(i) To check if h is one – one, we assume that h(b₁) = h(b₂)
Then we get 2.47b₁ + 54.10 = 2.47 b₂ + 54.10
2.47b₁ = 2.47b₂ ⇒ b₁ = b₂
Thus, h(b₁) = h(b₂) ⇒ b₁ = b₂, So, the function h is one – one.
(ii) If the length of the thigh bone b = 50, then the height is
h(50) = (2.47 x 50) + 54.10 = 177.6 cms
(iii) If the height of a person is 147.96 cms, then h(b) =147.96 and so the length of the thigh bone is given by 2.47b + 54.10 = 147.96.
b = \(\frac { 93.86 }{ 2.47 } \)
Therefore, the length of the thigh bone is 38 cms.

An arrow diagram
 

A table

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

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)}

A.P. is given by a + d, a + 2d, a + 3d,.....
In this case \(\frac{3}{4}, \frac{3}{4}+\frac{1}{2}, \frac{3}{4}+2\left(\frac{1}{2}\right), \frac{3}{4}+3\left(\frac{1}{2}\right), \ldots\)
\(\frac{3}{4}, \frac{3+2}{4}, \frac{3+4}{4}, \frac{3+6}{4}, \ldots \)
\(\frac{3}{4}, \frac{5}{4}, \frac{7}{4}, \frac{9}{4}, \ldots \)
The required A.P. is \(\frac{3}{4}, \frac{5}{4}, \frac{7}{4}, \frac{9}{4}, \ldots \)

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

\(\frac { 12{ t }^{ 2 }-22t+8 }{ 3t } \div \frac { 3{ t }^{ 2 }+2t-8 }{ 2{ t }^{ 2 }+4t } \)
\(=\frac { { 12 }^{ 2 }-2t+8 }{ 3t } \times \frac { 2{ t }^{ 2 }+4t }{ { 3t }^{ 2 }+2t-8 } \)

\(\frac { 4(2t-1) }{ 3 } \)

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

P(A) = 0.5
P(B) =  0.3
Since A and B are mutually exclusive events
\(P(A \cap B)=0\)
\( \mathrm{P}(\text { either } \mathrm{A} \text { or } \mathrm{B}) =\mathrm{P}(A \cup B) \)
\(\mathrm{P}(A \cup B) =\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(A \cap B) \)
\(\mathrm{P}(A \cup B) =0.5+0.3-0=0.8\)
P(neither A nor B) = \(\mathrm{P}(\overline{A \cup B})\)
\(\mathrm{P} \overline{(A \cup B)}=1-\mathrm{P}(A \cup B)=1-0.8=0.2\) 
Probability of neither A nor B happen = 0.2

Since A and B are mutually exclusive events
\(\mathrm{P}(A \cap B)=0\)
P(not A) = 0.45
P(A) = 1 - P(not A)
P(A) = 1 - 0.45 = 0.55
\(\mathrm{P}(A \cup B)=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(A \cap B)\)
0.65 = 0.55 + P(B) - 0
P(B) = 0.65 - 0.55 = 0.1

The standard deviation of a given data is 4.5. If we subtract some fixed constant from all the data, the standard deviation will not change.
Each value of the data decreased by 5, the new standard deviation will not change.
New standard deviation = 4.5

x² - 13x + k = 0 here, a = 1, b = -13, c = k
Let α, β be the roots of the equation. Then
α + β = \(\frac {-b}{a} = \frac {-(-13)}{1}\) = 13  ...... (1) also α - β = 17 ......(2)
(1) + (2) we get, 2α = 30 gives α = 15
Therefore, 15 + β = 13 (from (1)) gives β = -2
But, αβ = \(\frac {c}{a} = \frac {k}{1}\) gives 15 x (-2) = k we get, k = -30

Let AB = x be the distance between foot of the house and the observer at the median of the road.
DB = 2x is the width of the road.
Height of the house BC = \(4 \sqrt{3} m\)
From the right triangle \(\triangle\) ABC
\( \therefore \tan 30^{\circ} =\frac{B C}{A B} \)
\(\frac{1}{\sqrt{3}}=\frac{4 \sqrt{3}}{x} \)
\(x =4 \sqrt{3} \times \sqrt{3}=4 \times 3=12 \mathrm{~m} \)
\(\text { Width of the road } =2 \times x=2 \times 12=24 \mathrm{~m}\)
Width of the road = 24 m.

