An arrow diagram
 

It is not a function as the vertical line PQ cuts the graph at two points

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)}

4 x 4

\(\sqrt { 2 } { x }^{ 2 }+7x+5\sqrt { 2 } =0\) 
\(\sqrt{2} x^{2}+2 x+5 x+5 \sqrt{2}=0\)
\( \sqrt{2} x^{2}+\sqrt{2} \sqrt{2} x+5 x+5 \sqrt{2} =0 \)
\(\sqrt{2} x(x+\sqrt{2})+5(x+\sqrt{2}) =0 \)
\((x+\sqrt{2})(\sqrt{2} x+5) =0 \)
\( x+\sqrt{2} =0 \)                   \( \sqrt{2 x+5} =0 \)
\(x =-\sqrt{2}\)                             \(x =-\frac{5}{\sqrt{2}}\)
Solution is x \(=-\sqrt{2},-\frac{5}{\sqrt{2}}\)

\(\frac { 1 }{ 2 } \), 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.

4x + 3y + 12 = 0
4x + 3y = -12
Dividing by - 12
\(\Rightarrow \ \frac{4 x}{-12}+\frac{3 y}{-12} =1 \)
\(\Rightarrow \ \frac{x}{(-3)}+\frac{y}{(-4)} =1 \)
\(\frac{x}{a}+\frac{y}{b} =1 \) 
x intercept = a = -3
y intercept = b = -4.

Again in \(\triangle A B C, \angle A E D=\angle A E B=90^{\circ}\)
By Pythagoras theorem.
AB² = AE² + EB²
= AD² - DE² + (BD - DE)²
= AD² - DE² + BD² + DE² -2BD.DE
= AD² + BD² - 2BD.DE
\( \mathrm{AB}^{2} =\mathrm{AD}^{2}+\left(\frac{1}{2} B C\right)^{2}-2 \cdot \frac{1}{2} \mathrm{BC} \cdot \mathrm{DE} \)
\(\mathrm{AB}^{2} =\mathrm{AD}^{2}+\frac{1}{4} B C^{2}-\mathrm{BC} \cdot \mathrm{DE} \)
\(\mathrm{AB}^{2} =\mathrm{AD}^{2}-\mathrm{BC} \cdot \mathrm{DE}+\frac{1}{4} B C^{2} \ldots(2) \)
\(\mathrm{c}^{2} =\mathrm{p}^{2}-\mathrm{ax}+\frac{a^{2}}{4}\)  

We see that the numerators of n^th term is n - 1 and the denominators of n^th term is n.
\(a_{n}=\frac{n-1}{n}\)

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 of first n natural numbers
\(=\sqrt{\frac{n^{2}-1}{12}}\)
SD of first 21 natural numbers
\( =\sqrt{\frac{21^{2}-1}{12}} \)
\(=\sqrt{\frac{441-1}{12}}=\sqrt{\frac{440}{12}} \)
\(=\sqrt{36.6666}=6.05\)
Standard deviation of first 21 natural numbers = 6.05

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.

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

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

f o f(k) = f(f(k))
= 2(2k - 1) -1 = 4k - 3
Thus, f o f(k) = 4k - 3
But, it is given that f o f(k) = 5
Therefore 4k - 3 = 5 ⇒ k = 2

Given that, radius r = 7 cm
Now, total surface area of the cone = \(\pi\)r(l + r)sq. units
T.S.A = 704 cm²
704 = \(\frac{22}{7}\times7(l+7)\)
32 = l + 7 implies l = 25 cm
Therefore, slant height of the cone is 25 cm.

Given that, height of the cylinder h = 20 cm ; radius r =14 cm
Now, C.S.A. of the cylinder = 2p\(\pi\)h sq. units
C.S.A. of the cylinder = \(2\times \frac { 22 }{ 7 } \times 14\times 20=2\times 22\times 2\times 20\)
T.S.A. of the cylinder \(=2\pi r(h+r)\)sq.units
\(=2\times \frac { 22 }{ 7 } \times 14\times (20+14)=2\times \frac { 22 }{ 7 } \times 14\times 34\)
= 2992 cm²
Therefore, C.S.A. = 1760 cm² and T.S.A. = 2992 cm²

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.

