3 } \); a₃ = \(\frac { 3 }{ 4 } \)
We see that the numerator of n^th term is n, and the denominator is one more than the numerator. Hence, \({ a }_{ n }=\frac { n }{ n+1 } \) n \(\in\) N

78 + x ≡ 3 (mod 5)
78 + x - 3 = 5n for some integer n.
75 + x = 5n for some integer n
75 + x is a multiple of 5.
Least positive value of x = 5.

f(k) - 2k - 1
f o f(k) = 5
f(f(k)) = f(2k - 1) = 5
⇒ 2(2k-1) -1 = 5
4k - 2 -1 = 5 ⇒ 4k = 8
k = 2

\(\frac { { x }^{ 2 }-16 }{ { x }^{ 2 }+8x+16 } =\frac { \left( x+4 \right) \left( x-4 \right) }{ { \left( x+4 \right) }^{ 2 } } =\frac { x-4 }{ x+4 } \)

f(x) = 3 + x, g(x) = x - 4
fog(x) = f(g(x)) =f(x - 4) = 3 + x - 4  = x - 1
gof(x) = g(f(x)) = g(3 + x) = 3 + x - 4 = x - 1
f o g = g o f

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

\(A=\left[\begin{array}{ll} 2 & 5 \\ 4 & 3 \end{array}\right], B=\left[\begin{array}{cc} 1 & -3 \\ 2 & 5 \end{array}\right]\)
\(A B=\left[\begin{array}{ll} 2 & 5 \\ 4 & 3 \end{array}\right]\left[\begin{array}{cc} 1 & -3 \\ 2 & 5 \end{array}\right]\)
\(=\left[\begin{array}{cc} 2+10 & -6+25 \\ 4+6 & -12+15 \end{array}\right]=\left[\begin{array}{cc} 12 & 19 \\ 10 & 3 \end{array}\right]\)
\(\mathrm{BA}=\left[\begin{array}{cc} 1 & -3 \\ 2 & 5 \end{array}\right]\left[\begin{array}{ll} 2 & 5 \\ 4 & 3 \end{array}\right]\)
\(=\left[\begin{array}{cc} 2-12 & 5-9 \\ 4+20 & 10+15 \end{array}\right]=\left[\begin{array}{cc} -10 & -4 \\ 24 & 25 \end{array}\right]\)
\(\therefore A B \neq B A\)

The total marks in both the examinations for all the three groups is the sum of the given matrices.
A + B = \(\left[ \begin{matrix} 22+20 \\ 50+18 \\ 53+81 \end{matrix}\begin{matrix} 15+38 \\ 62+12 \\ 80+47 \end{matrix}\begin{matrix} 14+15 \\ 21+17 \\ 32+52 \end{matrix}\begin{matrix} 23+40 \\ 30+80 \\ 40+18 \end{matrix} \right] =\left[ \begin{matrix} 42 \\ 68 \\ 134 \end{matrix}\begin{matrix} 53 \\ 74 \\ 127 \end{matrix}\begin{matrix} 29 \\ 38 \\ 84 \end{matrix}\begin{matrix} 63 \\ 110 \\ 58 \end{matrix} \right] \)

A + B = \(\left[ \begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{matrix} \right] +\left[ \begin{matrix} 1 & 7 & 0 \\ 1 & 3 & 1 \\ 2 & 4 & 0 \end{matrix} \right] =\left[ \begin{matrix} 1+1 & 2+7 & 3+0 \\ 4+1 & 5+3 & 6+1 \\ 7+2 & 8+4 & 9+0 \end{matrix} \right] =\left[ \begin{matrix} 2 & 9 & 3 \\ 5 & 8 & 7 \\ 9 & 12 & 9 \end{matrix} \right] \)

\( \mathrm{P}(A \cup B) =\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(A \cap B) \)
\(\frac{1}{3} =\frac{2}{3}+\frac{2}{5}-\mathrm{P}(A \cap B) \)
\(\mathrm{P}(A \cap B) =\frac{2}{3}+\frac{2}{5}-\frac{1}{3} \)
\(\mathrm{P}(A \cap B) =\frac{10+6-5}{15}=\frac{16-5}{15}=\frac{11}{15} \)

\(\left[ \begin{matrix} 12 & 3 \\ x & \frac { 3 }{ 2 } \end{matrix} \right] =\left[ \begin{matrix} y & z \\ 3 & 5 \end{matrix} \right] \)
x = 3
y = 12
z = 3