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

Since A x A has 9 elements,
A would have 3 elements 
A x A contains (- 1, 0) and (0, 1)
-1, 0 \(\in \) A
Similarly (0, 1) is in A x A
So, 0,1 \(\in \) A
From (1) and (2). = -1, 0, 1 \(\in \) A
A = {-1, 0 , 1}
A x A = {-1, 0, 1} x {1, 0, -1}
= {(-1, 1), (-1, 0), (-1, -1), (0, -1),(0,0), (0, 1), (1, -1), (1, 0),(1, 1)}
The remaining elements of A x A are 
{(- 1, - 1), (- 1, 1), (0, - 1), (0, 0), (1, - 1), (1,0),(1, 1)}

Let r₁, r₂ be the radii of two spheres
\(\text { Given } \frac{r_{1}}{r_{2}}=\frac{4}{7} \Rightarrow \mathrm{r}_{1}=\frac{4 r_{2}}{7}\)
Ratio of the volumes = \(\frac{V_{1}}{V_{2}}=\frac{\frac{4}{3} \pi r_{1}^{3}}{\frac{4}{3} \pi r_{2}^{3}}\)
\(=\frac{\left(\frac{4 r_{2}}{7}\right)^{3}}{r_{2}^{3}}=\frac{4^{3}}{7^{3}}\)
Ratio of volumes \(V_{1}: V_{2}=64: 343=\frac{64}{343}\)

Given, (x₁, y₁) = (3 , − 4) and m = \(\frac { -5 }{ 7 } \)
The equation of the point-slope form of the straight line is y - y₁ = m(x - x₁)
we write it as y + 4 = − \(\frac { 5 }{ 7 } \) (x - 3)
gives us 5x + 7y + 13 = 0

Let r and h be the radius and height of the cone respectively.
Given that, volume of the cone = 11088 cm³
\(\frac { 1 }{ 3 } { \pi r }^{ 2 }h=11088\)
\(\frac { 1 }{ 3 } \times \frac { 22 }{ 7 } \times { r }^{ 2 }\times 24=11088\)
\({ r }^{ 2 }=441\)
Therefore, radius of the cone r = 21 cm.

Distance between the insect and the foot of the lamp post BD = 8 m
The height of the lamp post, AB = 6 m
After moving a distance of x m, let the insect be at C
Let, AC = CD = x . Then BC = BD − CD = 8 − x
In \(\triangle\)ABC, \(\angle\)B = 90^o
AC² = AB² + BC² gives x² = 62 + (8 - x)2
x² = 36 + 64 − 16x + x²
16x = 100 then x = 6.25
Then, BC = 8 − x = 8 − 6.25 = 1.75m
Therefore the insect is 1.75 m away from the foot of the lamp post.

The equation of any straight line parallel to X axis is y = b.
Since it passes through (5, 7), b = 7 .
Therefore, the required equation of the line is y = 7.

Let r and h be the radius and height of the cone respectively.
Given that, radius r = 7 m and height h = 24 m
Hence, l = \(\sqrt { { r }^{ 2 }+{ h }^{ 2 } } \)
\(=\sqrt { 49+576 } \)
\(l=\sqrt { 625 } =25m\)
C.S.A. of the conical tent = \(\pi\)rl sq. units
Area of the canvas \(=\frac { 22 }{ 7 } \times 7\times 25={ 550 }m^{ 2 }\)
Now, length of the canvas \(\frac{Area\ of\ the\ canvas}{width}=\frac{550}{4}=137.5m\)
Therefore, the length of the canvas is 137.5 m