In \(\Delta BAC\) and \(\Delta PRQ,\quad \frac { AB }{ RQ } =\frac { 3 }{ 6 } =\frac { 1 }{ 2 } \)
\(\frac { BC }{ QP } =\frac { 6 }{ 12 } =\frac { 1 }{ 2 } ;\frac { CA }{ PR } =\frac { 3\sqrt { 3 } }{ 6\sqrt { 3 } } =\frac { 1 }{ 2 } \)
Therefore, \(\frac { AB }{ QP } =\frac { BC }{ QP } =\frac { CA }{ PR } \)
By SSS similarity, we have \(\triangle\)BAC~\(\triangle\)QRB
\(\angle\)P =\(\angle\)C (since the corresponding parts of similar triangle)
\(\angle\)P =\(\angle\)C 180^o -\((\angle A+\angle B)={ 180 }^{ 0 }-({ 90 }^{ 0 }+{ 60 }^{ 0 })\)
\(\angle\)P = 180^o - 150^o = 30^o

      

cot 85° + cos 75°
= cot(90° - 5°) + cos(90° - 15°)
= tan 5° + sin 15°

Radius of the cylinder be r
Height of the cylinder be h
Total surface area of the article
= CSA of cylinder + CSA of 2 hemispheres
= 2ㅠrh + 2πr² = 2πr(h + 2r)
\(=2\times \frac { 22 }{ 7 } \times 3.5\times (10+2\times 3.5)\)
= 22 x 17 = 374 cm²

Team A
x1	d1 = x - 28	d12
50	22	484
20	-8	64
10	-18	324
30	2	4
30	2	4
140	Σd = 0	880

\(\bar { { x }_{ 1 } } \) =\(\frac { 140 }{ 5 } \) =28
σ₁ =\(\sqrt { \frac { \Sigma d^{ 2 } }{ n } } =\sqrt { \frac { 880 }{ 5 } } \)
= \(\sqrt { 176 } \)
= 13.27
CV₁ = \(\frac { \sigma }{ \bar { x_{ 1 } } } \) x 100
CV₁ = \(\frac { 13.27 }{ 28 } \) x 100
= 47.39%
CV₁ < CV₂

Team B
x2	d2 = x - 28	d2
40	10	100
60	30	900
20	-10	100
20	-10	100
10	-20	400
150	Σd = 0	1600

\(\bar { { x }_{ 2 } } =\frac { 150 }{ 5 } \) =30
σ₂ =\(\sqrt { \frac { \Sigma d^{ 2 } }{ n } } =\sqrt { \frac { 1600 }{ 5 } } \)
= \(\sqrt { 320 } \)
 =17.89
CV₁ = \(\frac { { \sigma }_{ 2 } }{ \bar { { x }_{ 2 } } } \) x 100
CV₂  = \(\frac { 17.89 }{ 30 } \) x 100
= 59.63%
∴ Team A is more consistent.

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.

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.

The number of rose plants in the 1st, 2nd, 3rd, . . . rows are
23,21, 19, ... 5
It forms an A.P.
Let the number of rows in the flower bed be n.
Then a = 23, d = 21 - 23 = -2/a = 5.
As, a_n = a + (n - 1)d i.e. t_n = a + (n - 1)d
We have 5 = 23 + (n - 1)(-2)
i.e. -18 = (n - 1)(-2)
n = 10
ஃ There are 10 rows in the flower bed.

