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

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

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}

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

\(\frac { { x }^{ 2 }+6x+8 }{ { x }^{ 2 }+x-2 } \) is undefined when x²+x-2 = 0 i.e.
(x+2) (x-1) = 0.
∴ The excluded values are -2.1.

Sample Space
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)}
Total number of outcomes = 36

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) = \(\frac { n(A) }{ n(S) } =\frac { 2 }{ 4 } =\frac { 1 }{ 2 } \).

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

in the right triangle ABC [see figure. (a)]
tan \(\theta \) =\(\frac { opposite\ side\ }{ adjacent\ side\ } =\frac { 4 }{ 5 } \)
= tan^-1(0.8)
\(\theta \) = \(38.7°\)(since tan \(38.7°\) = 0.8011)
\(\angle \)BAC = \(38.7°\)

Let the given points be A(1, 4) , B(2, 5) and C(4, 7).
Slope of line BC = \(\frac { 7-5 }{ 4-2 } =\frac { 2 }{ 2 } =1\)
Let m be the slope of the required line.
Since the required line is perpendicular to BC,
m x 1 = −1
m = −1
The required line also pass through the point A(1, 4).
The equation of the required straight line is y − y₁ = m(x - x₁)
y - 4 = − 1(x  − 1)
y - 4 = − x + 1
we get, x + y − 5 = 0

External radius of hollow cylinder R = 16 cm
length h = 13 cm
Thickness R - r = 4
16 - r = 4
r = 12 cm
Total surface area of hollow cylinder \(=2 \pi(\mathrm{R}+\mathrm{r})(\mathrm{R}-\mathrm{r}+\mathrm{h}) \text { sq. units }\)
\(=2 \times \frac{22}{7} \times(16+12)(4+13) \)
\(=2 \times \frac{22}{7} \times 28 \times 17 \)
= 2992 sq. cm

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

We set f₂(x) = 2x² - 5x + 3 and f₁(x) =\(\sqrt { x } \)
Then, f(x) = \(\\ \sqrt { 2x^{ 2 }-5x+3 } =\sqrt { { f }_{ 2 }(x) } \)
= f₁{f₂(x)} = f₁f₂(x)

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

Let BD =  x cm, then DC = (6 – x)cm
AD is bisector of\(\angle\) A
Therefore by Angle Bisector Theorem
\(\frac { AB }{ AC } =\frac { BD }{ DC } \)
\(\frac { 10 }{ 14 } =\frac { x }{ 6-x } \quad \frac { 5 }{ 7 } =\frac { x }{ 6-x } \)
So, 12x = 30 we get, \(x=\frac { 30 }{ 12 } =2.5\)
Therefore, BD = 2.5 cm, DC = 6−x = 6−2.5 = 3.5 cm

(1,–1), (–4, 6) and (–3, –5)

 A(-4, 6), B(-3, -5), C(1, -1)
Area of triangle ABC \( =\frac{1}{2}\left[\left(x_{1} y_{2}+x_{2} y_{3}+x_{3} y_{1}\right)\right. \left.-\left(x_{2} y_{1}+x_{3} y_{2}+x_{1} y_{3}\right)\right] \)
\(=\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)\right. \left.+x_{3}\left(y_{1}-y_{2}\right)\right] \text { sq. units } \)
\(=\frac{1}{2}[-4(-5+1)-3(-1-6)+1(6+5)] \)
\(=\frac{1}{2}[-4 \times(-4)-3 \times(-7)+1 \times(11)] \)
\(=\frac{1}{2}[16+21+11] \)
\(=\frac{1}{2}(48)=24 \text { sq. units. } \)

Given number of flower pots = 532
By Euclid's Division Lemma
a = bq + r where 0 < r < lbl
532 = 21 x 25 + 7
where 0 < r < l21l
Number of completed rows = 25
Number of flower pots left over = 7

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

tan² \(\theta \) - sin ²\(\theta \) = tan² \(\theta \) -\(\frac { si{ n }^{ 2 }\theta }{ co{ s }^{ 2 }\theta } \),cos²\(\theta \) 
= tan² \(\theta \) (1-cos² \(\theta \) ) = tan² \(\theta \) sin^{2 \(\theta \)}

(a) Capital employed - Goodwill + current liabilities

(i) To check if h is one – one, we assume that h(b₁) = h(b₂)
Then we get 2.47 b₁ + 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.

