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)

σ =\(\sqrt { \frac { \Sigma x^{ 2 } }{ n } -\left( \frac { \Sigma x }{ n } \right) ^{ 2 } } \)
=\(\\ \sqrt { \frac { 20300 }{ 7 } -\left( \frac { 350 }{ 7 } \right) ^{ 2 } } \)
=\(\sqrt { 400 } \) = 20

Let us start with taking a, where a is a +ve odd integer.
We apply the division algorithm with 'a' and 'b' = 4.
Since 0  ≤ r < 4, the possible remainders are 0,1,2,3.
That is, a can be 4q, or 4q + 1, or 4q + 2 or 4q + 3, where 1 is the quotient. However, since a is odd, a cannot be 4q or 4q + 2 (since they are both divisible by 2).
Any odd integer is of the form 4q + 1 or 4q + 3

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}

The general form of Geometric progression is a, ar, ar²,...
a = 256, ar = 256 x 0.5 = 128 , ar² = 256 x (0.5)² = 64
Therefore the required Geometric progression is 256,128, 64,....   

Given points are (a, 2 – 2a), (– a + 1, 2a) and (– 4 – a, 6 – 2a)
Since the points are collinear, Area of triangle is zero.
\(\text { i.e., } \frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right]=0\)
a(2a - 6 + 2a) + (- a + 1) (6 - 2a - 2 + 2a) + (- 4 - a) (2 - 2a - 2a) = 0
a (4a - 6) + (- a + 1) (4) + (- 4 - a) (2 - 4a) = 0
4a² - 6a - 4a + 4 - 8 + 16 a - 2a + 4a² = 0
8a² + 4a - 4 = 0
2a² + a - 1 = 0
(2a - 1)(a + 1) = 0
2a - 1 = 0,a + 1 = 0
a = \(\frac{1}{2},-1\)

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

Common ratio = 4 > 1, sum of first 6 terms S₆ = 4095
Hence , S₆ = \(\frac { a\left( { r }^{ n }-1 \right) }{ r-1 } =4095\)
Since, r = 4,\(\frac { a\left( 4^{ 6 }-1 \right) }{ 4-1 } \) = 4095 gives a x \(\frac { 4095 }{ 3 } =4095\)
First term a = 3.

\(\angle\)OPQ = 90^o - 50^o = 40^o (angle between the radius and tangent is 90^o)
OP = OQ (Radii of a circle are equal)
\(\angle\)OPQ = \(\angle\)OQP = 40^o (\(\triangle\)OPQ is isosceles)
\(\angle POQ={ 180 }^{ 0 }-\angle OPQ-\angle OQP\)
\(\angle\)POQ  = 180^o - 40^o- 40^o = 100^o.

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

XY + CZ = 17cm
XZ + YW = 26cm
We know that diagonals if a rectangle bisect each other and the diagonals have equal length.
\(\therefore \text { Each diagonal }=\frac{26}{2}=13 \mathrm{~cm}\)
i.e., XZ = 13 cm and YW = 13 cm
Also given XY + YZ = 17 cm
Squaring on both sides (XY + YZ)² = 17²
\((\mathrm{XY})^{2}+(\mathrm{YZ})^{2}+2 \times(\mathrm{XY}) \times(\mathrm{YZ})=289\)
By Pythagoras theorem (XY)² + (YZ)² = XZ²
\(\therefore[\mathrm{XZ}]^{2}+2(\mathrm{XY}) \times(\mathrm{YZ})=289\)
132 + 2 x length x breadth = 289
2 x Area = 289 - 169
\(\text { Area }=\frac{289-169}{2}=\frac{120}{2}\)
The possible length and breadth are
(1,60) (2,30) (3,20) (4, 15), (5, 12) (6, 10).
In this pair the length and breadth should satisfy Pythagoras theorem for diagonal.
5,12 is the possible length and breadth.

Let r₁ and h₁ be the radius and height of the cone - I and let r₂ and h₂ be the radius and height of the cone-II.
Given h₂ = 2h₁ = 2 and \(\frac { Volume\ of\ the\ cone\ I }{ Volume\ of\ the\ cone\ II } =\frac { 2 }{ 3 } \)
\(\frac { \frac { 1 }{ 3 } { \pi r }_{ 1 }^{ 2 }{ h }_{ 1 } }{ \frac { 1 }{ 3 } { \pi r }_{ 2 }^{ 2 }{ h }_{ 2 } } =\frac { 2 }{ 3 } \)
\(\frac { { r }_{ 1 }^{ 2 } }{ { r }_{ 2 }^{ 2 } } \times \frac { { h }_{ 1 } }{ 2{ h }_{ 2 } } =\frac { 2 }{ 3 } \)
\(\frac { { r }_{ 1 }^{ 2 } }{ { r }_{ 2 }^{ 2 } } =\frac { 4 }{ 3 } \text {gives} \frac { { r }_{ 1 } }{ { r }_{ 2 } } =\frac { 2 }{ \sqrt { 3 } } \)
Therefore, ratio of their radii = 2 : \(\sqrt3\)

