A Graph f = {(x, f(x) / x \(\epsilon \) A}
{(0, 1), (1, 3), (2, 5), (3, 7)}

A table

Domain of R = {1,2,3,4}
Co-domain of R = B = {-1, 2, 3,4,5,6, 7, 9, 10, 11,12}
Range of R = {3, 6,10, 9}

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

kx² - k²x - 2k³ = 0
a = k, b = -k², c = -2k³
α + β = \(-\frac {b}{a}\) = \(\frac {-(-k^{2})}{k}\) = k and αβ = \(\frac {c}{a}\) = \(\frac {-2k^3}{k}\) = -2k²

\(\frac{t_{2}}{t_{1}}=\frac{0.05}{0.5}=\frac{0.5}{5}=\frac{5}{50}=\frac{1}{10} \)
\(\frac{t_{3}}{t_{2}}=\frac{0.005}{0.05}=\frac{5}{50}=\frac{1}{10} \)
The ratios between the successive terms are equal 0.5,  0.05, 0.005,... are in G.P.

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%

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

2x² - x - 1 = 0
\({ x }^{ 2 }-\frac { x }{ 2 } -\frac { 1 }{ 2 } \) = 0   (÷2 make co-efficient of x² as 1)
\({ x }^{ 2 }-\frac { x }{ 2 } =\frac { 1 }{ 2 } \)
\({ x }^{ 2 }-\frac { x }{ 2 } +{ \left( \frac { 1 }{ 4 } \right) }^{ 2 }=\frac { 1 }{ 2 } +{ \left( \frac { 1 }{ 4 } \right) }^{ 2 }\)
\({ \left( x-\frac { 1 }{ 4 } \right) }^{ 2 }=\frac { 9 }{ 16 } ={ \left( \frac { 3 }{ 4 } \right) }^{ 2 }\)
\(x-\frac { 1 }{ 4 } =\pm \frac { 3 }{ 4 } \) ⇒ x = 1, \(\frac {-1}{2}\)

General form of the quadratic equation when the roots are given is
x² - (sum of the roots)x + product of the roots = 0
x² - 9x + 14 = 0

Slope of a line passing through (- 2,3) and (8, 5) is 
\(\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\frac{3-5}{-2-8}=\frac{-2}{-10}=\frac{1}{5}=m_{1}\)
Slope of the line y = ax + 2 is 'a' = m₂
Given that the lines are perpendicular
\(m_{1} \times m_{2} =-1 \)
\(\frac{1}{5} \times a =-1 \)
a = -5

Radius of well = 5 m
Depth of well = 14 m
Volume of earth taken out \(=\pi r^{2} h \)
\(=\frac{22}{7} \times(5)^{2} \times 14 \)
= 1100 m³
Now, it is spread to form an embankment, which is in the form of hollow cylinder
Inner radius = 5m
Width of embankment = 5 m
Outer radius = 5 + 5 = 10 m
height = h
Volume of hollow cylinder = \(\pi h\left(\mathrm{R}^{2}-\mathrm{r}^{2}\right)\)
\(\therefore \pi h\left(\mathrm{R}^{2}-\mathrm{r}^{2}\right)=1100 \)
\(\frac{22}{7} \times h\left(10^{2}-5^{2}\right)=1100 \)
height of the embankment
\(h=\frac{1100 \times 7}{22 \times 75}=4.67 \mathrm{~m}\)

The equation of a straight line passing through the two points (x₁, y₁) and (x₂, y₂) is \(\frac { y-{ y }_{ 1 } }{ { y }_{ 2 }-{ y }_{ 1 } } =\frac { x-{ x }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } \)
Substituting the points we get, \(\frac { y+3 }{ -4+3 } =\frac { x-5 }{ 7-5 } \)
gives 2y + 6 = − x + 5
Therefore, x + 2y + 1 = 0

f(x) = x - 6, g(x) = x²
fog(x) = f(g(x)) = f(x²) = x² - 6 ...(1)
gof(x) = g(f(x)) = g(f(x)) = g(x - 6) = (x - 6)²
= x² - 12x + 36
fog(x) ≠ gof