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

Since
cos²A + sin²A = 1 therefore
cos²A = 1 - sin²A
i.e., cos A = 土 \(\sqrt { 1-{ sin }^{ 2 }A } \)
This gives cos A = \(\sqrt { 1-{ sin }^{ 2 }A } \)
Hence, \(tanA=\frac { sinA }{ cosA } =\frac { sinA }{ \sqrt { { 1-sin }^{ 2 }A } } \)
and \(secA=\frac { 1 }{ cosA } =\frac { 1 }{ \sqrt { 1-{ sin }^{ 2 }A } } \)

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

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 a natural number be x.
The next number = x + 1
x (x + 1) = 20
X²+x-20 = 0
(x + 5)(x - 4) = 0
x=-5,4
∴ x=4
(∵ x≠-5, x is a natural number
The next number = 4 + 1= 5
Two consecutive numbers are 4,5.

Let the ten's digit of the number be x. It is given that the product of the digits is  12.
Unit's digit \(\frac{12}{x}\)
Number =10x+\(\frac{12}{x}\)
It 36 is added to the number the digits interchange their places.
\(\therefore 10x+\frac { 12 }{ x } +36=10\times \frac { 12 }{ x } +x\)
\(\Rightarrow 10x+\frac { 12 }{ x } +36=\frac { 120 }{ x } +x\)
\(\Rightarrow 9x-\frac { 108 }{ x } +36=0\)
⇒9x² - 108 + 36x = 0
⇒X²+ 4x - 12 = 0
⇒ (x + 6)(x - 2) = 0 (∵ (x + 6) ≠ 0 as x >0)
x=-6,2
But a number can never be (-ve). So, x = 2. The
number is 10x2+\(\frac{12}{2}\)=26

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.

A = {1,2,3,4,5}
B = {1,2,3,4, ... }
f: A \(\rightarrow\)B, f(x) = x²
\(\therefore\)  f(1) = 1² = 1
f(2)  = 2²= 4
f(3) = 3² = 9
f(4)  = 4²= 16
f(5) = 5²= 25
\(\therefore\) Range of f = {1, 4, 9, 16, 25}
Elements in A have been connected with different elements in B. Therefore it is one-one function. But not all the elements in B have preimages in A. Therefore it is not on-to function.

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.

2f(-4) + 3f(2)
f(-4) = x + 5 = -4 + 5 = 1
2f(-4) = 2 x 1 = 2
f(2) = x + 5 = 2 + 5 = 7
3f(2) = 3(7) = 21
∴ 2f(-4) + 3f(2) = 2 + 21 = 23

\(\frac { 10{ x }^{ 3 }-25{ x }^{ 2 }+4x-10 }{ -4-10{ x }^{ 2 } } \)
\(=\frac { { 5x }^{ 2 }\left( 2x-5 \right) +2\left( 2x-5 \right) }{ -2({ 5x }^{ 2 }+2) } \)=
\(=\frac { (2x-5) }{ -2 } =-x+\frac { 5 }{ 2 } \)

First let us factorize the polynomials
\(\sqrt { 15 } { x }^{ 2 }+\left( \sqrt { 3 } +\sqrt { 10 } \right) x+\sqrt { 2 } \) = \(\sqrt { 15 } { x }^{ 2 }+\sqrt { 3 } x+\sqrt { 10 } x+\sqrt { 2 } \)
= \(\sqrt { 3 } x\left( \sqrt { 5 } x+1 \right) +\sqrt { 2 } \left( \sqrt { 5 } x+1 \right) \)
= \(\left( \sqrt { 5 } x+1 \right) \times \left( \sqrt { 3 } x+\sqrt { 2 } \right) \)
\(\sqrt { 5 } { x }^{ 2 }+\left( 2\sqrt { 5 } +1 \right) x+2=\sqrt { 5 } { x }^{ 2 }+2\sqrt { 5 } x+x+2\)
= \(\sqrt { 5 } x\left( x+2 \right) +1\left( x+2 \right) =\left( \sqrt { 5 } x+1 \right) \left( x+2 \right) \)
\(\sqrt { 3 } { x }^{ 2 }+\left( \sqrt { 2 } +2\sqrt { 3 } \right) x+2\sqrt { 2 } =\sqrt { 3 } { x }^{ 2 }+\sqrt { 2 } x+2\sqrt { 3 } x+2\sqrt { 2 } \)
= \(x\left( \sqrt { 3 } x+\sqrt { 2 } \right) +2\left( \sqrt { 3 } x+\sqrt { 2 } \right) =\left( x+2 \right) \left( \sqrt { 3 } x+\sqrt { 2 } \right) \)
Therefore,
\(\left[ \sqrt { 15 } { x }^{ 2 }+\left( \sqrt { 3 } +\sqrt { 10 } \right) x+\sqrt { 2 } \right] \left[ \sqrt { 5 } { x }^{ 2 }+\left( 2\sqrt { 5 } +1 \right) x+2 \right] \left[ \sqrt { 3 } { x }^{ 2 }+\left( \sqrt { 2 } +2\sqrt { 3 } \right) x+2\sqrt { 2 } \right] \)
= \(\left| \left( \sqrt { 5 } x+1 \right) \left( \sqrt { 3 } x+\sqrt { 2 } \right) \left( x+2 \right) \right| \)