f(x) = 3 x -2, g(x) = 2x + k
f o g = f(g(x)) = f(2x + k) = 3(2x + k) - 2 = 6x + k - 2
Thus, f o g(x) = 6x + 3k - 2
g o f(x) = g(3x - 2) = 2(3x - 2) + k
Thus, g o f(x) = 6x - 4 + k
Given that f o g = g o f
Therefore, 6x + 3k-2 = 6z - 4 + k
6x - 6x + 3k - k = -4 + 2 ⇒ -1

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

2x − 3y = 6 … (1)
x + y = 1 … (2)


Substituting, y = \(\frac {-4}{5}\) in (2), x - \(\frac {4}{5}\) = 1 we get, x = \(​​\frac {9}{5}\)
Therefore, x = \(​​\frac {9}{5}\), y = \(\frac {-4}{5}\).

\(\sqrt { \frac { 1+cos\theta }{ 1-cos\theta } } \)=\(\sqrt { \frac { 1+cos\theta }{ 1-cos\theta } \times \frac { 1+cos\theta }{ 1+cos\theta } } \) [multiply numerator and denominator by the conjugate of 1 - cos\(\theta \)]
=\(\sqrt { \frac { (1+cos\theta { ) }^{ 2 } }{ (1-cos\theta { ) }^{ 2 } } } \) =\(\frac { 1+cos\theta }{ \sqrt { si{ n }^{ 2 }\theta } } \) [since sin²\(\theta \) + cos²\(\theta \) = 1]
=\(\frac { 1+cos\theta }{ sin\theta } =cosec\theta +cot\theta \)

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

Let BC be the height of the tower and CD be the height of the pole
Let ‘A’ be the point of observation.
Let BC = x and AB = y.
From the diagram,
ㄥBAD = 60° and ㄥXCA = 45° = ㄥBAC
In right triangle ABC, tan 45^o = \(\frac{BC}{AB}\)
gives 1 = \(\frac{x}{y}\) so, x = y         ...(1)
In right triangle ABD, tan60° = \(\frac{BC}{AB}\) = \(\frac{BC+CD}{AB}\)
gives \(\sqrt3\)  = \(\frac{x+5}{y}\) so, \(\sqrt3\)y = x + 5
we get \(\sqrt3\) x = x + 5 [From (1)]
so, \(\frac { 5 }{ \sqrt { 3 } -1 } =\frac { 5 }{ \sqrt { 3 } -1 } \times \frac { \sqrt { 3 } +1 }{ \sqrt { 3 } +1 } =\frac { 5(1.732+1) }{ 2 } \) = 6.83
Hence, height of the tower is 6.83 m.

From \(\Delta ADC\) we have
AC² = AD² + CD²       ....(1)
(Pythagoras theorem)
From \(\Delta ADB\) we have
AB² = AD² + BD²     ...(2)
(Pythagoras theorem)
Subtracting (1) from (2) we have,
AB² - AC² = BD² - CD²
AB² + CD² = BD² + AC²

No. of coins required \(=\frac{Volume\ of\ the\ cylinder}{Volume\ of\ 1\ coin}\)
\(=\frac { \pi { r }_{ 1 }^{ 2 }{ h }_{ 1 } }{ \pi { r }_{ 2 }^{ 2 }{ h }_{ 2 } } =\frac { \pi \times \frac { 42 }{ 20 } \times \frac { 45 }{ 20 } \times 10 }{ \pi \times \frac { 15 }{ 20 } \times \frac { 15 }{ 20 } \times \frac { 2 }{ 10 } } \)

= 450

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

We have
a₃ = a + (3 - 1)d = a + 2d = 5  (1)
a₇ = a + (7 - 1)d = a + 6d = 9 (2)
(1) - (2) ⇒ -4d -4 ⇒ d = 1.
Sub, d = 1 in (1), we get
a + 2(1) = 5
a = 3
Hence the required A.P. is 3, 4, 5, 6, 7.

R = {(a, -2), (-5, b), (8, c), (d, -1)} represents the identity function.
a = -2, b = -5, c = 8, d = -1.

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.