Mid point of line segment joining (4, 2) and (-6, 4)
\(\text { Mid point } =\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right) \)
\(=P\left(\frac{4-6}{2}, \frac{2+4}{2}\right) \)
\(=P\left(-\frac{2}{2}, \frac{6}{2}\right)=\mathrm{P}(-1,3) \)
Now, slope of a line joining A (5, 1) and P (- 1, 3)
\(\mathrm{m}=\frac{y_{1}-y_{2}}{x_{1}-x}=\frac{1-3}{5+1}=\frac{-2}{6}=-\frac{1}{3}\)
Slope of a perpendicular to the line joining A and P
\(=-\frac{1}{m}=-\frac{1}{\left(-\frac{1}{3}\right)}=3\)

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

67 + x \(\equiv \) 1 (mod 4)
67 + x  - 1  = 4n, for some integer n,
66 + x  = 4n
66 + x is a multiple of 4
Therefore, the least positive value of x must be 2, since 68 is the nearest multiple of 4 more than 66.

\(\sqrt { \frac { 1+cos\theta }{ 1-cos\theta } } \)=\(\sqrt { \frac { 1+cos\theta }{ 1-cos\theta } \times \frac { 1+cos\theta }{ 1+cos\theta } } \) [multiply numerator and denominator by the conjugate of 1 - cos\(\theta \)]
=\(\sqrt { \frac { (1+cos\theta { ) }^{ 2 } }{ (1-cos\theta { ) }^{ 2 } } } \) =\(\frac { 1+cos\theta }{ \sqrt { si{ n }^{ 2 }\theta } } \) [since sin²\(\theta \) + cos²\(\theta \) = 1]
=\(\frac { 1+cos\theta }{ sin\theta } =cosec\theta +cot\theta \)

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}

In \(\triangle\)PST and \(\triangle\)PQR,
\(\frac { PS }{ PQ } =\frac { 2 }{ 2+1 } =\frac { 2 }{ 3 } ,\frac { PT }{ PR } =\frac { 4 }{ 4+2 } =\frac { 2 }{ 3 } \)
Thus, \(\frac { PS }{ PQ } =\frac { PT }{ PR } \) and \(\angle\)P is common
Therefore, by SAS similarity,
\(\triangle\) PST~\(\triangle\)PQR

We are given that sin 3A = cos (A - 26°) ...(1)
Since sin 3A = cos(90° - 3A) we can write (1) as
cos (90° - 3A) = cos (A - 26°)
Since 90° - 3A and A - 26° are both acute angles
90° - 3A = A - 26°
which gives A = 29°

Seven years ago, let Swathi's age be x years .
Seven years ago, let Varun's age was 5x² years.
Swathi's present age = x + 7 years
Varun's present age = (5x² + 7) years
3 years hence, we have
Swathi's age = x + 7 + 3 years
=x + 10 years
Varun's age = 5x² + 7 + 3 years
= 5x² + 10 years
It is given that 3 years hence Swathi's age will
be \(\frac{2}{5}\) of Varun's age.
∴ x+10=\(\frac{2}{5}\)(5x²+10)
⇒ x+10=2x²+4
⇒ 2x²-x-6=0
⇒ 2x(x-2)+3(x-2)=0
⇒(2x+3)(x-2)=0
⇒ x-2=0
⇒ x=2(∵2x+3≠0 as x>0)
Hence Swathi's present age = (2 + 7) years
= 9 years
Varun's present age = (5 x 2² + 7) years
= 27 years

1,-1,-3, -5, ...
t₂ - t₁ = -1 - 1 = -2
t₃ - t₂ = -3 - (-1)= -2
t₄ - t₃ = -5 - (-3) = -2
The given list of numbers form an A.P with the common difference -2.
The next two terms are (-5 + (-2)) = -7, -7 + (-2) = -9.

