An arrow diagram
 

A = {1,2,3},B =  {1,3,5, 7,9}
f(x) = 2x + 1
f(0)) = 2(0) + 1 = 1
f(1) =  2(1) + 1=3
f(2) = 2(2) + 1 = 5
f(3) = 2(3) + 1 = 7
(i) A set of ordered pairs.
f = {(0, 1), (1, 3), (2, 5), (3, 7)}
(ii) A table

x	0	1	2	3
f(x)	1	3	5	7

(ii) An arrow diagram

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

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

 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.

Given points (1, - 5) and (4, 2)
Mid-point of the line joining the points (1, - 5), (4, 2)
\(=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)=\left(\frac{1+4}{2}, \frac{-5+2}{2}\right) \)
\(=\left(\frac{5}{2},-\frac{3}{2}\right) \)
Equation of a straight line passing through
\(\left(\frac{5}{2},-\frac{3}{2}\right) \) and parallel to Y-axis is x = a
\(\text { i.e., } x=\frac{5}{2} \Rightarrow 2 x-5=0\)

This sequence is generated by a_n \(=\frac{n}{(n+1)^{2}}\)
The numerator is increased by 1 in successive terms.
The denominators are 2², 3², 4²,..........
\(\therefore a_{4}=\frac{4}{5^{2}}=\frac{4}{25} \)
\(a_{5}=\frac{5}{6^{2}}=\frac{5}{36} \)
\(a_{6}=\frac{6}{7^{2}}=\frac{6}{49} \)
The next three terms are \(\frac{4}{25}, \frac{5}{36}, \frac{6}{49}\)

Here each term is decreased by 3. So the next three terms are -7, -10, -13.

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

f(x) = \(\frac{2}{x}\), g(x) = 2x² - 1
fog(x) = f(g)x)) = f(2x² - 1) = \(\frac { 2 }{ 2x^{ 2 }-1 } \) ..(1)
gof(x) = g(f(x)) = \(g\left( \frac { 2 }{ x } \right) =2\left( \frac { 2 }{ x } \right) ^{ 2 }\) -1
= 2\(\left( \frac { 4 }{ { x }^{ 2 } } \right) -1=\frac { 8 -x^2}{ { x }^{ 2 } } \) 
fog ≠ gof

P(A) = 0.37, P(B) = 0.42, P(A∩B) = 0.09
P(AUB) = P(A) + P(B) - P(A∩B)
P(AUB) = 0.37 + 0.42 - 0.09 = 0.7

kx² - (8k + 4) + 81 = 0
Since the equation has real and equal roots, Δ = 0
That is, b² - 4ac = 0
Here, a = k, b = -(8k + 4), c = 81
That is, [-(8k + 4)]² - 4(k)(81) = 0
64k² + 64k + 16 - 324k = 0
64k² - 260k + 16 = 0
dividing by 4 we get 16k² - 65k + 4 = 0
(16k - 1)(k - 4) = 0 then, k = \(\frac {1}{16}\) or k = 4

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

Let r and h be the radius and height of the cylinder respectively.
Given that, height h = 2 m, base area = 250 m²
Now, volume of a cylinder = \(\pi\)r h ² cu. units
= base area x h
= 250 x 2 = 500 m³
Therefore, volume of the cylinder = 500 m³

f = {(-1, 3), (0, -1), 2, -9)
Since 'f' is a linear function
f(x) = (ax) + b
in (0 , -1) , when x = 0 , f(0) = -1
a(0) + b = -1 , b = -1
in (-1 , 3) , when x = -1 , f(-1) = 3
a( -1) + b = 3 ,a - 1 = 3
-a = 4, a = -4
f(x) = - 4x - 1

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

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
\(=-\frac{1}{m}=-\frac{1}{\left(-\frac{1}{3}\right)}=3\)

4x²y, 8x³y²
LCM of (4, 8) = 8
LCM of (x²y, x³y²) = x³y²
LCM of (4x²y, 8x³y²) = 8x³y²

(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
\(p=\frac{104}{8}=13\)

