(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.

We are given a right triangle ABC right angled at B.
We need to prove that AC² = AB² + BC²
Let us draw \(BD\bot AC\)
Now,\(\Delta ADB\sim \Delta ABC\)
\(\cfrac { AD }{ DB } =\cfrac { BC }{ AC } \)
(sides are proportional)
Also,
\(\Delta BDC\sim \Delta ABC\)
\(\cfrac { CD }{ BC } =\cfrac { BC }{ AC } \)
CD·AC = BC²     ..(2)
Adding (1) and (2)
AD .AC + CD . AC = AB²+ BC²
AC(AD + CD) = AB² + BC²
AC.AC = AB² + BC²
AC = AB² + BC²

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

Volume of a cylinder \(=\pi { r }^{ 2 }h\)
Volume of a cone \(=\frac { 1 }{ 3 } \pi { r }^{ 2 }h\)
Volume of a sphere \(=\frac { 4 }{ 3 } \pi { r }^{ 3 }\)
Their ratio V₁ : V₂ : V₃

\(h:\frac { h }{ 3 } :\frac { 4r }{ 3 } \)
3h: h: 4r
3h : h : 2(2r)    (where 2r = h)
∴  V₁ : V₂ : V₃ = 3: 1 : 2

CV =\(\frac { \sigma }{ \bar { x } } \) x 100 ⇒ \(\bar { x } =\frac { \sigma }{ CV } \) x 100
\(\bar { x } =\frac { 15.6 }{ 6.9 } \) x 100  = 22.6

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.

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.

f(3) =. 2x - 1
= 2(3) - 1 = 6 - 1 = 5

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

Let P and Q be the points of trisection at AB.
i.e., AP = PQ = QB

Therefore, P divides AB internally in the ratio 1:2. Therefore, the coordinates at P, by applying the section formula, are
\(\left[ \frac { 1(-7)+2(2) }{ 1+2 } ,\frac { 1(7)+2(-2) }{ 1+2 } \right] \) i.e., (-1,10)
Now, Q also divides AB internally in the ratio 2:1, so, the coordinates at Q are
\(\left[ \frac { 2(-7)+1(2) }{ 2+1 } ,\frac { 2(4)+(-2) }{ 2+1 } \right] \) i.e., (-4,2)
Therefore, the coordinates at the points at trisection of the line segment joining A and B are (-1, 0) and (-4, 2).

\(\frac{\cot \theta-\cos \theta}{\cot \theta+\cos \theta}=\frac{\operatorname{cosec} \theta-1}{\operatorname{cosec} \theta+1}\)
\( \text { LHS } =\frac{\cot \theta-\cos \theta}{\cot \theta+\cos \theta} \)
\(=\frac{\frac{\cos \theta}{\sin \theta}-\cos \theta}{\frac{\cos \theta}{\sin \theta}+\cos \theta} \)
\( \frac{\frac{\cos \theta-\sin \theta \cos \theta}{\sin \theta}}{\frac{\cos \theta+\sin \theta \cos \theta}{\sin \theta}} \)
\(=\frac{(\cos \theta-\sin \theta \cos \theta)}{\sin \theta} \times \frac{\sin \theta}{(\cos \theta+\sin \theta \cos \theta)} \)
\(=\frac{\cos \theta(1-\sin \theta)}{\cos \theta(1+\sin \theta)} \)
\( \mathrm{LHS} =\frac{1-\sin \theta}{1+\sin \theta} \)           ....(1)
\(\mathrm{RHS} =\frac{\operatorname{cosec} \theta-1}{\operatorname{cosec} \theta+1}\)
\(=\frac{\frac{1}{\sin \theta}-1}{\frac{1}{\sin \theta}+1}=\frac{\frac{1-\sin \theta}{\sin \theta}}{\frac{1+\sin \theta}{\sin \theta}}=\frac{1-\sin \theta}{\sin \theta} \times \frac{\sin \theta}{1+\sin \theta}\)
\(\text { RHS }=\frac{1-\sin \theta}{1+\sin \theta}\)
From (1) and (2)
LHS = RHS

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.