We have AB = 56.cm, AD = 14. cm, AC = 72. cm and AE = 18.cm.
BD = AB - AD = 5.6 –1.4 = 4.2 cm
and EC = AC – AE = 7.2–1.8 = 5.4 cm
\(\frac { AD }{ DB } =\frac { 1.4 }{ 4.2 } =\frac { 1 }{ 3 } \) and \(\frac { AE }{ EC } =\frac { 1.8 }{ 5.4 } =\frac { 1 }{ 3 } \)
\(\frac { AD }{ DB } =\frac { AE }{ EC } \)
Therefore, by converse of Basic Proportionality Theorem, we have DE is parallel to BC. Hence proved.

Given first two terms of the sequence.
a₁ = 1, a₂ = 1,
For n > 3 a_n = 2a_{n - 1} + a_{n - 2} 
When n = 3 a₃ = 2a_{(3 - 1)} + a_{(3 - 2)} = 2a₂ + a₁
a₃ = 2 + 1 = 3
When n = 4 a₄ = 2a_(4-1) + a_(4-2) = 2a₃ + a₁
a₄ = 2 x 3 + 1 = 6 + 1 = 7
When n = 5 a₅ = 2a_(5-1) + a_(5-2) = 2a₄ + a₃
a₅ = 2 x 7 + 3 = 14 + 3 = 17
When n = 6 a₆ = 2a_(6-1) + a_(6-2) = 2a₅ + a₄
a₆ = 2 x 17+ 7 = 34 + 7 = 41
The first six terms of the sequence are 1, 1, 3, 7, 17, 41, ....

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

           

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 \(=10\times 9+\frac { 18 }{ 9 } =92\)

Let the length of the side of the chess board be x cm. Then
Area of 64 squares = (x - 4)²
(x - 4)² = 64 x 6.25
⇒ x²-8x+ 16=400
⇒X²- 8x - 384 =0
⇒ X²- 24x + 16x - 384 = 0
⇒(x - 24)(x + 16) = 0
⇒ x=24 cm.

a_n = 3 +2n
a₁ = 3 + 2 = 5
a₂ = 3 + 2 x 2 = 7
a₃ = 3 + 2 x 3 = 9
List of numbers becomes 5, 7, 9, 11,.........
Here, 7 - 5 = 9 - 7 = 11 - 9 = 2 and so on. So, it forms an A.P. with common difference d = 2.
To find S₂₄ we have n = 24, a = 5, d = 2.
S₂₄ = \(\frac{24}{2}\) [2 x 5 + (24 - 1) x ]
= 12 [10 + 46] = 672.
So, sum of first 24 terms of the list of numbers is 672.

The area of the triangle formed by the given points is equal to
= \(\frac { 1 }{ 2 } \) [-1.5 (-2 - 4) + 6 (4 - 3) + (-3) (3 + 2)]
= \(\frac { 1 }{ 2 } \) [9 + 6 - 15] = 0
We can have a triangle at area 0 square units? What does this mean?
If the area of a triangle is 0 square units, then its vertices will be collinear.

f(x) = 1 + x
g(x) = (2x - 1)
fog(x) = f(g(x) =  f(2x - 1)
= 1 + 2x - 1 = 2x ...(1)
gof(x) = g(f(x) = g(1 + x) = 2(1 + x) - 1
= 2 + 2x - 1
= 2x + 1      ...(2)
(1) \(\neq \)(2)
\(\therefore\) fog(x) \(\neq \) gof(x)
It is verified

\(\cfrac { 4f(-3)+2f(4) }{ f(-6)-3f(1) } \)
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

4, 10, 16,22, ...
We have a² - a² = 10 - 4 = 6
a³ - a² = 16 -10 = 6
ஃ It is an A.P. with common difference 6
ஃ The next two terms are, 