Since the ratio of area of two similar triangles is equal to the ratio of the squares of any two corresponding sides, we have
\(\frac { Area(\Delta ABC) }{ Area(\Delta DEF) } =\frac { { BC }^{ 2 } }{ { EF }^{ 2 } } \) gives \(\frac { 54 }{ Area(\Delta DEF) } =\frac { { 3 }^{ 2 } }{ { 4 }^{ 2 } } \)
\(Area(\Delta DEF)=\frac { 16\times 54 }{ 9 } =96{ cm }^{ 2 }\)

sec\(\theta \) - cos\(\theta \)  = \(\frac { 1 }{ cos\theta } -cos\theta =\frac { 1-co{ s }^{ 2 }\theta }{ cos\theta } \)
= \(\frac { si{ n }^{ 2 }\theta }{ cos\theta } \) [since 1 - cos²\(\theta \) = sin²\(\theta \)]
= \(\frac { sin\theta }{ cos\theta } \times sin\theta =tan\theta sin\theta \) 

        

A and B represent points on the bank on opposite sides at the river, so that AB is the width of the river. P is a point on the bridge at a height of 3m i.e., DP = 3 m. We are interested to determine the width at the river which is the length at the side AB of the ΔAPB.

Now, AB = AD + DB
In right ΔAPD,
ㄥA = 30^o
So, tan 30° = \(\frac { 1 }{ \sqrt { 3 } } =\frac { 3 }{ AD } \)
i.e., \(\frac { 1 }{ \sqrt { 3 } } =\frac { 3 }{ AD } \) (or) AD = 3\(\sqrt { 3 } \)
Also, in right ΔPBD,
B = 45^o
So, BD = PD = 3m
Now, AB = BD + AD
= 3 + 3\(\sqrt { 3 } \) = 3(1+\(\sqrt { 3 } \))m
Therefore, the width at the river is 3(+\(\sqrt { 3 } \))m

No. of balls required \(=\frac { \frac { 4 }{ 3 } \pi { r }^{ 3 } }{ \frac { 4 }{ 3 } \pi \times { \left( \frac { r }{ 4 } \right) }^{ 3 } } \)
\(={ r }^{ 3 }\times \frac { 4 }{ r } \times \frac { 4 }{ r } \times \frac { 4 }{ r } =64\)

Total number of cards = 52
Event of selecting a spade card = A
Event of selecting a heart card = B
n(A) = 13, n(B) = 13
P(A) =\(\frac { 13 }{ 52 } =\frac { 1 }{ 4 } \), P(B) =\(\frac { 13 }{ 52 } =\frac { 1 }{ 4 } \).
A ⋂ B refers to a card is both spade and heart. It is not possible

A, B are exclusive events.
P(A,B) = P(A) + P(B) =\(\frac { 1 }{ 4 } +\frac { 1 }{ 4 } =\frac { 1 }{ 2 } \).

Let the tens digits be x. Then the uints digits=\(\frac{18}{x}\)
∴ Number =10x+\(\frac{18}{x}\)
and number obtained by interchanging the digits =10x+\(\frac{18}{x}\)
\(\\ \therefore \left( 10x+\frac { 18 }{ x } \right) -\left( 10\times \frac { 18 }{ x } +x \right) =63\)
\(\Rightarrow 10x+\frac { 18 }{ x } -\frac { 180 }{ x } -x=0\)
\(\Rightarrow 9x-\frac { 162 }{ x } -63=0\)
⇒ 9x²-63x-162=0
⇒ x²-7x-18=0
⇒ (x-9)(x+2)=0⇒ x=9,-2
But a digit can never be (-ve), so x = 9.
So, the required number \(=10\times 9+\frac { 18 }{ 9 } =92\)

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

Here a = 24, d = 21-24 = -3, S_n = 78. We need to find n.
We know that,
S_n= \(\frac{n}{2}\) (2a + (n - 1)d)
78 = \(\frac{n}{2}\) 48 + (n -1)(-3))
78 = \(\frac{n}{2}\) 2(51-3n)
or 3n² - 51n + 156 = 0
n² -17n + 52 = 0
(n - 4)(n - 13) = 0
n = 4 or 13
The number of terms are 4 or 13.