When three coins are tossed, the sample space
S = {HHH, HHT HTH, HTT, THH, THT, TTH, TTT}
n(S) = 3
Let A be the event of getting atmost 2 tails (i.e., 0 tail, 1 tail,2 tails)
A = {HHH, HHT, FITH, HTT THH, THT, TTH}
n(A) = 7
\(\mathrm{P}(\mathrm{A})=\frac{n(A)}{n(S)}=\frac{7}{8}\)
Let B be the event of getting atleast 2 heads. (i.e.,2 heads,3 heads)
B = {HHT, HTH, THH, HHH}
n(B) = 4
\(\mathrm{P}(\mathrm{B})=\frac{n(B)}{n(S)}=\frac{4}{8}\)
A n B = {HHT, HTH, THH, HHH}
\(\mathrm{n}(A \cap B)=4 \)
\(\mathrm{P}(A \cap B)=\frac{n(A \cap B)}{n(\mathrm{~S})}=\frac{4}{8} \)
\(\mathrm{P}(A \cup B)=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(A \cap B) \)
\(\mathrm{P}(A \cup B)=\frac{7}{8}+\frac{4}{8}-\frac{4}{8}=\frac{7+4-4}{8}=\frac{7}{8} \)
Probability of getting atmost two tails or atleast 2 heads = \(\frac{7}{8}\)

The height of the tower AB = 50 m
Let the height of the tree CD = y and BD = x
From the diagram,\(\angle \)XAC = 30° = \(\angle \)ACM and\(\angle \) = XAD = 45° =\(\angle \) ADB
In right triangle ABD,
tan45° = \(\frac { AB }{ BD } \)
1 = \(\frac { 50 }{ x } \) gives x = 50 m
In right triangle AMC,
tan30° = \(\frac { AM }{ CM } \)
\(\frac { 1 }{ \sqrt { 3 } } =\frac { AM }{ 50 } \)[since DB = CM]
AM = \(\frac { 50 }{ \sqrt { 3 } } =\frac { 50\sqrt { 3 } }{ 3 } =\frac { 50\times 1.732 }{ 3 } \) = 28.87 m.
Therefore, height of the tree = CD = MB = AB − AM = 50 – 28.87 = 21.13 m

Sample space S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}; n(S) = 12
(i) Let A be the event of resting in 7. n(A) = 1
P(A) = \(\frac { n(A) }{ n(S) } =\frac { 1 }{ 12 } \)
(ii) Let B be the event that the arrow will come to rest in a prime number.
B = {2 , 3, 5, 7, 11}; n(B) = 5
P(B) = \(\frac { n(B) }{ n(S) } =\frac { 5 }{ 12 } \)
(iii) Let C be the event that arrow will come to rest in a composite number.
C = {4, 6, 8, 9, 10, 12}; n(C) = 6
P(C) = \(\frac { n(C) }{ n(S) } =\frac { 6 }{ 12 } =\frac { 1 }{ 2 } \)