f(2) - f(4)
f(2) = 2x - 1
= 2(2) - 1 = 3
f(4) = 3x² - 10
= 3(42) - 10 = 38
\(\therefore\) f(2) - f(4) = 3 - 38 = 35

f(-7) = x² + 2x + 1
= (-7)² + 2(-7) + 1
= 49 - 14 + 1 = 36
f(3) = x + 5 = -3 + 5 = 2
f(-7) - f(-3) = 36 + 2 = 38

36y² - 12ay + (a² - b²) = 0
      a          b          c
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
Here \(y=\frac { -(-12)\pm \sqrt { { (-12) }^{ 2 }-4\times 36\times \left( { a }^{ 2 }-{ b }^{ 2 } \right) } }{ 2\times 36 } \)
\(=\frac { 12a\pm \sqrt { { 144a }^{ 2 }-{ 144a }^{ 2 }+{ 144b }^{ 2 } } }{ 72 } \)
\(=\frac { 12a\pm 12b }{ 72 } \Rightarrow \frac { a\pm b }{ 6 } \)
\(=\frac { a+b }{ 6 } ,\frac { a-b }{ 6 } \)

Σ(x - 5) = 3
Σ(x - 5)² = 43
N = 18, standard deviation.
σ = \(\sqrt { \left( \frac { \sum { { (x-5) }^{ 2 } } }{ N } \right) { -\left( \frac { \sum { (x-5) } }{ N } \right) }^{ 2 } } \)
= \(\sqrt { \frac { 43 }{ 18 } -{ \left( \frac { 3 }{ 18 } \right) }^{ 2 } } \)
= \(\sqrt { 2.389-0.278 } \)
= \(\sqrt { 2.111 } \)
⇒ 1.45

Total number of possible outcomes n(S) = 5 + 4 = 9
(i) Let A be the event of getting a blue ball.
Number of favourable outcomes for the event A. Therefore, n(A) = 5
Probability that the ball drawn is blue. Therefore, P(A) = \(\frac { n(A) }{ n(S) } =\frac { 5 }{ 9 } \)
(ii) \(\bar { A } \) will be the event of not getting a blue ball. So P(\(\bar { A } \)) = 1 - P(A) = \(1-\frac { 5 }{ 9 } =\frac { 4 }{ 9 } \).

Total pages = 60 
Completed pages by the students are
32, 35, 37, 30, 33, 36, 35, 37
Let the pages yet to be completed by 8 students be x_i

Standard deviation \(\sigma=\sqrt{\frac{\sum x_{i}^{2}}{n}-\left(\frac{\sum x_{i}}{n}\right)^{2}}\)
\(=\sqrt{\frac{9497}{8}-\left(\frac{275}{8}\right)^{2}} \)
\(=\sqrt{1187.125-(34.375)^{2}}=\sqrt{1187.125-1181.64063} \)
\(=\sqrt{5.48437}=2.34\)

Let  AB = 1800 m be the height where the
aeroplane is flying D and E are positions of two boats
o
o
In right triangle BAE
\(\tan 60^{\circ} =\frac{B A}{E A} \)
\(\sqrt{3} =\frac{1800}{E A} \)
\(\mathrm{EA} =\frac{1800}{\sqrt{3}} \)
In right triangle BDA
\(\tan 30^{\circ} =\frac{A B}{A D} \)
\(\frac{1}{\sqrt{3}} =\frac{1800}{D E+E A} \)
\(\mathrm{DE}+\mathrm{EA} =1800 \sqrt{3} \)
\(\mathrm{DE} =1800 \sqrt{3}-E A \)
\(\mathrm{DE} =1800 \sqrt{3}-\frac{1800}{\sqrt{3}} \)
\(=\frac{1800 \sqrt{3} \sqrt{3}-1800}{\sqrt{3}} \)
\(=\frac{1800 \times 3-1800}{\sqrt{3}} \)
\(=\frac{5400-1800}{\sqrt{3}}=\frac{3600}{\sqrt{3}} \)
\(=\frac{3600 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}=\frac{3600 \sqrt{3}}{3}=1200 \sqrt{3}\)
= 1200 x 1.732
DE = 2078.4 m
Distance between the boats = 2078.4 m