Here S₁₄ = 1050
n = 14
a = 10
S_n = \(\frac{n}{2}\) 2(2a + (n -1)d)
1050 = \(\frac{14}{2}\)(20 + 13d)
= 140 + 91d
910 = 91d
d = 10
a₂₀ = 10 + (20 - 1) x 10
= 20
∴ 20th term = 200.

We have Area of the quadrilateral

=\(\frac { 1 }{ 2 } \) [(-12 - 30 - 28 -10) - (+ 10 + 28 + 30 + 12)]
\(\frac { 1 }{ 2 } \) [-80 - (80)]
\(\frac { 1 }{ 2 } \)[-160] = -80 = 80 square units.
(∵ Area is always +ve).

\(\cfrac { 4f(-3)+2f(4) }{ f(-6)-3f(1) } \)
For f(1) & f(3)
Since 1 & 3 lies between -5 ≤ x ≤2, We take, f(x) = x +5
f(1) = 1 + 5 = 6
f(-3) = -3 +5 = 2
For f(4): Since 4 lies between 2 < x < 6 We take, f(x) = x -1
f(4) = 4 -1 = 3
For f(-6): Since -6 lies between -7≤ x < -5 ,We take, f(x) = x² + 2x +1
f(-6) = (-6)² + 2(-6) + 1 = 36 -12 + 1 = 24 +1 = 25
Now ,\(\cfrac { 4f(-3)+2f(4) }{ f(-6)-3f(1) } \) =\(\frac{ 4(2) +2(3) }{ 25 - 3(6)}\)
= \(\frac{ 8 + 6 }{ 25 -18}\)
= \(\frac{14 }{ 7}\) = 2
Hence, the value of \(\cfrac { 4f(-3)+2f(4) }{ f(-6)-3f(1) } \) = 2

It is both one-one and onto function

All the elements in A have their separate images in B. All the elements in B have their preimage in A. Therefore it is one-one and onto function.

Given that intercepts are equal.
a = b
Equation of the line in intercepts form is
\(\frac{x}{a}+\frac{y}{b}=1 \)
\(\frac{x}{a}+\frac{y}{a}=1 \)
x + y = a
This passes through (- 8, 4)
-8 + 4 = a
a = -4
b = -4
Equation of the straight line is
\(\frac{x}{a}+\frac{y}{b}=1 \)
\(\frac{x}{-4}+\frac{y}{-4}=1 \)
x + y + 4 = 0.

King spade, Queen spade, Jack spade are removed
∴ total number of cards = 52 - 3 = 49.
(i) Probability (diamond)= \(\frac { 13 }{ 49 } \)
(ii) Probability (queen) \(\frac { 4-1 }{ 49 } =\frac { 3 }{ 49 } \)
(iii) Probability (spade) = \(\frac { 13-3 }{ 49 } =\frac { 10 }{ 49 } \)
(iv) Probability (heart bearing number 5) = \(\frac { 13-5 }{ 49 } =\frac { 8 }{ 49 } \)

When we roll two dice, the sample space is given by
S = \(\{ (1,1),(1,2),(1,3),(1,4),(1,5),(1,6)\\ (2,1),(2,2),(2,3),(2,4),(2,5),(2,6)\\ (3,1),(3,2),(3,3),(3,4),(3,5),(3,6)\\ (4,1),(4,2),(4,3),(4,4),(4,5),(4,6)\\ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6)\\ (6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\} \)
n(S) = 36
(i) Let A be the event of getting the sum of outcome values equal to 4.
Then A = {(1, 3),(2, 2),(3, 1)}; n(A) = 3.
Probability of getting the sum of outcomes equal to 4 is  P(A) = \(\frac { n(A) }{ n(S) } =\frac { 3 }{ 36 } =\frac { 1 }{ 12 } \)
(ii) Let B be the event of getting the sum of outcome values greater than 10.
Then B = {(5,6),(6,5),(6,6)}; n(B) = 3
Probability of getting the sum of outcomes greater than 10 is P(B) = \(\frac { n(B) }{ n(S) } =\frac { 3 }{ 36 } =\frac { 1 }{ 12 } \)
(iii) Let C be the event of getting the sum of outcomes less than 13. Here all the outcomes have the sum value less than 13. Hence C = S
Therefore, n(C) = n(S) = 36
Probability of getting the sum value less than 13 is P(C) = \(\frac { n(C) }{ n(S) } =\frac { 36 }{ 36 } \) = 1

In the right triangle GOH, cos 24° = \(\frac { OG }{ GH } \)
0.9135 = \(\frac { OG }{ 250 } \); OG = 228.38 km
Distance of H to the north of G = 228.38 km.