Use the slope and y intercept to write an equation 
The equation is y =  \(\frac { 9 }{ 5 } x\) + 32

Number of non defective bulbs = 20.
Let x be the number of defective bulbs
Then total number of bulbs n(s) = 20 + x
Let 'A' be the event of getting defective bulbs
\(\mathrm{P}(\mathrm{A})=\frac{x}{20+x}=\frac{3}{8}\)
8x = 3 (20 + x)
8x = 60 + 3x
8x - 3x = 60
5x = 60
\(x=\frac{60}{5}\)
x = 12
Number of defective bulbs = 12

Radius of hemisphere = 6 cm
Volume of hemisphere \(=\frac{2}{3} \pi r^{3} \text { cu. units }\)
\(=\frac{2}{3} \pi(6)^{3} \)
\(=\frac{2}{3} \pi(216) \)
\(=144 \pi \mathrm{cm}^{3} \)
base of cone = 6 cm
Height of the cone = 12 cm
Volume of the cone \(=\frac{1}{3} \pi r^{2} h \text { cu. units } \)
\(=\frac{1}{3} \pi(6)^{2}(12) \)
= 144 cm³
volume of the solid = Volume of cone + Volume of hemisphere
\(=144 \pi+144 \pi=288 \pi\)
Volume of water displaced
= Volume of the solid placed in the cylinder
\(=288 \pi=288 \times \frac{22}{7}\)
= 905.14 cm³

Given that radius and height of a cylinder are in the ratio 5 : 7
\(\text { i.e., } \frac{r}{h}=\frac{5}{7} \Rightarrow \mathrm{h}=\frac{7 r}{5}\)
Curved surface area = 5500 sq. cm
\(2 \pi r h =5500 \)
\(2 \times \frac{22}{7} \times r \times \frac{7 r}{5} =5500 \)
\(r^{2} =\frac{5500 \times 5}{2 \times 22} \)
\(r^{2} =625 \Rightarrow r=25 \)
\(=\frac{7(25)}{5}=35 \)
radius = 25 cm, height = 35 cm

Let the A.P. be t₁, t₂ , t₃, t₄,....
It is given that t₇ = -1 and t₁₆ = 17
a + (7 -1)d = -1 and a + (16 - 1)d = 17
a + 6d = -1 .....(1)
a + 15d = 17 ......(2)
Subtracting equation (1) from equation (2), we get 9d = 18 gives d = 2
putting d = 2 in equation (1), we get a + 12 = -1 so a = -13
Hence, General term t_n  = a + (n - 1)d
= -13 + (n -1) x 2 = 2n - 15 

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.

From the diagram,
The solid is a cuboid volume of cuboid = length x breadth x height
where l = 24 - 2x, b - 24 - 2x,. h = x
Volume V (x) = (24 - 2x) (24 - 2x) x
V(x) = x(24 - 2x)², x > 0
= 4x³ - 96x² + 576x, x > 0
So, the domain is 0 < x < 12

A x B = {(3,1), (3,7), (3,10), (4,1), (4,7), (4,10), (7,1), (7,7), (7,10), (8,1), (8,7), (8,10)}
Here, (7,8) \(\in \) R₃, but (7,8) \(\notin \) A x B. So, R₃ is not a relation from A to B.