When we toss three coins the outcome will be

The sample space = { HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
Total number of outcomes = 8

leap year has 366 days. So it has 52 full weeks and 2 days. 52 Saturdays must be in 52 full weeks.
The possible chances for the remaining two days will be the sample space.
S = {(Sun-Mon, Mon-Tue, Tue-Wed, Wed-Thu, Thu-Fri, Fri-Sat, Sat-Sun)}
n(S) = 7
Let A be the event of getting 53^rd Saturday.
Then A = {Fri-Sat, Sat-Sun}; n(A) = 2
Probability of getting 53 Saturdays in a leap year is P(A0 = \(\frac { n(A) }{ n(S) } =\frac { 2 }{ 7 } \).

To find the coefficient of variation we need standard deviation
\(\sigma =\sqrt{\frac{\Sigma x_{i}^{2}}{n}-\left(\frac{\Sigma x_{i}}{n}\right)^{2}} \)
\(\frac{\Sigma x^{2}}{n} =\frac{765}{5}=153 \)
\(\left(\frac{\Sigma x}{n}\right)^{2} =(\bar{x})^{2}=6^{2}=36 \)
\(\sigma =\sqrt{(153)-36}=\sqrt{117} \)
\(=\sqrt{3 \times 3 \times 13} \)
\(\sigma =3 \sqrt{13} \)
Coefficient of variation \(=\frac{\sigma}{x} \times 100 \%\)
\(=\frac{3 \sqrt{13}}{6} \times 100 \%=\frac{\sqrt{13}}{2} \times 100 \%=\frac{3.60555}{2} \times 100 \% \)
\(=1.80277 \times 100 \%=180.277 \% \)
Coefficient of variation = 180.28 %

For comparing two data, first we have to find their coefficient of variations
Mean \(\bar { { x }_{ 1 } } \) = 155 cm, variance σ₁² = 72.25 cm²
Therefore standard deviation σ₁ = 8.5
Coefficient of variation C.V₁ = \(\frac { { \sigma }_{ 1 } }{ \bar { { x }_{ 1 } } } \) x 100%
C.V₁ = \(\frac { 8.5 }{ 15.5 } \) x 100% = 5.48% (for heights)
Mean \(\bar { { x }_{ 2 } } \) = 155 cm, variance σ₂² = 72.25 kg²
Standard deviation σ₂ = 5.3 kg
Coefficient of variation CV₂ = \(\frac { { \sigma }_{ 2 } }{ \bar { { x }_{ 2 } } } \) x 100%
C.V₂ = \(\frac { 5.3 }{ 46.50 } \) x 100% = 11.40% (for weights)
C.V₁ = 5.48% = and CV₂ = 11.40%
Since C.V₂ > C.V₁,  the weight of the students is more varying than the height.

2x² - x - 1 = 0 here, a = 2, b = -1, c = -1
α + β = \(\frac {-b}{a} = \frac {-(-1)}{2} = \frac {1}{2}\), αβ = \(\frac {c}{a} = -\frac {1}{2}\)
Given roots are \(\frac { 1 }{ \alpha } ,\frac { 1 }{ \beta } \)
Sum of the roots = \(\frac { 1 }{ \alpha } +\frac { 1 }{ \beta } =\frac { \alpha +\beta }{ \alpha \beta } =\frac { \frac { 1 }{ 2 } }{ -\frac { 1 }{ 2 } } \) = -1
Product of the roots = \(\frac { 1 }{ \alpha } \times \frac { 1 }{ \beta } =\frac { 1 }{ \alpha \beta } =\frac { 1 }{ -\frac { 1 }{ 2 } } \) = -2
The required equation is x² – (Sum of the roots)x + (Product of the roots) = 0
x² - (-1)x - 2 = 0 gives x² + x - 2 = 0

Let AB be the rock.
C be the position of the car.
In right triangle ABC
\(\tan 30^{\circ} =\frac{A B}{B C} \)
\(\frac{1}{\sqrt{3}} =\frac{50 \sqrt{3}}{B C} \)
\(B C =50 \sqrt{3} \times \sqrt{3} \)
= 50 x 3 = 150 m
Distance of the car from the rock = 150 m.