Given AB || CD || EF
AB = 6 cm, BD = 5 cm, EF = 4 cm, CD = x cm, DE = y cm
\(\text { In } \triangle E C D \text { and } \triangle E A B\)
\( \angle C E D=\angle A E B \)            [common]
\(\angle E C D=\angle E A B \)             [corresponding angles]
\(\triangle E C D \sim \triangle E A B\)
[By AA similarity criteria]      ...(1)
\(\therefore \ \frac{E C}{E A}=\frac{C D}{A B}\)
[ Corresponding parts of similar triangles are proportional]
\(\frac{E C}{E A}=\frac{x}{6}\)            ....(2)
In \(\triangle A C D\ and\ \triangle A E F \)
\(\angle C A D=\angle E A F\)          [Common]
\( \angle A C D =\angle A E F \)        [Corresponding angles]
\(\triangle A C D \sim \triangle A E F \)      [By AA similarity]
\(\frac{A C}{A E} =\frac{C D}{E F}\)
[Corresponding parts of similar triangle are proportional]
\(\therefore \quad \frac{A C}{A E}=\frac{x}{4}\)          ....(3)
Adding (2) and (3)
\( \frac{E C}{E A}+\frac{A C}{A E} =\frac{x}{6}+\frac{x}{4} \)
\(\frac{E C+A C}{A E} =\frac{4 x+6 x}{24} \)
\(\frac{A E}{A E} =\frac{10 x}{24} \)
\(1 =\frac{10 x}{24} \)
\(\mathrm{x} =\frac{24}{10}=\frac{12}{5} \mathrm{~cm}\)
From (1) \(\triangle E C D \sim \triangle E A B\)
\( \frac{D C}{A B} =\frac{E D}{E B} \)
\(\frac{x}{6} =\frac{y}{5+y} \)
\(\because \mathrm{x}=\frac{12}{5} \Rightarrow \frac{12 / 5}{6} =\frac{y}{5+y} \)
\(\frac{12}{5 \times 6} =\frac{y}{5+y} \)
12(5 + y) = 30 y
60 + 12y = 30 y
60 = 30y - 12y
18y = 60
\( y=\frac{60}{18} \)
\(y=\frac{10}{3} \mathrm{~cm}\)

Volume of a cone = 1005 \(\frac{5}{7}\) cu.cm
\(\text { i.e., } \frac{1}{3} \pi r^{2} h=1005 \frac{5}{7}\)
area of base = area of circle
\( =201 \frac{1}{7} \text { sq. units } \)
\(i.e \ \pi r^{2}=201 \frac{1}{7} \Rightarrow r^{2}=64 \)
Substituting in (1), r = 8 cm
\(\frac{1}{3}\left(201 \frac{1}{7}\right) \mathrm{h}=1005 \frac{5}{7} \)
\(\frac{1}{3}\left(\frac{1408}{7}\right) \mathrm{h}=\frac{7040}{7} \)
\(h=\frac{7040}{7} \times \frac{7}{1408} \times 3=15 \mathrm{~cm}\)
Slant height of cone \(l =\sqrt{h^{2}+r^{2}} \)
\(=\sqrt{15^{2}+8^{2}}=\sqrt{225+64} \)
\(=\sqrt{289}=17 \mathrm{~cm} \)

