Through O, draw PQIIBC so that P lies on AB and Q lies on DC
Now, PQ II BC
\(PQ\bot AB\quad PQ\bot OC\)
\(\left( \because \angle B={ 90 }^{ 0 }and\angle C={ 90 }^{ 0 } \right) \)
So, \(\angle BPQ={ 90 }^{ 0 }\quad \angle CQP={ 90 }^{ 0 }\)
Therefore BPQC and APQD are both rectangles. Now from \(\Delta OPB\)
OB² = BP² + OP²
Similarly from \(\Delta OQD\)
OD² = OQ² + DQ²
From \(\Delta OQC\)
OC²  = OQ² + CQ²
\(\Delta OAP\) we have
OA² = AP² +OP²
Adding (1) and (2)
OB² + OD² = BP² + OP² + OQ² + DQ²
(As BP = CQ and DQ = AP)
= CQ² + OP² + OQ² + AP²
= CQ² + OQ² + OP² + AP²
= OC²+ OA
[From (3) and (4)]

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

We know:
cot A = tan(90^o - A)
So,
cot 25° = tan (90° - 25°) = tan 65°
\(\frac { tan{ 65 }^{ o } }{ tan{ 25 }^{ o } } \) = \(\frac { tan{ 65 }^{ o } }{ tan{ 25 }^{ o } } \) = 1

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) =\(\frac { 1 }{ 4 } +\frac { 1 }{ 4 } =\frac { 1 }{ 2 } \).

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.

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

Let the numbers be ∝, β
Sum of the roots = ∝ + β = 15          ...(1)
\(\frac { 1 }{ \alpha } +\frac { 1 }{ \beta } =\frac { 3 }{ 10 } \quad \quad \quad ...(2)\)
\(\\ \frac { +\alpha }{ \alpha \beta } =\frac { 3 }{ 10 } \)
10(∝+ β)= 3∝β    ....(3)
30∝β=10x15=150
Products of the roots =∝β=50 ....(4)
∴ From (1) & (4), we have
x²-15x+50=0
(x-10)(x-5)=0⇒x=10,5

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.

-2, 2, -2, 2, -2
t₂ - t₁ = 2-(-2) = 4
t₃ - t₂ = -2 -2 = -4
t₄ - t₃ = 2 - (-2) = 4
It is not an A.P.

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

F(5) = 3x² - 10
= 3(5)²- 10 = 75 - 10 = 65

Since the given points are collinear, the area a the triangle formed by them must be 0, i.e.,
= \(\frac { 1 }{ 2 } \) [2(k + 3) + 4(-3 - 3) + 6(3 - k)] = 0
= \(\frac { 1 }{ 2 } \) [-4k] = 0
k = 0 
Area of ΔABC
= \(\frac { 1 }{ 2 } \)[2(0 + 3) + 4(-3 - 3) + 6(3 - 0)] = 0

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).

5² + 10² + 15² +...+ 105² = 52(1² + 2² + 3² +...+ 21²) 
= \(25\times \frac { 25\times \left( 21+1 \right) \left( 2\times 21+1 \right) }{ 6 } \)
= \(\frac { 25\times 21\times 22\times 43 }{ 6 } =82775\)

3 + 33 + 333 + ... to n terms.
Let S_n = 3 + 33 + 333 + ...upto n terms
= 3( 1 + 11 + 111 + ... to n terms)
= \(\frac { 3 }{ 9 } \) (9 + 99 + 999 + ... to n terms)
(multiply and divide by 9)
= \(\frac { 1 }{ 3 } \)3[(10 - 1) + (100 - 1) + (1000 - 1) + to n terms]
= \(\frac { 1 }{ 3 } \) [(10 - 1) + (10² - 1) + (10³ - 1) +... + n upto n terms]
= \(\frac { 1 }{ 3 } \) {[10 + 10² + 10³ +...upto n terms] - n}
10 + 10² +... is a G.P. with a = 10, r = 10.
\(\therefore S_{n} =\frac{a\left(r^{n}-1\right)}{r-1} \)
\(S_{n} =\frac{1}{3}\left\{\left[\frac{10\left(10^{n}-1\right)}{10-1}\right]-n\right\} \)
\(=\frac{1}{3}\left[\frac{10\left(10^{n}-1\right)}{9}-n\right] \)
\(=\frac{10}{27}\left(10^{n}-1\right)-\frac{n}{3}\)
3 + 33 + 333 + ... to n terms \(=\frac{10}{27}\left(10^{n}-1\right)-\frac{n}{3}\)

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) = \(\frac { 2 }{ 2\times 6 } =\frac { 1 }{ 18 } \)

If the roots are given, the quadratic equation is X² - (sum of the roots) x + product the roots =0. For the given equation.
x² - 6x -4 = 0
∝+β = -6
∝β = -4
∝²+β² = (∝+β)²-2∝β
= (-6)²-2(-4) = 36 + 8 = 44
∝²β² = (∝β)² = (-4)2 = 16
∴ The required equation = x²-44x+16 = 0