Clearly bucket forms frustum of a cone such that the radii of its circular ends are r₁ = 28 cm, r₂ = 7 cm, h = 45 cm
Capacity of the bucket = volume of the frustum
\(\Rightarrow \frac { 1 }{ 3 } \times \pi h[{ r }_{ 1 }^{ 2 }+{ r }_{ 2 }^{ 2 }+{ r }_{ 1 }{ r }_{ 2 }]\)
\(\Rightarrow \frac { 1 }{ 3 } \times \frac { 22 }{ 7 } \times 45[{ 28 }^{ 2 }+{ 7 }^{ 2 }+28\times 7)]\)
= 22 x 15 x (28 x 4 + 7 + 28)
\(\Rightarrow 330\times 147{ cm }^{ 2 }\Rightarrow 48510{ cm }^{ 2 }\)

OA· OB = OC . OD (Given)

so, \(\cfrac { OA }{ OC } =\cfrac { OD }{ OB } \)
Also we have 
(vertically opposite angles) ...(2)
From (1) and (2)
\(\Delta AOD\sim \Delta COB\)( SAS similarity criterion)
So

(corresponding angles of similar triangles)

Let assumed mean A = 30
C = 5

σ =\(\sqrt { \left( \frac { \Sigma d'^{ 2 } }{ n } \right) -\left( \frac { \Sigma d' }{ n } \right) ^{ 2 } } \) x c
=\(\sqrt { \frac { 10 }{ 5 } -0 } \) x 5
=\(\sqrt { 2 } \) x 5
= 5\(\sqrt { 2 } \)
To convert the values for 100, all the values will be multiplied by 2. Therefore the new values are 40, 50, 60, 70, 80.
Let A = 60, C = 10

σ =\(\sqrt { \left( \frac { \Sigma d'^{ 2 } }{ n } \right) -\left( \frac { \Sigma d' }{ n } \right) ^{ 2 } } \) x c
=\(\sqrt { \frac { 10 }{ 5 } -0 } \) x 10 
=\(\sqrt { 2 } \) x 10
= 10\(\sqrt { 2 } \)
S.D. also be multiplied by 2. It is also true for the division also.

9x²-9(a+b)x+(2a²+5ab+2b²)=0
Comparing this with ax² + bx + c = O.
a =9
b = -9(a + b)
c = (2a² + 5ab + 2b²)
∴ ∆=B²-4AC
⇒ 81(a+b)²-36(2a²+5ab+2b²)
⇒ 9a² + 9b² - 18ab
⇒ 9(a - b)2> 0
∴ the roots are real and given by
\(\beta =\frac { -B-\sqrt { D } }{ 2A } =\frac { 9(a+b)+3(a-b) }{ 18 } \)
\(=\frac { 12a+6b }{ 18 } =\frac { 2a+b }{ 3 } \)
\(\beta =\frac { -B-\sqrt { D } }{ 2A } =\frac { 9(a+b)+3(a-b) }{ 18 } \)
\(=\frac { 6a+12b }{ 18 } =\frac { a+2b }{ 3 } \)

Let us assume the opposite, (1) \(\sqrt { 3 } \) is irrational.
Hence \(\sqrt { 3 } =\frac { p }{ q } \)
Where p and q (q ≠ 0) are co-prime (no common factor other than 1)
Hence, \(\sqrt { 3 } =\frac { p }{ q } \)
\(\sqrt { 3 } \)q = p
Squaring both side
\({ (\sqrt { 3 }q ) }^{ 2 }={ p }^{ 2 }\)
3q² = p²
\({ q }^{ 2 }=\frac { p }{ 3 } \)
Hence, 3 divides p² So 3 divides p also   .....(1)
Hence we can say
\(\frac{p}{3}\) = c where c is some integer
s, p =p²
Putting p = 3c
3q² = (3c)²
3q² = 9c²
q2 = \(\frac13\) x 9c²
q² = 3c²
\(\frac{9^2}{3}\) = c²
Hence 3 divides q²
So, 3 divides q also  ...(2)
By (1) and (2) 3 divides both p and q
By contradiction \(\sqrt { 3 } \) is irrational.