We know that diagonals at a parallelogram bisect each other.
So, the coordinates of the mid-point at AC = coordinates of the mid-point at BD.
i.e., \(\left[ \frac { 6+9 }{ 2 } ,\frac { 1+4 }{ 2 } \right] =\left[ \frac { 8+P }{ 2 } ,\frac { 2+3 }{ 2 } \right] \)
\(\left[ \frac { 15 }{ 2 } ,\frac { 5 }{ 2 } \right] =\left[ \frac { 8+P }{ 2 } ,\frac { 5 }{ 2 } \right] \)
\(\frac { 15 }{ 2 } =\frac { 8+P }{ 2 } \)
P = 7

We have x sin³\(\theta \) + ycos³\(\theta \) = sin\(\theta \) cos\(\theta \)
\( \Rightarrow \ (x \sin \theta)\left(\sin ^{2} \theta\right)+(y \cos \theta) \cos ^{2} \theta \)
\(= \sin \theta \cos \theta \)
\(\Rightarrow \ x \sin \theta\left(\sin ^{2} \theta\right)+(x \sin \theta) \cos ^{2} \theta \)
\(= \sin \theta \cos \theta \quad[\because x \sin \theta=y \cos \theta] \mathrm{S} \)
\(\Rightarrow \ x \sin \theta\left(\sin ^{2} \theta+\cos ^{2} \theta\right)=\sin \theta \cos \theta \)
\(\Rightarrow x \sin \theta=\sin \theta \cos \theta \)
\( \mathrm{x}=\cos \theta\)            ...(1)
\(\text { Now, } x \sin \theta=y \cos \theta \)
\( \Rightarrow \cos \theta \sin \theta=y \cos \theta \)
\(\Rightarrow [\because x=\cos \theta \text { from (1) }] \)
y = sin\(\theta \)          ...(2)
From (1) and (2)
\( x^{2}+y^{2}=\cos ^{2} \theta+\sin ^{2} \theta \)
= 1
x² + y² = 1.

Given, n = 5

Standard deviation
σ = \(\sqrt { \frac { { \Sigma d }_{ i }^{ 2 } }{ n } -\left( \frac { { \Sigma d }_{ i } }{ n } \right) ^{ 2 } } \)
σ = \(\sqrt { \frac { 151 }{ 5 } -\left( \frac { 25 }{ 5 } \right) ^{ 2 } } =\sqrt { 30.2-25 } =\sqrt { 5.2 } \) ≃ 2.28
When we multiply each data by 4, we get the new values as 8, 12, 20, 28, 32.

 Standard deviation
σ = \(\sqrt { \frac { { \Sigma d }_{ i }^{ 2 } }{ n } -\left( \frac { { \Sigma d }_{ i } }{ n } \right) ^{ 2 } } \)
= \(\sqrt { \frac { 2416 }{ 5 } -\left( \frac { 100 }{ 5 } \right) ^{ 2 } } =\sqrt { 483.2-400 } =\sqrt { 83.2 } \)
σ = \(\sqrt { 16\times 5.2 } =4\sqrt { 5.2 } \) ≃ 9.12

Let r be the radius of the hemisphere.
Given that, diameter of the hemisphere = side of the cube = l
Radius of the hemisphere = \(\frac{l}{2}\)
TSA of the remaining solid = Surface area of the cubical part + C.S.A. of the hemispherical part − Area of the base of the hemispherical part
= 6 x (Edge)² + 2\(\pi\)r²−\(\pi\)r²
= 6 x (Edge)² +  \(\pi\)r²
\(=6{ \times (l) }^{ 2 }+\pi { \left( \frac { l }{ 2 } \right) }^{ 2 }=\frac { 1 }{ 4 } (24+\pi ){ l }^{ 2 }\)
Total surface area of the remaining solid \(=\frac { 1 }{ 4 } (24+\pi ){ l }^{ 2 }\)sq. units

(i) To check if h is one – one, we assume that h(b₁) = h(b₂)
Then we get 2.47b₁ + 54.10 = 2.47b₂ + 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.
2. 47 = 147. 96 - 54. 10 = 93. 86
b = \(\frac { 93.86 }{ 2.47 } \)
Therefore, the length of the thigh bone is 38 cm.

x² + 2x + 5 = 0
Let y = x² + 2x + 5
Step 1 Prepare a table of values for the equation y = x² + 2x + 5

Step 2: Plot the above ordered pairs(x, y) on the graph using suitable scale.

Step 3: Join the points by a free-hand smooth curve this smooth curve is the graph of  y = x² + 2x + 5
Step 4: The solutions of the given quadratic equation are the x coordinates of the intersecting points of the parabola the X axis.
Here the parabola doesn’t intersect or touch the X axis.
So, we conclude that there is no real root for the given quadratic equation.