Let the initial position of the bird be A and after two seconds its position is at D.
AC = DE = 80m
∠ABC = 45°
∠DBC = 30°
In right ABC
\(\tan 45^{\circ} =\frac{A C}{B C} \)
\(1 =\frac{80}{B C} \)
BC = 80 m
In right triangle DBE
\(\tan 30^{\circ} =\frac{D E}{B E} \)
\(\frac{1}{\sqrt{3}} =\frac{80}{B C+C E} \)
\(\frac{1}{\sqrt{3}} =\frac{80}{80+C E} \)
\(80+\mathrm{CE} =80 \sqrt{3} \)
\(\mathrm{CE} =(80 \sqrt{3}-80) \)
\(=80(\sqrt{3}-1) \)
= 80 (1.732 - 1)
= 80 x 0.732
CE = 58.56 m
The bird travelled 58.56 m in 2 seconds.
Speed of the bird \(=\frac{\text { Distance travelled }}{\text { Time taken }} \)
\(=\frac{58.56}{2} \)
= 29.28 m/s
Speed of flying bird = 29.28 m/s.

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

Completed Pages (xi)	\(\left(\mathbf{x}_{\mathrm{i}}^{2}\right)\)
32	1024
35	1225
37	1369
30	900
33	1089
36	1296
35	1225
37	1369
\(\Sigma x_{i}=275\)	\(\Sigma x_{i}^{2}=9497\)

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 be the height of the kite above the ground. Then, AB = 75.
Let AC be the length of the string.
In right triangle ABC,\(\angle \)ACB = \(60°\)
\(sin\theta =\frac { AB }{ AC } \)
\(sin60°=\frac { 75 }{ AC } \)
gives \(\frac { \sqrt { 3 } }{ 2 } =\frac { 75 }{ AC } \) so, AC = \(\frac { 150 }{ \sqrt { 3 } } =50\sqrt { 3 } \) 
Hence, the length of the string is 50\(\sqrt { 3 } m\)

Given intercepts are 4, - 6
a = 4, b = - 6
Equation of the line in the intercepts form is
\(\frac{x}{a}+\frac{y}{b}=1 \)
\(\frac{x}{4}+\frac{y}{-6}=1 \)
3x - 2y - 12 = 0

Let the radius of hemispherical bowl = r
Volume of hemispherical bowl
\(=\frac{2}{3} \pi r^{3} \text { cu. units }\)
Let the height of cylindrical vessel = h
Given \(r=h+h \frac{50}{100} \Rightarrow \mathrm{r}=\mathrm{h}\left(1+\frac{50}{100}\right) \)
\(\mathrm{h}= \frac{2}{3} r \)
Now, Volume of cylindrical vessel
\(=\pi r^{2}\left(\frac{2 r}{3}\right)=\frac{2}{3} \pi r^{3}\)
Hence, Volume of juice in the cylindrical vessel
\(=\frac{\frac{2}{3} \pi r^{3}}{\frac{2}{3} \pi r^{3}} \times 100 \%=100 \%\)

Consider\(\triangle\)ABC. Then D, E and F are respective points on the sides CA, AB and BC. By construction D, E, F are collinear
By Menelaus’ theorem \(\frac { AE }{ EB } \times \frac { BF }{ FC } \times \frac { CD }{ DA } =1\)
By assumption, AE = EB = 2, DA = 1 and 
FC = FB + BC = BF + 3
By Pythagoras theorem,AC²=AB²+BC²=16+9=25. Therefore AC=5
and So, CD = AC – AD = 5 – 1 = 4.
Substituting the values of FC, AE, EB, DA, CD in (1)
we get, \(\frac { 2 }{ 2 } \times \frac { BF }{ BF+3 } \times \frac { 4 }{ 1 } =1\)
4BF=BF+3
4F-BF=8 therefore BF=1

From the figure, radius of cylinder = \(\frac{3}{2}=1.5 \mathrm{~cm}\)
Height = 8 cm
Volume of cylinder \(=\pi r^{2} h \text { cu. units } \)
\(=\frac{22}{7} \times 1.5 \times 1.5 \times 8 \)
Radius of cone = 1.5 cm
Height of cone = 2 cm
Volume of 2 cones \(=2\left(\frac{1}{3} \pi r^{2} h\right)\ cu. units \)
\(=\frac{2}{3} \times \frac{22}{7} \times 1.5 \times 1.5 \times 2 \)
Volume of the model = Volume of cylinder + Volume of 2 cones
\(=\frac{22}{7} \times(1.5)^{2}\left[8+\frac{4}{3}\right] \)
\(=\frac{22}{7} \times 2.25 \times \frac{28}{3} \mid \)
\(=\frac{1386}{21}=66 \mathrm{~cm}^{3} \)