By joining B to D, you will get too triangles ABD and BCD.
Now, the area of ΔABD
=\(\frac { 1 }{ 2 } \)[ -5(-5 - 5) + (-4)(5 -7) + 4(7 + 5)]
=\(\frac { 1 }{ 2 } \)(50 + 8 + 48)
=\(\frac { 106 }{ 2 } \) = 53 square units.
Also, the area of ΔBCD
=\(\frac { 1 }{ 2 } \) = [-4(-6 - 5) - 1(5 + 5) + 4(-5 + 6)]
=\(\frac { 1 }{ 2 } \)(44 - 10 + 4)
= 19 square units.
So, the area of quadrilateral ABCD
= 53 + 19 = 72 square units.

(i) Given P(A) = 0.42
P(not A) = 1 - P(A)
\(\mathrm{P}(\bar{A})=1-0.42=0.58\)
(ii) Given P(B) = 0.48
P(not B) = 1 - P(B)
\(\mathrm{P}(\bar{B})=1-0.48=0.52\)
(iii) P(A or B) = \(P(A \cup B)\)
\(=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(A \cap B)\)
= 0.42 + 0.48 - 015
= 0.90 - 0.16
P(A or B) = 0.74

The height of the first building AB = 60 m. Now, AB = MD = 60 m
Let the height of the second building CD = h. Distance BD = 140 m
Now, AM = BD = 140 m
From the diagram,
\(\angle \)XCA = 30° =\(\angle \)CAM
In right triangle AMC, tan30° = \(\frac { CM }{ Am } \)
\(\frac { 1 }{ \sqrt { 3 } } =\frac { CM }{ 140 } \)
CM=\(\frac { 140 }{ \sqrt { 3 } } =\frac { 140\sqrt { 3 } }{ 3 } \)
\(=\frac { 140\times 1.732 }{ 3 } \)
CM = 80.78
Now, h = CD = CM + MD = 80.78 + 60 = 140.78
Therefore the height of the second building is 140.78 m

Given r = 5 and S₆ = 46872
Sum upto n terms of a G.P \(\mathrm{S}_{\mathrm{n}}=\frac{a\left(r^{n}-1\right)}{r-1}\)
\(46872 =\frac{a\left(5^{6}-1\right)}{5-1} \)
\(46872 =a \frac{(15625-1)}{4}=a \times \frac{15624}{4} \)
46872 = a x 3906
\(\frac{46872}{3906}=a\)
a = 12
First term of the G.P., a = 12

OA = 4 cm, OB = 5 cm; also OA\(\bot \)BC.
OB² = OA² + AB²
5² =  4² + AB² gives AB² = 9
Therefore AB = 3 cm
BC = 2AB hence BC = 2 x 3 = 6 cm

Given vertices A(1, - 4) , B(2, - 3) and C(4, - 7)
Slope of the line = \(\frac{y_{1}-y_{2}}{x_{1}-x_{2}}\)
Slope of AB = \(\frac{-4+3}{1-2}=\frac{-1}{-1}=1 \)
Slope of BC = \(\frac{-3+7}{2-4}=\frac{4}{-2}=-2 \)
Slope of AC = \(\frac{-4+7}{1-4}=\frac{3}{-3}=-1 \)
(Slope of AB) x (Slope of AC) = - 1
AB is perpendicular to AC.
Hence, the given vertices form a right angled triangle
Distance between the points (x₁, y₁) and (x₂, y₂) is \(\sqrt{\left.\left(x_{1}-x_{2}\right)^{2}+y_{1}-y_{2}\right)^{2}}\) units
\(A B =\sqrt{(1-2)^{2}+(-4+3)^{2}}=\sqrt{1+1}=\sqrt{2} \)
\(A B^{2} =(\sqrt{2})^{2}=2 \)
\(B C =\sqrt{(2-4)^{2}+(-3+7)^{2}}=\sqrt{4+16}=\sqrt{20} \)
BC² = 20
\(A C=\sqrt{(1-4)^{2}+(-4+7)^{2}}=\sqrt{9+9}=\sqrt{18}\)
AC² = 18
Now, AB² + AC² = BC²
Hence, the Pythagoras theorem is satisfied.

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