If A(-2, -1), B(a, 0), C(4, b) and D(1, 2) are the vertices of a parallelogram, find the values of a and b.
We know that the diagonals of a parallelogram bisect each other. Therefore the co-ordinates of the midpoint of AC are same as the co-ordinates of the mid-point of BD. i.e.
\(\left( \frac { -2+4 }{ 2 } ,\frac { -1+b }{ 2 } \right) =\left( \frac { a+1 }{ 2 } ,\frac { 0+2 }{ 2 } \right) \)
⇒ \(\left( 1,\frac { b-1 }{ 2 } \right) =\left( \frac { a+1 }{ 2 } ,1 \right) \)
⇒ \(\frac { a+1 }{ 2 } \) = 1 ⇒ a + 1 = 2 ⇒ a = 1
⇒ \(\frac { b-1 }{ 2 } \) = 1 ⇒ b - 1 = 2 ⇒ b = 3

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

\(f(x)=\frac { 1{ x }^{ 2 } }{ a } -\frac { 2x }{ b } +\frac { 3 }{ c } \)
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 sin \(\theta \) (1+ sin² \(\theta \)) = cos² \(\theta \)
Squaring on both the sides.
sin² \(\theta \)(1+ sin² \(\theta \))² = cos⁴ \(\theta \)
\(\Rightarrow\left(1-\cos ^{2} \theta\right)\left\{1+\left(1-\cos ^{2} \theta\right)\right\}^{2}=\cos ^{4} \theta\)
\(\Rightarrow\left(1-\cos ^{2} \theta\right)\left\{1+\left(1-\cos ^{2} \theta\right)^{2}+2(1)\left(1-\cos ^{2} \theta\right)\right\} =\cos ^{4} \theta \)
\(\Rightarrow\left(1-\cos ^{2} \theta\right)\left(1+1+\cos ^{4} \theta-2(1) \cos ^{2} \theta\right. \left.+2-2 \cos ^{2} \theta\right)=\cos ^{4} \theta \)
\(\Rightarrow\left(1-\cos ^{2} \theta\right)\left(4-4 \cos ^{2} \theta+\cos ^{4} \theta\right)=\cos ^{4} \theta \)
\(\Rightarrow 4-4 \cos ^{2} \theta+\cos ^{4} \theta-4 \cos ^{2} \theta+4 \cos ^{4} \theta-\cos ^{6} \theta \ =\cos ^{4} \theta \)
\(\Rightarrow \ 4-8 \cos ^{2} \theta+5 \cos ^{4} \theta-\cos ^{6} \theta=\cos ^{4} \theta \)
\(4-8 \cos ^{2} \theta+4 \cos ^{4} \theta-\cos ^{6} \theta=0 \)
\(4=8 \cos ^{2} \theta-4 \cos ^{4} \theta+\cos ^{6} \theta \)
\(\Rightarrow \cos ^{6} \theta-4 \cos ^{4} \theta+8 \cos ^{2} \theta=4 \)

Given intercepts are  -5, \(\frac 34\)
\(a=-5, b=\frac{3}{4}\)
Equation of the line in the intercepts form is
\(\frac{x}{a}+\frac{y}{b}=1 \)
\(\frac{x}{-5}+\frac{y}{\left(\frac{3}{4}\right)}=1 \)
\(\frac{x}{-5}+\frac{4 y}{3}=1 \)
3x - 20y = -15
3x - 20y + 15 = 0

Range = L- S = 20
Co-efficient of range = \(\frac { L-S }{ L+S } \) = 0.2
L - S = 20 ...(1)
L - S = 0.2(L + S)
(L + S) 0.2 = 20   ...(2)

L = \(\frac{24}{0.4}\) = 60
Substitute L = 60 in (1)
60 - S = 20
-S = 20 - 60 =-40
S = 40
∴ The largest is 60, the smallest is 40°.