Arranging the values in ascending order we get, 4, 7, 8, 10, 11 and n = 5

xi	xi2
4	16
7	49
8	64
10	100
11	121
Σxi = 40	Σxi2 = 350

Standard deviation
σ = \(\sqrt { \frac { { \Sigma d }_{ i }^{ 2 } }{ n } -\left( \frac { { \Sigma d }_{ i } }{ n } \right) ^{ 2 } } \)
= \(\sqrt { \frac { 350 }{ 5 } -\left( \frac { 40 }{ 5 } \right) ^{ 2 } } \)
σ = \(\sqrt { 6 } \) ⋍ 2.45
When we add 3 to all the values, we get the new values as 7, 10, 11, 13, 14.

xi	xi2
7	9
10	100
11	121
13	169
14	196
Σxi = 55	Σxi2 = 635

 Standard deviation
σ = \(\sqrt { \frac { { \Sigma d }_{ i }^{ 2 } }{ n } -\left( \frac { { \Sigma d }_{ i } }{ n } \right) ^{ 2 } } \)
= \(\sqrt { \frac { 635 }{ 5 } -\left( \frac { 55 }{ 5 } \right) ^{ 2 } } \)
σ = \(\sqrt { 6 } \) ⋍ 2.45

Let AB the lighthouse. Let C and D be the positions of the two ships.\(\times \)
Then, AB = 200m.
\(\angle ACB=30°,\angle ADB=45°\)
In right triangles BAC, tan30°= \(\frac { AB }{ Ac } \) 
\(\frac { 1 }{ \sqrt { 3 } } =\frac { 200 }{ AC } \ gives\ AC=200\sqrt { 3 } \)      ...(1)
In the right triangle BAD,tan45°= \(\frac { AB }{ AD } \)
\(1=\frac { 200 }{ AD } \) gives AD = 200     ...(2)
Now, CD = AC + AD = \(200\sqrt { 3 } +200\) [by(1) and (2)]
CD = 200\((\sqrt { 3 } +1)\) = 200 x 2.732 = 546.4
Distance between two ships is 546.4m

\(f(x)=\frac { x-1 }{ x+1 } ,x\neq 0\)
\(f(f(x))=f\left( \frac { x-1 }{ x+1 } \right) =\frac { \left( \frac { x-1 }{ x+1 } \right) -1 }{ \left( \frac { x-1 }{ x+1 } \right) +1 } \)
\(=\frac{\frac{\not x-1-x-1}{(\not x+1)}}{\frac{\not x-1+x+1}{(\not x+1)}}=\frac{-2}{2 x}=\frac{-1}{x}\)
Hence it is proved.

\(\sqrt { 16{ x }^{ 2 }+9{ y }^{ 2 }-24xy+24x-18y+9 } \)
= \(\sqrt { { \left( 4x \right) }^{ 2 }+{ \left( -3y \right) }^{ 2 }+{ \left( 3 \right) }^{ 2 }+2{ \left( 4x \right) }\left( -3y \right) +2\left( -3y \right) \left( 3 \right) +2\left( 4x \right) \left( 3 \right) } \)
= \(\sqrt { { \left( 4x+-3y+3 \right) }^{ 2 } } \) = |4x - 3y + 3|

Let r and h be the radius and height of the cylinder respectively.
Given that, volume of the tank = 1.078 x 10⁶ = 1078000 litre
1078 m³ (since 1l = \(\frac{1}{1000}m^3\))
diameter = 7m gives radius = \(\frac{7}{2}\)m
volume of the tank = \(\pi\)r h 2 cu. units
1078 = \(\frac { 22 }{ 7 } \times \frac { 7 }{ 2 } \times \frac { 7 }{ 2 } \times h\)
Therefore, height of the tank is 28 m