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

Let Sarah's house is at A and James's house is at 'B' from the picture.
Distance between Sarah's house to James house through Street B and C
= 1.5 miles + 2 miles = 3.5 miles
Distance through street C is AC² = AB² + BC²

AC² = = (1.5)² + (2),
= 2.25 + 4
= 6.25
\(A C=\sqrt{6.25}=2.5\)
AC = 2.5 miles
Difference between two paths = 3.5 - 2.5 = 1 mile
Direct path along C street is 1 mile shorter

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.

i) f: R ⟶ R and f(x) = 2x + 1 
when x = -1,f(- 1) = 2(- 1) + 1 =- 1 \( \in\) R
when x =  0, f(0) = 2 (0) + 1 = 1 \( \in\) R
when x = 1,f (1) = 2(1) + 1 = 3 \( \in\) R and soon
For every value of x \( \in\) R, f (x) also \( \in\) R.
The function is well defined and it is one-to-one function (Injective)
For f (x) : R ⟶ R , the domain and range are also well defined. So it is an onto function (Surjective)
Thus, the function is one - to one onto i.e. Bijective function.
(ii) f : R ⟶ R , f(x) = 3 - 4x²
when x = 0, f (0) = 3 - 4(0) = 3 \( \in\) R
when x = 1, f (1) = 3 - 4(1)² = -1  \( \in\) R
when x = 2, f (2) = 3 - 4(2)² = 13  \( \in\) R
when x = -1, f (-1) = 3 - 4(-1)² = -1  \( \in\) R
when x = -2, f (-2) = 3 - 4(-2)² = -13  \( \in\) R
From this, it is clear that two or more elements having same image in the co-domain So, it is not one-to-one and it is many-to-one function. Hence it is not Bijective.

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

\(f(x)=\frac { 1{ x }^{ 2 } }{ a } -\frac { 2x }{ b } +\frac { 3 }{ c } \)
Sum of the roots \((\alpha +\beta )=\frac { -b }{ a } =-\frac { (-2) }{ 1 } \)
Product of the roots \((\alpha \beta )=\frac { c }{ a } =\frac { 3 }{ 1 } =3\)
∝+2, β+2 are the roots (given)
Sum of the roots = ∝ + 2 + β + 2
= ∝ + β + 4
= 2 + 4 = 6
Product of the roots = (∝ + 2) (β + 2)
= ∝β + 2∝ + 2β + 4
= ∝β + 2(∝ + β) + 4
= 3 + 2 x 2 + 4
= 3 + 4 + 4 = 11
∴ The required equation x² - 6x + 11 = 0.

Construction:
Step (1) Draw a line segment QR = 5 cm.
Step (2) At Q, draw QE such that \(\angle RQE\) = 40°.
Step (3) At Q, draw QF such that \(\angle EQF\) = 90^o
Step (4)Drawn a perpendicular bisector to QR, which intersects QF at 'O' and QR at G.
Step (5) With O as centre and OQ as radius, draw a circle
Step (6) From G marked arcs of radius 4.4 cm on the circle. Marked them as P and S.
Step (7) Joined QP and PR. Now \(\triangle\)PQR is the required triangle
Step (8) From P draw a line PN which is \(\bot \) to LR. LR meets PN at M.
Step (9) The length of the altitude is PM = 2.1cm

Given a triangle PQR, we are required to construct another triangle whose sides are \(\frac { 7 }{ 4 } \) of the corresponding sides of the triangle PQR.
Steps of construction
1. Construct a DPQR with any measurement.
2. Draw a ray QX making an acute angle with QR on the side opposite to vertex P.
3. Locate 7 points (the greater of 7 and 4 in \(\frac { 7 }{ 4 } \))
Q_1,Q_2,Q_3,Q_4,Q_5,Q₆ and Q₇ on QX so that 
QQ₁ = Q₁Q₂ = Q₂Q₃ = Q₄Q₅ = Q₅Q₆ = Q₆Q₇
4. Join Q₄ (the 4th point, 4 being smaller of 4 and 7 in \(\frac { 7 }{ 4 } \)) to R and draw a line through Q₇ parallel to Q₄R, intersecting the extended line segment QR at R'.
5. Draw a line through R' parallel to RP intersecting the extended line segment QP at P'.
Then \(\triangle\)P'QR' is the required triangle each of whose sides is seven-fourths of the corresponding sides of \(\triangle\)PQR.