\(\text { Given } P(A \cap B)=\frac{1}{6} \)
\(P(B \cap C) =\frac{1}{4} \)
\(P(A \cap C) =\frac{1}{8} \)
\(P(A \cup B \cup C)=\frac{9}{10} \)
\(P(A \cap B \cap C)=\frac{1}{15} \)
Also given that P(B) = 2 P(A)
P(C) = 3 P(A)
Now \(P(A \cup B \cup C)= P(A)+P(B)+P(C)- P(A \cap B)-P(B \cap C) -P(A \cap C)+P(A \cap B \cap C) \)
\(\frac{9}{10}= \mathrm{P}(\mathrm{A})+2 \mathrm{P}(\mathrm{A})+3(\mathrm{P}(\mathrm{A}))- \frac{1}{6}-\frac{1}{4}-\frac{1}{8}+\frac{1}{15} \)
\(\frac{9}{10} =6 \mathrm{P}(\mathrm{A})-\frac{1}{6}-\frac{1}{4}-\frac{1}{8}+\frac{1}{15} \)
\(6 \mathrm{P}(\mathrm{A}) =\frac{9}{10}+\frac{1}{6}+\frac{1}{4}+\frac{1}{8}-\frac{1}{15} \)
\(6 \mathrm{P}(\mathrm{A}) =\frac{108+20+30+15-8}{120} \)
\(6 \mathrm{P}(A) =\frac{165}{120} \)
\(\mathrm{P}(\mathrm{A}) =\frac{165}{120 \times 6}=\frac{11}{48} \)
\(\mathrm{P}(\mathrm{B}) =2 \times \frac{11}{48}=\frac{11}{24} \)
\(\mathrm{P}(\mathrm{C}) =3 \times \frac{11}{48}=\frac{11}{16} \)
\(\mathrm{P}(\mathrm{A}) =\frac{11}{48} ; \mathrm{P}(\mathrm{B})=\frac{11}{24} ; \mathrm{P}(\mathrm{C})=\frac{11}{16} \)

City A			City B
x1	\({ d }_{ 1 }=x-\bar { x } \)	\({ d }_{ 1 }^{ 2 }\)	x1	\(d=x-\bar { x } \)	\({ d }_{ 1 }^{ 2 }\)
18	-4	16	11	-4	16
20	-2	4	14	-1	1
22	0	0	15	0	0
24	2	4	16	3	4
26	4	16	18	3	9
110	0	40	75		20

\(\bar { { x }_{ 1 } } =\frac { { \Sigma x }_{ 1 } }{ n } \)
= \(\frac { 110 }{ 5 } \)
= 22
\({ \sigma }_{ 1 }=\sqrt { \frac { { \Sigma d }_{ 1 }^{ 2 } }{ n } } \)
= \(\sqrt { \frac { 40 }{ 5 } } \)
= \(\sqrt { 8 } \)
\(\bar { { x }_{ 2 } } =\frac { { \Sigma x }_{ 1 } }{ n } \)
= \(\frac { 75 }{ 5 } \)
= 15
\({ \sigma }_{ 2 }=\sqrt { \frac { { \Sigma d }_{ 2 }^{ 2 } }{ n } } \)
= \(\sqrt { \frac { 20 }{ 5 } } \)
= \(\sqrt { 4 } \)
= \(2\sqrt { 2 } \)
\(CV_{ 1 }=\frac { { { \sigma }_{ 1 } } }{ { x }_{ 1 } } \times 100\)
= \(\frac { 2.83 }{ 22 } \times 100\)
= 12.86
\({ CV }_{ 2 }=\frac { { \sigma }_{ 2 } }{ \bar { { x }_{ 2 } } } \times 100\)
= \(\frac { 2 }{ 15 } \times 100\)
= 13.33
\(\therefore\)Co-efficient of variation of City A is less than C.V of City B.
\(\therefore\) City A is more consistent.