From the right \(\triangle\)ABC
\( \tan \theta =\frac{\text { Opposite side }}{\text { Adjacent side }}=\frac{A C}{B C} \)
\(=\frac{10 \sqrt{3} m}{30 m}=\frac{\sqrt{3}}{3} \)
\(=\frac{\sqrt{3}}{\sqrt{3} \sqrt{3}}=\frac{1}{\sqrt{3}} \)
\(\tan \theta =\frac{1}{\sqrt{3}} \)
\(\theta =\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)\)
0 = 30^o
Angle of elevation is 30^o

Here the largest value = 650
The smallest value = 400
 \(\therefore\) Range = L- S
= 650 - 400
= 250
Range R = 250

63,89,98, 125,79,108, 117,68
Largest value L = 125
Smallest value S = 63
\(\therefore\) R = L - S = 125- 63 = 62
Co-efficient of range = \(\frac { L-S }{ L+S } \)
=\(​​​​\frac {125-63 }{ 125+63 } \)
\(=\frac{62}{188}=0.329=0.33\)
Range = 62; coefficient of range = 0.33

The vertices Q(3, - 2) and R(- 5, 4)
slope of the line QR \(=\frac{y_{1}-y_{2}}{x_{1}-x_{2}} \)
\(=\frac{-2-4}{3+5}=\frac{-6}{8}=\frac{-3}{4} \)
Slope of the line parallel to QR is \(-\frac{3}{4}\)
Equation of the line passing through
P(- 5, 2) and having slope \(-\frac{3}{4}\) is
y - y₁ = m(x - x₁)
y - 2 = \(-\frac{3}{4}(x+5)\)
4y - 8 = -3x - 15
3x + 4y + 7 = 0

\(\Delta AEC\quad \Delta ADB\)
\(\angle AEC=\angle ADB={ 90 }^{ 0 }\) 
\(\angle C A E=\angle B A D\)
[common By AA similarity criteria] 
 \(\Delta AEC\sim \Delta ADB\)
(ii) Their corresponding sides are Proportional
\(\frac{C A}{A B}=\frac{C E}{D B}\)
Hence proved.

1² + 2² +...+ 19² = \(\frac { 19\times \left( 19+1 \right) \left( 2\times 19+1 \right) }{ 6 } =\frac { 19\times 20\times 39 }{ 6 } =2470\)

Radius of hemisphere = r units
Radius of cone = Radius of hemisphere
Height of cone = Radius of hemisphere
Maximum volume of cone = \(\frac{1}{3} \pi r^{2} h \text { cu. units }\)
\(=\frac{1}{3} \pi\left(r^{2}\right) r=\frac{1}{3} \pi r^{3} \text { cu. units }\)

1 + 3 + 5 + ...40 terms = 40² = 1600

We can express the number 0.6666 ....as follows
0.6666 ..... = 0.6 + 0.06 + 0.006 + 0.0006 + .....
We now see that numbers 0.6,0.06,0.006....forms an G.P Whose first term a = 0.6 and common ration r = \(\frac { 0.06 }{ 0.6 } \) = 0.1
Also  - 1 < r = 0.1  < 1
Using the infinite G.P. formula, we have
0.6666.... = 0.6 + 0.06 + 0.006 + 0.0006 + .... =\(\frac { 0.6 }{ 1-0.1 } =\frac { 0.6 }{ 0.9 } =\frac { 2 }{ 3 } \)
Thus the rational number equivalent of 0.6666..is \(\frac { 2 }{ 3 } \) 

Let PQ the height of the tower.
Take PQ = h and QR is the distance between the tower and the point R.in right triangle PQR,\(\angle \)PRQ=30°
tan\(\theta =\frac { PQ }{ QR } \)
tan30° = \(\frac { h }{ 48 } \) gives,\(\frac { 1 }{ \sqrt { 3 } } =\frac { h }{ 48 } \) so, h =\(16\sqrt { 3 } \)
Therefore the height of the tower \(16\sqrt { 3 } \) m

Given a, b, c are in A.P
t₂ - t₁ = t₃ - t₂
b - a = c - b
b + b = c + a
2b = c + a
If we multiply both the sides by same number value will not change
3^2b = 3^a+c
3^{b + b} = 3^a+c
3^b . 3^b = 3^a . 3^c
\(\frac{3^{b}}{3^{a}}=\frac{3^{c}}{3^{b}}\)
Thus 3^a, 3^b, 3^c are in G.P.

a = 6, r = 3
The three terms of G.P. are a, ar, ar²
= 6, 6 x 3,6 (3)²
= 6, 18, 54
First three terms are 6, 18, 54