\(f(x)=\frac { 1{ x }^{ 2 } }{ a } -\frac { 2x }{ b } +\frac { 3 }{ c } \)
Sum of the roots \((\alpha +\beta )=\frac { -b }{ a } =-\frac { (-2) }{ 1 } \)
Product of the roots \((\alpha \beta )=\frac { c }{ a } =\frac { 3 }{ 1 } =3\)
\(\frac { \alpha -1 }{ \alpha +1 } ,\frac { \beta -1 }{ \beta +1 } \)
⇒ sum of the roots=\(\frac { \alpha -1 }{ \alpha +1 } ,\frac { \beta -1 }{ \beta +1 } \)
Sum\(=\frac { (\alpha -1)(\beta +1)+(\beta -1)(\alpha +1) }{ (\alpha +1)(\beta +1) } \)

\(=\frac { \alpha \beta +\alpha \beta -2 }{ \alpha \beta +(\alpha +\beta )+1 } =\frac { 2\alpha \beta -2 }{ \alpha \beta +(\alpha +\beta )+1 } \)

Product=\(\frac { \alpha -1 }{ \alpha +1 } \times \frac { \beta -1 }{ \beta +1 } =\frac { (\alpha -1)(\beta -1) }{ (\alpha +1)(\beta +1) } \)
\(=\frac { \alpha \beta -\beta -\alpha +1 }{ \alpha \beta ++\alpha +1 } =\frac { \alpha \beta -(\alpha +\beta )+1 }{ \alpha \beta +(\alpha +\beta )+1 } \)
\(=\frac { 3-2+1 }{ 3+2+1 } \)
\(=\frac { 2 }{ 6 } =\frac { 1 }{ 3 } \)
∴ Required equation= \({ x }^{ 2 }-\frac { 2 }{ 3 } x+\frac { 1 }{ 3 } =0\)
⇒3x² - 2x + 1 = 0

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.

Product of face values 6: { (1, 6), (2, 3), (6, 1), (3,2)}
Difference of face value 5: {(1, 6), (6, 1)}
P(product 6) = \(\frac { 4 }{ 6\times 6 } =\frac { 4 }{ 36 } =\frac { 1 }{ 9 } \)
p( difference 5) = \(\frac { 2 }{ 2\times 6 } =\frac { 1 }{ 18 } \)

Total number of balls = blue balls + red balls
n(S) = 12 + x
(i) The probability that it will be a red ball:
Let A be the event of selecting a red ball
n(A) = x
\(\mathrm{P}(\mathrm{A})=\frac{n(A)}{n(S)} \)
\(\mathrm{P}(\mathrm{A})=\frac{x}{12+x} \)
(ii) If 8 more red balls are put in the bag then the number of red balls = x + 8
Total number of balls = 12 + x + 8
Probability of drawing a red ball
\(\mathrm{P}(\mathrm{B}) =\frac{n(B)}{n(S)} \)
\(=\frac{x+8}{12+x+8}=\frac{x+8}{x+20} \)
If the probability of drawing a red ball will be twice that of the probability (i), then
\(\frac{x+8}{x+20}=2 \times\left[\frac{x}{12+x}\right]\)
(x + 8)(12 + x) = 2x (x + 20)
12x + 96 + x² + 8x = 2x² + 40x
2x²  - x² + 40x - 12x - 8x - 96 = 0
x² + 20x - 96 = 0
(x - 4) (x + 24) = 0
x = 4 and x = - 24 (not possible)
By applying the value of x in P(A)
we get P(A) \(=\frac{4}{12+4}=\frac{4}{16}\)
\(P(A)=\frac{1}{4}\)
x = 4

Let CD is the tree of height 13 m.
AB is another tree.
\( \angle A D E=45^{\circ} \)
\(\angle E D B=\angle D B C=30^{\circ}\)
CB = DE and CD = EB = 13m
In right triangle \(\triangle\)AED
\( \tan 45^{\circ} =\frac{A E}{D E} \)
\(1 =\frac{A E}{D E}\)
AE = DE        ....(1)
In the right triangle \(\triangle\)DBC
\( \tan 30^{\circ} =\frac{D C}{B C} \)
\(\frac{1}{\sqrt{3}} =\frac{13}{B C} \)
\(B C =13 \sqrt{3} \mathrm{~m}\)
From (1) AE \( =13 \sqrt{3} \mathrm{~m}\)
Height of the tree = AB = AE + EB
(\(13 \sqrt{3} \mathrm{~m}\) + 13) m
= (13 x 1.732 + 13) m
= (22.516 + 13) m = 35.516 m - 35.52 m
Height of the second tree = 35.52 m