If a cos θ - b sin θ = c
Squaring on both the sides.
(a cos θ - b sin θ)² = c²
a² cos² θ + b² sin² θ - 2ab cos θ sin θ = c²
\(a^{2}\left(1-\sin ^{2} \theta\right)+b^{2}\left(1-\cos ^{2} \theta\right)-2 a b \sin \theta \cos \theta=c^{2} \)
\(a^{2}-a^{2} \sin ^{2} \theta+b^{2}-b^{2} \cos ^{2} \theta-2 a b \sin \theta \cos \theta=c^{2} \)
\(a^{2}+b^{2}-c^{2}=a^{2} \sin ^{2} \theta+b^{2} \cos ^{2} \theta+2 a b \sin \theta \cos \theta \)
\(a^{2}+b^{2}-c^{2}=(a \sin \theta+b \cos \theta)^{2} \)
\(\pm \sqrt{a^{2}+b^{2}-c^{2}}=a \sin \theta+b \cos \theta \)
\(\therefore \ a \sin \theta+b \cos \theta=\pm \sqrt{a^{2}+b^{2}-c^{2}} \)

Distance travelled by the first train in 2 hours
= 2 x 20 = 40km
Distance travelled by the second train in 2 hours
= 2 x 30 = 60km
Let the distances are represents by OB and OA respectively
Now applying Pythagoras theorem,
Distance between the trains after 2 hours is AB.
we have AB² = OA² + OB² = 60² + 40²
= 3600 + 1600
= 5200
\( A B =\sqrt{5200} \)
\(=\sqrt{2^{2} \times 2^{2} \times 5 \times 5 \times 13} \)
\(=2^{2} \times 5 \sqrt{13} \)
\(=20 \sqrt{13} \mathrm{~km}\)
Distance between the trains after 2 hrs
\(=20 \sqrt{13} \mathrm{~km}\)

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)

If a and b are the intercepts then a + b = 7 or b = 7 − a
By intercept form \(\frac { x }{ a } +\frac { y }{ b } =1\)...(1)
We have \(\frac { x }{ a } +\frac { y }{ 7-a } =1\)
As this line pass through the point (-3,8), we have
\(\frac { -3 }{ a } +\frac { 8 }{ 7-a } =1\) gives –3(7–a) + 8a = a(7 – a)
−21 + 3a + 8a = 7a − a²
So, a² + 4a − 21 = 0
Solving this equation (a − 3) (a + 7) = 0
a = 3 or a = −7
Since a is positive, we have a = 3 and b = 7 – a = 7 – 3 = 4.
Hence \(\frac { x }{ 3 } +\frac { y }{ 4 } =1\)
Therefore, 4x + 3y − 12 = 0 is the required equation.

Given
\( \frac{\cos \alpha}{\cos \beta}=m \)
\(\frac{\cos \alpha}{\sin \beta} =n \)
\(\text { LHS } =\left(m^{2}+n^{2}\right) \cos ^{2} \beta \)
\(=\left(\frac{\cos ^{2} \alpha}{\cos ^{2} \beta}+\frac{\cos ^{2} \alpha}{\sin ^{2} \beta}\right) \cos ^{2} \beta \)
\(=\frac{\left(\cos ^{2} \alpha \sin ^{2} \beta+\cos ^{2} \alpha \cos ^{2} \beta\right)}{\cos ^{2} \beta \sin ^{2} \beta} \cos ^{2} \beta \)
\(=\frac{\cos ^{2} \alpha\left(\sin ^{2} \beta+\cos ^{2} \beta\right)}{\sin ^{2} \beta} \)
\(=\frac{\cos ^{2} \alpha}{\sin ^{2} \beta}(1) \)
\(=\left(\frac{\cos \alpha}{\sin \beta}\right)^{2} \)
= n² = RHS

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

Let r and R be the inner and outer radii of the hollow sphere.
Given that, inner diameter d = 14 cm; inner radius r = 7 cm; thickness = 1 mm = \(\frac{1}{10}\)cm
Outer radius R = 7 + \(\frac { 1 }{ 10 } =\frac { 71 }{ 10 } =7.1cm\)
Volume of hollow sphere \(=\frac { 4 }{ 3 } \pi \left( { R }^{ 3 }-{ r }^{ 3 } \right) cu.cm\)
\(=\frac { 4 }{ 3 } \times \frac { 22 }{ 7 } (357.91-343)=62.48cm^{ 3 }\)
But, weight of brass in 1 cm³ = 17.3 gm
Total weight = 17362 x 62.48 = 1080.90 gm
Therefore, total weight is 1080.90 grams.