QS = 2 PR
\(\frac { { x }_{ 2 }+1 }{ 2 } =2\Rightarrow { x }_{ 2 }=3\)
\(\frac { { y }_{ 2 }+1 }{ 2 } =2\Rightarrow { y }_{ 2 }=3\)
\(\Rightarrow Q=(3,-3)\)
\(\because QS=2PR\)
\(QS=\sqrt { { (-3-1) }^{ 2 }+{ (3-1) }^{ 2 } } \)
\(=\sqrt { { (-4) }^{ 2 }+{ (2) }^{ 2 } } =\sqrt { 20 } \)
\(\Rightarrow PR=\frac { \sqrt { 20 } }{ 2 } \)
\(\Rightarrow \sqrt { { { (x }_{ 3 }-{ x }_{ 1 }) }^{ 2 }+{ { (y }_{ 3 }-{ y }_{ 1 }) }^{ 2 } } =\frac { \sqrt { 20 } }{ 2 } \)
form (3),
\(\Rightarrow \sqrt { { \left[ 2({ y }_{ 3 }-{ y }_{ 1 }) \right] }^{ 2 }+({ { y }_{ 3 }-{ y }_{ 1 }) }^{ 2 } } =\frac { \sqrt { 20 } }{ 2 } \)
\(\Rightarrow \sqrt { 5{ ({ y }_{ 3 }-{ y }_{ 1 }) }^{ 2 } } =\frac { \sqrt { 20 } }{ 2 } \)
\(\Rightarrow \left(y_{3}-y_{1}\right) \times \sqrt{\not 5}=\frac{\sqrt{\not5} \times \sqrt{4}}{2}\)
\(\Rightarrow { y }_{ 3 }-{ y }_{ 1 }=1\rightarrow (4)\)
\(\therefore { x }_{ 3 }-{ x }_{ 1 }=2\rightarrow (5)\)
∵ Diagram bisect each other at right angle in rhombus, \(\frac { { x }_{ 1 }+{ x }_{ 2 } }{ 2 } =2\) and \(\frac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } =-1\)
\(\left( \frac { { y }_{ 3 }-{ y }_{ 1 } }{ { x }_{ 3 }-{ x }_{ 1 } } \right) \times \left( 3-1\frac { -3-1 }{ 3-1 } \right) =-1\)
\(\{ { m }_{ 1 }\times { m }_{ 2 }=-1\} \)
\(\Rightarrow \frac { { y }_{ 3 }-{ y }_{ 1 } }{ { x }_{ 3 }-{ x }_{ 1 } } =\frac { 1 }{ 2 } \Rightarrow { x }_{ 3 }-{ x }_{ 1 }=2({ y }_{ 3 }-{ y }_{ 1 })\quad ...(3)\)
From (2), y₁ + y₃ = -2
(3), x₁ + x₃ = 4
Solving (4) and (2),
(4) + (2) ⇒ 2y₃ = -1 ⇒ y_{3 =} \(\frac { -1 }{ 2 } \)
(4) - (2) ⇒ -2y₁ = 3 ⇒ y₁= \(\frac { -3 }{ 2 } \)
Solving (5) and (1),
(5) + (1) ⇒ 2x₃ = 6 ⇒ x₃ = 3
(5) - (1) ⇒ -2x₁ = -2 ⇒ x₁ = +1
\(\therefore P=\left( +1,\frac { -3 }{ 2 } \right) \)

From the figure
r = 6 cm
R = 12 cm
h = 8 cm
\(l =\sqrt{h^{2}+(\mathrm{R}-\mathrm{r})^{2}} \)
\(=\sqrt{8^{2}+(12-6)^{2}} \)
\(=\sqrt{64+36} \)
\(=\sqrt{100}=10 \mathrm{~cm} \)
Area to be painted = C.S.A + area of top circular region
\(=\pi(R+r) l+\pi r^{2} \)
\(=\frac{22}{7}(12+6)(10)+\frac{22}{7}(6)^{2} \)
\(=\frac{22}{7}(180)+\frac{22}{7}(36) \)
\(=\frac{22}{7}(180+36) \)
\(=\frac{22}{7}(216)=\frac{4752}{7}=678.86 \)
Cost of painting per sq. cm = Rs. 2
Total cost = 678.86 x 2 = Rs. 1357.72