Let AC be the pole and let point 'B' divide it in the ratio 
\(\not x: 9 \not x=1: 9\)
Let 'D' be the point 25 m.
\(tan\alpha =\frac { x }{ 25 } \) \(tan2\alpha =\frac { 10x }{ 25 } \)
\(tan2\alpha =\frac { 2tan\alpha }{ 1-{ tan }^{ 2 }\alpha } \)
\(\frac{\not 10 x^{5}}{\not 25}=\frac{2 \times \frac{x}{25}}{1-\frac{x^{2}}{625}}\)

Height of pole = 10x
= \(100\sqrt { 5 } \)
x = \(10\sqrt { 5 } \) m

Let AB be the height of the biggest tree and CD be the highest of the smaller tree and x is  the point on the ground.
In the right triangle XCD, cos40° =\(\frac { CX }{ XD } \)
Therefore the distance between  X and top of the smaller tree = XD =10.44m.

Hollow metallic cylinder
External radius R = 4.3 cm
Internal radius r = 1.1 cm
Length = height = h = 4 cm
Volume \(=\pi\left(\mathrm{R}^{2}-\mathrm{r}^{2}\right) \mathrm{h} \text { cu. units } \)
\(=\pi\left((4.3)^{2}-(1.1)^{2}\right)(4) \)
\(=\pi(18.49-1.21) 4 \)
\(=69.12 \pi \mathrm{cm} 3 \)
Solid cylinder
height h = 12 cm
radius r = ?
volume \(=\pi r^{2} h\ sq. units \)
\(=\pi r^{2}(12) \)
Given, Hollow cylinder is melted to form solid cylinder
Volume of cylinder = Volume of hollow cylinder
\(\pi r^{2}(12) =69.12 \pi \)
\(r^{2} =\frac{69.12}{12}=5.76 \)
r = 2.4 cm
Diameter of cylinder  = 2r = 4.8 cm

Diameter of the bowl = 14 cm
Radius r = 7 cm
Volume of hemisphere  \(=\frac{2}{3} \pi r^{3} \text { cu. units } \)
\(=\frac{2}{3} \times \frac{22}{7} \times 7 \times 7 \times 7 \)
\(=\frac{2156}{3}=1718.67 \mathrm{~cm}^{3} \)
Radius of cylinder 'r' = 7 cm
Height 'h' = 6 cm
Volume of cylinder \(=\pi r^{2} h \text { cu. units } \)
\(=\frac{22}{7} \times 7 \times 7 \times 6 \)
= 924 cm³
capacity of the vessel = Volume of hemisphere + Volume of cylinder
= 718.67 + 924
= 1642.67 cm³

Given radius of lower end r = 8 cm
radius of upper end R = 20 cm
height h = 16 cm
Volume \(=\frac{\pi h}{3}\left(\mathrm{R}^{2}+\mathrm{Rr}+\mathrm{r}^{2}\right) \mathrm{cu} . units \)
\(=\frac{22 \times 16}{7 \times 3}\left((20)^{2}+(20)(8)+(8)^{2}\right) \)
\(=\frac{22 \times 16}{21}[400+160+64] \)
\(=\frac{22 \times 16}{21}(624)=10459.43 \mathrm{~cm}^{3} \)
\(=\frac{10459.43}{1000}\left[\because 1000 \mathrm{~cm}^{3}=1\right. litre ]\)
= 10.45943 litre
Cost of milk per litre = Rs. 40
Total cost = 10.459 x 40
= Rs. 418.36