Let r, R and h be the internal radius, external radius and height of the hollow cylinder respectively.
Given that, r = 21cm, R = 28 cm, h = 9 cm
Now, volume of hollow cylinder = \(\pi\)(R² − r²)h cu. units
\(=\frac { 22 }{ 7 } \left( { 28 }^{ 2 }-21^{ 2 } \right) \times 9\)
\(=\frac { 22 }{ 7 } (784-441)\times 9=9702\)
Therefore, volume of iron used = 9702 cm³

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)

Considering, \(\frac { x }{ 2 } -1=\frac { y }{ 6 } +1\)
\(\frac { x }{ 2 } -\frac { y }{ 6 } \) = 1 + 1  \(\frac { 6x-2y }{ 12 } \) = 2 we get, 3x - y = 12.... (1)
Considering \(\frac { x }{ 2 } -1=\frac { z }{ 7 } +2\)
\(\frac { x }{ 2 } -\frac { z }{ 7 } \) = 1 + 2 gives, \(\frac { 7x-2z }{ 14 } \) = 3 we get, 7x - 2z = 42... (2)
Also, from \(\frac { y }{ 3 } +\frac { z }{ 2 } \) = 13  \(\frac { 2y+3z }{ 6 } \) = 13 we get, 2y + 3z = 78 .(3)
Eliminating z from (2) and (3)

Substituting x = 10 in (1), 30 - y = 12 we get, y = 18
Substituting x = 10 in (2), 70 - 2x = 42 then, z = 14
Therefore, x = 10, y = 18, z = 14.

Vertices of one triangular tile are at (-3, 2), (-1, -1) and (1, 2)

(-3, 2), (-1, -1) and (1, 2)
Area of this tile = \(\frac12\) {(3 - 2 + 2) - (- 2 - 1 - 6)} sq. units
= \(\frac12\) (12) = 6 sq. units
Since the floor is covered by 110 triangle shaped identical tiles,
Area of floor = 110 x 6 = 660 sq. units

Let the 
100's digit be 'x'
10's digit be 'y'
Unit's digit be 'z'
Given 100y+ 10x + z - 54 = 3(100x + 10y + z)
Substituting 290x - 70y + 22 = -54 ( /2)
145x - 35y + z = -27          .........(1)
l00x + 10y + z + 198 = 100z + 10y + x
Substituting 99x - 992 = - 198 (99)
x - z = - 2              ...........(2)
y = x + 2 (y - z)
x + y - 2z = 0          ..........(3)
Consider (1) and (3)
145x - 35y + z = -27        .......(1)
35x + 35y - 70z = 0          .......(4)
180x - 69z = -27            .......(5)
Consider (5) and (2)

\(x=\frac{111}{111}=1\)
Substituting x = 1 in         .........(2)
1 - z = -z
z = 1 + 2 = 3
Substituting x - 1, z = 3 in (3)
1 + y - 6 = 0
y = 5
solution: x = 1, y = 5, z = 3
The number is 153.

Given the radius of the circle = 4 cm

Length of the tangents PA = PB = 10.2 cm
Construction:
Steps:
(1) With centre O, draw a circle of radius 4cm,
(2) Draw a line OP = 11 cm.
(3) Draw a \(\bot '\) bisector of OP, which cuts at 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 = 10.2 cm.

The distance between the point from the centre is 10 cm.

Length of the tangents PA - PB = 8.7 cm
Construction:
Steps:
(1) With O as centre, draw a circle of radius 5cm.
(2) Draw a line OP = 10 cm.
(3) Draw a perpendicular bisector of OP which cuts OP at 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 and PB = 8.7 cm