Largest value L = 67; Smallest value S =18
Range R = L = S = 67 - 18 = 49
Coefficient of range = \(\frac { L-S }{ L+S } \)
Coefficient of range = \(\frac { 67-18 }{ 67+18 } =\frac { 49 }{ 85 } \) = 0.576

Equation of the given line is 4x − 9y + 36 = 0
we write it as 4x - 9y = − 36 (bring it to the normal form)
Dividing by - 36 we get, \(\frac { x }{ -9 } +\frac { y }{ 4 } =1\) ...(1)
Comparing (1) with intercept form, we get x intercept a = −9 ; y intercept b = 4

Total amount to repay = Rs. 65000
He pays Rs. 400 in the first installment and increasing the payment by Rs. 300 every month.
This form an A.P.
i.e., 400, 700, 1000, .....
a = 400,.d = 300
Sum upto n terms
\(\mathrm{S}_{\mathrm{n}} =\frac{n}{2}[2 a+(n-1) d] \)
\(65000 =\frac{n}{2}[2(400)+(n-1) 300] \)
\(65000 =\frac{n}{2} \times 2[400+(n-1) 150] \)
= n[400 + 150n - 150]
= n[150 n + 250]
65000 = 150n² + 250n
Divided by 50
1300 = 3n² + 5n
3n² + 5n - 1300 = 0
\(\mathrm{n} =\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \)
\(=\frac{-5 \pm \sqrt{5^{2}-4(3)(-1300)}}{2(3)} \)
\(=\frac{-5 \pm \sqrt{25+15600}}{6}=\frac{-5 \pm \sqrt{15625}}{6} \)
\(=\frac{-5 \pm 125}{6} \)
\(n =\frac{-5-125}{6}, \frac{-5+125}{6} \)
\(\mathrm{n} =\frac{-130}{6}, \frac{120}{6} \)
\(=\frac{-130}{6}, 20 \)
n cannot be negative
n = 20.
He will clear the loan by 20 months

Each person requires 22 m² of floor area.
Required base area = 22 x 4 = 88 m²
\(\pi r^{2} =88 \)
\(r^{2} =\frac{88 \times 7}{22}=4 \times 7 \)
\(r =2 \sqrt{7} \mathrm{~m} \)
slant height = 19 m
height of the tent, h \(=\sqrt{l^{2}-r^{2}}\)
\(=\sqrt{(191)^{2}-(2 \sqrt{7})^{2}} \)
\(=\sqrt{361-28} =\sqrt{330}=18.25 \mathrm{~m} \)
Height of the tent = 18.25 m

Let l, R and r be the slant height, top radius and bottom radius of the frustum.
Given that, l = 5 cm, R = 4 cm, r = 1 cm
Now, C.S.A. of the frustum \(\pi\)(R + r)l sq.units
\(\frac { 22 }{ 7 } \times (4+1)\times 5\)
\(=\frac { 550 }{ 7 } \)
Therefore, C.S.A. = 78.57 cm²

Given vertices A(1, - 4) , B(2, - 3) and C(4, - 7)
Slope of the line = \(\frac{y_{1}-y_{2}}{x_{1}-x_{2}}\)
Slope of AB = \(\frac{-4+3}{1-2}=\frac{-1}{-1}=1 \)
Slope of BC = \(\frac{-3+7}{2-4}=\frac{4}{-2}=-2 \)
Slope of AC = \(\frac{-4+7}{1-4}=\frac{3}{-3}=-1 \)
(Slope of AB) x (Slope of AC) = - 1
AB is perpendicular to AC.
Hence, the given vertices form a right angled triangle
Distance between the points (x₁, y₁) and (x₂, y₂) is \(\sqrt{\left.\left(x_{1}-x_{2}\right)^{2}+y_{1}-y_{2}\right)^{2}}\) units
\(A B =\sqrt{(1-2)^{2}+(-4+3)^{2}}=\sqrt{1+1}=\sqrt{2} \)
\(A B^{2} =(\sqrt{2})^{2}=2 \)
\(B C =\sqrt{(2-4)^{2}+(-3+7)^{2}}=\sqrt{4+16}=\sqrt{20} \)
BC² = 20
\(A C=\sqrt{(1-4)^{2}+(-4+7)^{2}}=\sqrt{9+9}=\sqrt{18}\)
AC² = 18
Now, AB² + AC² = BC²
Hence, the Pythagoras theorem is satisfied.