Given lines 4x + 7y - 3 = 0 = 4x + 7y = 3         ..... (1)
2x - 3y + 1 = 0 = 2x - 3y = -1       ...(2)
Solving (1) and (2)

\(y=\frac{5}{13} \)
Substituting in (1) \(\Rightarrow \ 4 x+7\left(\frac{5}{13}\right)=3 \)
\(x=\frac{1}{13}\)
The point intersection of (1) and (2) is \(\left(\frac{1}{13}, \frac{5}{13}\right)\)
Given that the line has equal intercepts on the axes.
i.e., a = b
Now, intercept form of the line is \(\frac{x}{a}+\frac{y}{b}=1\)
\(\Rightarrow \quad \frac{x}{a}+\frac{y}{a}=1 \Rightarrow \mathrm{x}+\mathrm{y}=\mathrm{a}\)
This passes through \(\left(\frac{1}{13}, \frac{5}{13}\right)\)
\( \Rightarrow \ \frac{1}{13}+\frac{5}{13} =a \Rightarrow a=\frac{6}{13} \)
\(\therefore \ b =\frac{6}{13} \)
Hence the equation of a line is \(\frac{x}{\frac{6}{13}}+\frac{y}{\frac{6}{13}}=1\)
13x + 13y - 6 = 0

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.

As given there are x² swans.
As per the given data x² - 10x - \(\frac {1}{8}\)x² = 6 we get, 7x² - 80x - 48 = 0
x = \(\frac { 80\pm \sqrt { 6400-4\left( 7 \right) \left( -48 \right) } }{ 14 } =\frac { 80\pm 88 }{ 14 } \)
Therefore, x = 12, \(-\frac {4}{7}\)
Here x = \(-\frac {4}{7}\) is not possible as the number of swans cannot be negative.
Hence, x = 12. Therefore total number of swans is x² = 144.

Distance d = t²- 0.75t,
Given that d = 11.25 = t²- 0.75t.
\(t = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
\(=\frac { (+0.75)\pm \sqrt { { (-0.75) }^{ 2 }-4\times 1\times -11.25 } }{ 2\times 1 } \)
\(=\frac { +0.75\pm \sqrt { 0.5625+45 } }{ 2 } \)
\(=\frac { +0.75\pm \sqrt { 45.5625 } }{ 2 } \)
\(=\frac { +0.75\pm 6.75 }{ 2 } \)
\(=\frac { 7.50 }{ 2 } or\frac { -6 }{ 2 } \)
= 3.75 or-3 It is not possible
∴ t = 3.75 s.

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 \(\frac{\cos \theta}{1+\sin \theta}=\frac{1}{a} \)
\(\therefore a=\frac{1+\sin \theta}{\cos \theta} \)
\(\mathrm{LHS}=\frac{a^{2}-1}{a^{2}+1} \)
\(=\frac{\left(\frac{1+\sin \theta}{\cos \theta}\right)^{2}-1}{\left(\frac{1+\sin \theta}{\cos \theta}\right)^{2}+1}\)
\(=\frac{\frac{1^{2}+\sin ^{2} \theta+2 \sin \theta}{\cos ^{2} \theta}-1}{\frac{1^{2}+\sin ^{2} \theta+2 \sin \theta}{\cos ^{2} \theta}+1}\)
\(=\frac{\frac{1+\sin ^{2} \theta+2 \sin \theta-\cos ^{2} \theta}{\cos ^{2} \theta}}{\frac{1+\sin ^{2} \theta+2 \sin \theta+\cos ^{2} \theta}{\cos ^{2} \theta}}\)
\(=\frac{\left(1-\cos ^{2} \theta\right)+\sin ^{2} \theta+2 \sin \theta}{\cos ^{2} \theta} \times \frac{\cos ^{2} \theta}{1+\left(\sin ^{2} \theta+\cos ^{2} \theta\right)+2 \sin \theta}\)
\(=\frac{\sin ^{2} \theta+\sin ^{2} \theta+2 \sin \theta}{1+1+2 \sin \theta}\)
\(=\frac{2 \sin ^{2} \theta+2 \sin \theta}{2+2 \sin \theta} \)
\(=\frac{2 \sin \theta(\sin \theta+1)}{2(1+\sin \theta)} \)
= sin \(\theta \) = RHS