Given cosec\(\theta \) + cot\(\theta \) = p   ...(1)
cosec²\(\theta \) - cot²\(\theta \) = 1 (identity)
\(\operatorname{cosec} \theta-\cot \theta=\frac{1}{\operatorname{cosec} \theta+\cot \theta}\)
cosec\(\theta \) - cot\(\theta \) =\(\frac { 1 }{ { p } } \)  .... (2)
Adding(1) and (2) we get, 2cosec\(\theta \) = \(p+\frac { 1 }{ p } \) 
2cosec\(\theta \)\(\frac { { p }^{ 2 }+1 }{ p } \) ....(3) 
Subtracting (2) from (1), we get, 2cot\(\theta \) = \(p-\frac { 1 }{ p } \) 
2cot\(\theta \) = \(\frac { { p }^{ 2 }-1 }{ p } \)   ...(4)
Dividing (4) by (3) we get,\(\frac { 2cot\theta }{ 2cosec\theta } =\frac { { p }^{ 2 }-1 }{ p } \times \frac { p }{ { p }^{ 2 }+1 } gives,cos\theta =\frac { { p }^{ 2 }-1 }{ { p }^{ 2 }+1 } \)

Let the 100t digit be 'x', 10's be 'y' and Unit digit be 'z'.
The three digit number is 100x + 10y + z.
Now given, x + y + z = 11 .       ..(1)
100z +10y + x = 5 (100x + 10y + z) + 46
Simplifying
499x + 40y - 95z = - 46          .........(2)
x + 2y = z
x + 2y - z = 0    ....(3)

Substituting the value x = 1 in(4)
2(1) + 3y = 11
3y = 11 - 2 = 9
\(y=\frac{9}{3}=3\)
Substituting x = 1, y = 3 in (1)
1 + 3 + z = 11
z = 11 - 4 = 7
x = 1, y = 3, z = 7
The original three digit number is 137
i.e., 100(1) + 10(3) + 1(7) = 100+ 30 + 7 =137

Let AP and CR be the vertical poles of height 6 m and 3 m respectively
Let BQ = ym 
In CRA and BQA
o
By AA criterion of similarity
\(\triangle C R A \sim \triangle B Q A\)
Their corresponding sides are Proportional
\(\frac{A C}{A B}=\frac{C R}{B Q} \)
\(\frac{A C}{A B}=\frac{3}{y} \)
\(\mathrm{AB}=\frac{A C \times y}{3}\)
In CBQ and CAP
o
\(\therefore \triangle C B Q \sim \triangle C A P \)
\(\frac{C B}{C A}=\frac{B Q}{A P} \)
\(\frac{C B}{C A}=\frac{y}{6} \)
\(B C=\frac{y \times C A}{6} \)
\((1)+(2) \Rightarrow A B+B C =\frac{A C \times y}{3}+\frac{y \times C A}{6} \)
\(A C=y \times A C\left(\frac{1}{3}+\frac{1}{6}\right) \)
\(\frac{A C}{A C} =y\left(\frac{2+1}{6}\right) \)
\(1 =y\left(\frac{3}{6}\right) \)
\(1 =\frac{1}{2} y \)
y = 2m.

Construction:
Step (1) Draw a line segment PQ = 4.5 cm
Step (2) At P, draw PE such that \(\angle QPE={ 35 }^{ 0 }\)
Step (3) At P, draw PF such that \(\angle EPF={ 90 }^{ 0 }\)
Step (4) Draw \(\bot \) bisector to PQ which intersects PF at O.
Step (5) With O centre OP as radius draw a circle.
Step (6) From G, marked arcs of radius 6 cm on the circle marked them as R and S.
Step (7) Joined PR and RQ. Then \(\triangle\)PQR is the required triangle
Step (8) \(\triangle\)PQS is the required triangle

Given a triangle LMN, we are required to construct another triangle whose sides are \(\frac { 4 }{ 5 } \) of the corresponding sides of the \(\triangle\)LMN

Steps of construction:
1. Constructed a LMN with any measurement
2. Drawn a ray MX making an acute angle with MN on the side opposite to the vertex L.
3. Located 5 points M₁, M₂, M₃, M₄, M₅ on MX so that
MM₁ = M₁M₂ = M₂M₃ = M₃M₄ = M₄M₅
4. Joined, M₅N and drawn a line through M₄ parallel to M₅N to intersect MN at N.
5. Drawn a line through N, parallel to the line NL to intersect ML at L
Then L'MN' is the required triangle each of whose sides is four-fifth of the corresponding sides of LMN