Given slope 'm' = 0
tan θ = 0 = tan 0⁰
θ = 0⁰

f(x) = 3 x -2, g(x) = 2x + k
f o g = f(g(x)) = f(2x + k) = 3(2x + k) - 2 = 6x + k - 2
Thus, f o g(x) = 6x + 3k - 2
g o f(x) = g(3x - 2) = 2(3x - 2) + k
Thus, g o f(x) = 6x - 4 + k
Given that f o g = g o f
Therefore, 6x + 3k-2 = 6z - 4 + k
6x - 6x + 3k - k = -4 + 2 ⇒ -1

First term a = 5; common difference d = 6.
A.P.is given by., a + d, a + 2d, a + 3d,......
In this case 5, 5 + 6, 5 + 2(6),5 + 3 (6),...
5, 11, 17 ,23,...
The required A.P. is 5, 11,17,23,....

Let r be the radius of the sphere. Given that, surface area of sphere = 154 m²
4\(\pi\)r² = 154
\(4\times \frac { 22 }{ 7 } \times { r }^{ 2 }=154\)
gives \({ r }^{ 2 }=154\times \frac { 1 }{ 4 } \times \frac { 7 }{ 22 } \)
hence, \({ r }^{ 2 }=\frac { 49 }{ 4 } \)We get r = \(\frac{7}{2}\)
Therefore, diameter is 7 m

Given that, diameter d = 2.8 m and height = 3 m
radius r = 1.4 m
Area covered in one revolution = curved surface area of the cylinder
= 2\(\pi\)rh sq. units
\(2\times \frac { 22 }{ 7 } \times 1.4\times 3=26.4\)
Area covered in 1 revolution = 26.4 m²
Area covered in 8 revolutions = 8 x 26.4 = 211.2
Therefore, area covered is 211.2 m²

8x⁴y², 48x²y⁴
First let us find the LCM of the numerical coefficients.
That is, LCM (8, 48) = 2 x 2 x 2 x 6 = 48
Then find the LCM of the terms involving variables.
That is, LCM (x⁴y², x²y⁴) = x⁴y⁴
Finally find the LCM of the given expression.
We conclude that the LCM of the given expression is the product of the LCM of the numerical coefficient and the LCM of the terms with variables.
Therefore, LCM (8x⁴y², 48x²y⁴) = 48 x⁴y⁴

The curves in Fig.1.35(a) and Fig.1.35(c) represent a one – one function as the horizontal lines meet the curves in only one point P.
The curve in Fig.1.35(b) does not represent a one–one function, since, the horizontal line meet the curve in two points P and Q.

Given points are (2, 3), (4, a) and (6, - 3)
Since the points are colinear, Area of triangle is zero
\(\text { i.e., } \frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right]=0\)
2(a + 3) + 4(- 3 -3) + 6(3 - a) = 0
2a + 6 - 24 + 18 - 6a = 0
-4a + 0 = 0
-4a = 0
a = 0

(i) Given vertices are (0, 0), (P, 8) and (6,2)
Area of triangle = 20 sq. units.
Area of triangle = \(\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+\right. \left.x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right] \)
\(\frac{1}{2}\) [(8 - 2) + p (2 - 0) + 6 (0 - 8)] = 20
2p - 48 = 40
2P = 40 + 48
2P = 88
\(p=\frac{88}{2}=44\)
(ii) Given vertices are (p, p), (5,6) and (5, - 2)
Area of triangle = 32 sq. units
Area of triangle = \(\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+\right. \left.x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right] \)
\(\frac{1}{2}\) [p( 6 + 2) + 5(- 2 -p) + 5(P - 6) = 32
8p -10 - 5P + 5P - 30 = 64
8p - 40 = 64
8P = 64 + 40 = 104
\(p=\frac{104}{8}=13\)

Given o = \(\triangle\)ABC and \(\triangle\)ADE
By AA criteria for similarity
\(\triangle A B C \sim \triangle A D E\)
Their corresponding sides are proportional
\(\therefore \frac{B C}{D E} =\frac{A C}{A E}=\frac{A B}{A D} \)
\(\frac{12}{D E} =\frac{3+2}{A E}=\frac{A B}{3} \)
Also in right ABC , o
using Pythagoras theorem, we have
AB² = BC² + AC² = 12² + 5²
= 144 + 25 = 169
AB = 13 cm
Now from (1), we get
\(\frac{12}{D E}=\frac{13}{3} \)
\(\mathrm{DE}=\frac{12 \times 3}{13}=\frac{36}{13} \)
DE = 2.77 cm
Also from (1) \(\frac{5}{A E}=\frac{13}{3} \)
\(\mathrm{AE}=\frac{5 \times 3}{13}=\frac{15}{13}=1.15 \mathrm{~cm} \)