A = {1, 2),B = {1, 2, 3, 4}, C = {5, 6},D = {5, 6, 7, 8}
A x C = {1, 2} x {5, 6}
\(\mathrm{A} \times \mathrm{C}=\{(1,5),(1,6),(2,5),(2,6)\}\)
B x D = { 1, 2, 3, 4} x { 5, 6, 7, 8}
\(\begin{array}{r} \mathrm{B} \times \mathrm{D}=\{({1,5}),(1,6),(1,7),(1,8) ({2,5}),(2,6),(2,7),(2,8) (3,5),(3,6),(3,7),(3,8) (4,5),(4,6),(4,7),(4,8)\} \end{array}\)
(A x C) is a subset of (B x D)

Right triangle PQR, right angled at Q and
PR = 20 cm, QR = 16 cm
PQ² = PR² - QR²
= (20)² - (16)²
= 400 - 256 = 144
PQ = 12 cm

When right triangle PQR, rotates about QR, a right circular cone is formed with PQ = 12 cm as base radius and PR = 20 cm as
slant height.
C.S.A of the Cone = \(\pi r l\) sq. units
\(=\frac{22}{7} \times 12 \times 20=754.29 \mathrm{~cm}^{2}\)
When right triangle PQR, rotates about PQR, a right circular cone is formed with
QR = 16 cm as base radius and PR = 20 cm as slant height

C.S.A of the Cone \(=\pi r l \text { sq.units } \)
\(=\frac{22}{7} \times 16 \times 20 \)
= 1005.71 cm²
Hence, C.S.A of the cone when rotates about PQ is larger.

In \(\triangle\) ABC we have DE || BC.
By Thales theorem, we have \(\frac { AD }{ DB } =\frac { AE }{ EC } \)
\(\frac { x }{ x-2 } =\frac { x+2 }{ x-1 } \) gives x(x - 1) = (x - 2)(x + 2)
When x = 4, AD = 4, DB = x - 2, AE + x + 2 = 6, EC = x - 1 = 3 
Hence, AB = AD + DB = 4 + 2 = 6, AC = AE + EC = 6 + 3 = 9
Therefore, AB = 6, AC = 9

Given, radius = 4 cm
Construction
Step 1 : With O as the centre, draw a circle of radius 4 cm.
Step 2 : Take a point L on the circle. Through L draw any chord LM.
Step 3 : Take a point M distinct from L and N on the circle, so that L, M and N are in anti clockwise direction. Join LN and NM.
Step 4 : Through L draw a tangent TT' such that \(\angle\)TLM =\(\angle\)MNL
Step 5 : TT' is the required tangent.

Construction:
Steps (1) Draw a line segment BC = 5.6 cm
Steps (2) At B, draw BE such that \(\angle CBE={ 60 }^{ 0 }\)
Steps (3) At B draw BF such that \(\angle EBF={ 90 }^{ 0 }\)
Steps (4) Drawn the perpendicular bisector to BC, which intersects BF at O and BC at G.
Steps (5) With O as centre and OB as radius draw a circle
Steps (6) From B, marked an arc of 4 cm on BC at D.
Steps (7) The perpendicular bisector intersects the circle at I. Joined ID.
Steps (8) ID produced meets the circle at A. Now joined AB and AC. Then \(\triangle\)ABC is the required triangle.