Given that PS is the bisector of \(\angle\)P of \(\triangle\)PQR
\(\frac{P Q}{P R}=\frac{Q S}{S R}\) [By Angle Bisector Theorem]
Adding 1 on both the sides
\( \frac{P Q}{P R}+1 =\frac{Q S}{S R}+1 \)
\(\frac{P Q+P R}{P R}=\frac{Q S+S R}{S R} \)
\(\frac{P Q+P R}{P R} =\frac{Q R}{S R} \)           ..(2)
In \(\triangle\)RST and \(\triangle\)RQP we have
\( \angle S R T=\angle Q R P=\angle R \)
\(\angle Q P R=\angle S T R=90^{\circ}\)
By AA criterion for similarity, we have
\(\triangle R S T \sim \triangle R Q P \)
\(\frac{R S}{R Q}=\frac{S T}{Q P} \)
\(\frac{Q P}{S T}=\frac{Q R}{R S}\)             ...(2)
From (1) and (2)
\(\frac{Q P}{S T}=\frac{P Q+P R}{P R}\)
PQ \(\times\) PR = ST(PQ + PR)

Given vertices are (- 4, - 2),(- 3, k), (3, - 2) and (2,3) and area of quadrilateral is 28 sq. units.
Area of quadrilateral \(=\frac{1}{2}\left[\left(x_{1}-x_{3}\right)\left(y_{2}-y_{4}\right)-\right. \left.\left(x_{2}-x_{4}\right)\left(y_{1}-y_{3}\right)\right] \)
\(\frac{1}{2}\) [(- 4 - 3) (k - 3) - (- 3 - 2) (- 2 + 2)] = 28
(-7) (k - 3) - (- 5) (0) = 56
-7k + 21 = 56
-7k = 56 - 21 = 35
\(k=\frac{35}{-7}=-5\)
k = -5

Let the no of seats in each row be x

⇒ 2x²-10x = x²+375
⇒ x²-10x-375 = 0
⇒x²-25x +15x-375 = 0
⇒ x(x-25) +15(x-25) = 0
⇒(x-25)(x+15) = 0
⇒ x = 25, x = -15, x > 0
∴ 25 rows are in the hall

Diameter = 6 cm
Radius =\(\frac { 6 }{ 2 } =3cm\) 

Length of the tangents PA = PB = 4 cm
Construction:
Steps:
(1) With centre O, draw a circle of radius 3cm.
(2) Draw a line OP = 5 cm
(3) Draw a bisector of OP, which cuts OP and M
(4) With M as centre and MO as radius draw a circle which cuts previous circle at A and B
(5) Join AP and BP. AP and BP are the required tangents. Thus length of the tangents are PA = PB = 4 cm.

Construction:
Steps (1) Draw QR = 6.5 cm.
Steps (2) Draw \(\angle RQE={ 60 }^{ 0 }\)
Steps (3) Draw \(\angle FQE={ 90 }^{ 0 }\)
Steps (4) Drawn the perpendicular bisector XY to ER which intersects QF at O and ER at G.
Steps (5)  With O as center and OQ as radius drawn a circle
Steps (6) XY intersects QR at G. On XY, from G marked an arc at M, such that GM = 4.5 cm
Steps (7) Drawn AB through M which is parallel to QR
Steps (8)  AB meets the circle at P and S
Steps (9) Joined QP and RP Then \(\triangle\)PQR is the required triangle.
Steps (10) Here \(\triangle\)SQR is also another required triangle.

Given a triangle PQR we are required to construct another triangle whose sides are \(\frac{3}{5}\) of the corresponding sides of the triangle PQR.



Steps of construction
1. Construct a \(\triangle\) PQR with any measurement
2. Draw a ray QX making an acute angle with QR on the side opposite to vertex P.
3. Locate 5 (the greater of 3 and 5 in \(\frac { 3 }{ 5 } \)) points.
Q₁Q_2, Q₃, Q₄ and Q₅ on QX so that QQ₁ = Q₁Q₂ = Q₂Q₃ = Q₄Q₅
4. Join Q₅R  and draw a line through Q₃ (the third point, 3 being smaller of 3 and 5 in \(\frac { 3 }{ 5 } \)) parallel to Q₅R to intersect QR at R'.
5. Draw line through R' parallel to the line RP to intersect QP at P'.
Then, \(\triangle\)P'QR' is the required triangle each of whose sides is three-fifths of the corresponding sides of \(\triangle\) PQR.