Given = O
PR and SQ intersect at T.
In \(\triangle\)QPT and \(\triangle\)RST
< QSR = o
Given
[Vertically opposite angles]
By AA similarity criteria
\(\triangle P Q T \sim \triangle S R T\)
Their corresponding sides are proportional 
\(\frac{P T}{S T}=\frac{T Q}{T R}\)
PT x TR = ST x TQ
Hence proved

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...........}
Codomain = {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.

3x \(\equiv \) 1 (mod 15) can be written as
3x - 1 = 15k for some integer k
3x = 15k + 1
\(x=\frac { 15k+1 }{ 3 } \)
\(x=5k+\frac { 1 }{ 3 } \)
Since 5k is an integer , 5k + \(\frac { 1 }{ 3 } \) cannot be an integer 
So there is no integer solution

\(\frac { sec\theta }{ sin\theta } -\frac { sin\theta }{ cos\theta } = \frac { \frac { 1 }{ cos\theta } }{ sin\theta } -\frac { sin\theta }{ cos\theta } =\frac { 1 }{ sin\theta cos\theta } -\frac { sin\theta }{ cos\theta } \)
\(=\frac { 1-si{ n }^{ 2 }\theta }{ sin\theta cos\theta } =cot\theta \)

Consider 2ⁿ \(\times\) 5^m
Since the product has 2 as a factor
2ⁿ \(\times\) 5^m is even
But if a number ends with the digit 5, then the number is an odd number
It is impossible.
For no value of m, 2ⁿ \(\times\) 5^m ends in 5.

\(\Delta \ CAB\) and \(\Delta CED\),\(\angle C\) is common, \(\angle A=\angle CED\)
Therefore, \(\Delta CAB\sim \Delta CED\)
Hence, \(\frac { CA }{ CE } =\frac { AB }{ DE } =\frac { CB }{ CD } \)
\(\frac { AB }{ DE } =\frac { CB }{ CD } \quad \frac { 9 }{ x } =\frac { 10+2 }{ 8 } ,x=\frac { 8\times 9 }{ 12 } =6\) cm.

Since 6ⁿ = (2 x 3)ⁿ = 2ⁿ x 3ⁿ
2 is a factor of 6ⁿ. So, 6 n is always even.
But any number whose last digit is 5 is always odd.
Hence, 6ⁿ cannot end with the digit 5 

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.

Let h₁ and h₂ be the heights of a cone and cylinder respectively.
Also, let r be the radius of the cone.
Given that, height of the cone h₁ = 24 cm; radius of the cone and cylinder r = 6 cm
Since, Volume of cylinder = Volume of cone
\({ \pi r }^{ 2 }=\frac { 1 }{ 3 } { \pi r }^{ 2 }{ h }_{ 1 }\)
\({ h }_{ 2 }=\frac { 1 }{ 3 } \times { h }_{ 1 }\quad gives\quad { h }_{ 2 }=\frac { 1 }{ 3 } \times 24=8\)
Therefore, height of cylinder is 8 cm

 f(x) = 3x + 2, g(x) = 6x - k
f o g = f{g(x)) = f{6x - k)
= 3(6x - k) + 2
= 18x- 3k + 2
g o f = g(f(x)) = 6(3x + 2) - k
= 18x + 12 - k
Given f o g = g o f
18x - 3k + 2 = 18x + 12 - k
3k - k = -12 + 2
2k = -10
k = -5

Given total surface area of cylinder
= 1848 sq.m
(h + r) = 1848
It is given that C.S.A \(=\frac{5}{6}(\text { T.S.A })\)
\(\text { C.S.A }=\frac{5}{6}(1848)=1540 \)
\(\text { C.S.A }=\frac{5}{6}(\text { T.S.A }) \)
\(2 \pi r h=\frac{5}{6}(2 \pi r(h+r)) \)
h = 5r
We have C.S.A = 1540
2rh = 1540
\(2 \times \frac{22}{7} \times r \times 5 r =1540 \)
\(r^{2} =\frac{1540 \times 7}{44 \times 5}=49 \)
r = 7
h = 5r = 5(7) = 35
radius = 7 m , height = 35 m

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