(i) Sathya
Sum of marks in 5 subjects = 460
Mean marks of sathya \(\bar{x}=\frac{\sum x_{i}}{N}=\frac{460}{5}=92\)
Standard deviation \(\sigma=4.6\)
Co-efficient of variation C.V  \(=\frac{\sigma}{x} \times 100 \%\)
\(=\frac{4.6}{92} \times 100 \%=\frac{460}{92} \% \)
\(=\frac{10}{2} \%=5 \% \)
C.V of Sathya = 5%
(ii) Vidhya
Sum of the marks in 5 subjects = 480
Mean marks of vidhya \(\bar{x}=\frac{\sum x_{i}}{N}=\frac{480}{5}=96\)
Standard deviation \(\sigma=2.4\)
Co-efficient of variation C.V. \(=\frac{\sigma}{x} \times 100 \%\)
\(=\frac{2.4}{96} \times 100 \%=\frac{240}{96} \%\)
\(=\frac{10}{4} \%=2.5 \%\)
C. V. of Vidhya = 2.5%
C. V of Vidhya < C. V. of Sathya
Vidhya is more consistent in her performance.

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\(\sqrt { 3 } m\)

Hollow Hemisphere 
Internal diameter = 6 cm
Internal radius 'r' = 3 cm
External diameter = 10 cm
External radius 'R' = 5 cm
\(\left.\begin{array}{l} \text { Volume of hemisphere (or) } \\ \text {Volume of material used } \end{array}\right\}=\frac{2}{3} \pi\left(\mathrm{R}^{3}-\mathrm{r}^{3}\right) \text { cu. units }\)
\(=\frac{2}{3} \pi\left(5^{3}-3^{3}\right) \)
\(=\frac{2}{3} \pi(125-27)=\frac{196 \pi}{3} \mathrm{~cm}^{3} \)
Cylinder
Diameter = 14 cm
radius = 7 cm
height = h
Volume of cylinder \(=\pi r^{2} h\ cu. units \)
\(=\pi(7)^{2} h \)
\(=49 \pi h \mathrm{~cm}^{3} \)
Given that hollow hemisphere is melted and cast into a solid cylinder
Volume of cylinder = volume of hollow hemisphere
\(49 \pi h =\frac{196 \pi}{3} \)
\(h =\frac{196}{3 \times 49}=\frac{4}{3}=1.33 \)
Height of the cylinder = 1.33 cm.

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.

f: [-5,9] ⟶ R
(i) f(-3) + f(2)
= [6(-3) + 1 ] + [ 5(2)² - 1]
= ( -18 + 1) + ( 20 - 1)
 = -17 + 19 = 2.
(ii) f(7) - f(1)
= [ 3(7) - 4 ] - [6(1) + 1 ]
= (21 - 4) - (6 + 1)
=17 - 7 = 10
(iii) 2 f(4) + f(8)
= 2 [ 5(4)² - 1] + [3(8) - 4]
= 2[80 - 1] + [ 24 - 4]
= 158 + 20 = 178
(iv) \(\frac { 2f(-2)-f(6) }{ f(4)+f(-2) } \)
f(-2) = 6x + 1 = 6(-2) + 1 = -11
f(6) = 3x - 4 = 3(6) - 4 = 14
f(4) = 5x² - 1 = 5(4²) - 1 = 79
f(-2) = 6x + 1 = 6(-2) + 1 =-11
\(\frac { 2f(-2)-f(6) }{ f(4)+f(-2) } =\frac { 2(-11)-14 }{ 79+(-11) } =\frac { -22-14 }{ 68 } \)
= \(\frac { -36 }{ 68 } =\frac { -9 }{ 17 } \)

\(\frac { sinA }{ secA+tanA-1 } +\frac { cosA }{ cosecA+cotA-1 } =1\) 
\(=\frac { sinA(cosecA+cotA-1)+cosA(secA+tanA-1) }{ (secA+tanA-1)(cosecA+cotA-1) } \)
=\(\frac { sin\ A \ cosec \ A \ + \ sin \ A \ cot \ A \ - \ sin \ A\ +cos \ A \ sec \ A \ + cos \ A \ tan \ A \ - \ cos \ A }{ (sec \ A \ + \ tan \ A-1)(cosec \ A \ + \ cot \ A-1) } \)
=\(\frac { 1+cosA-sinA+1+sinA-cosA }{ \left( \frac { 1 }{ cosA } +\frac { sinA }{ cosA } -1 \right) \left( \frac { 1 }{ sinA } +\frac { cosA }{ sinA } -1 \right) } \)
=\(\frac { 2 }{ \left( \frac { 1+sinA-cosA }{ cosA } \right) \left( \frac { 1+cosA-sinA }{ sinA } \right) } \)
=\(\frac { 2sinAcosA }{ (1+sinA-cosA)(1+cosA-sinA) } \)
=\(\frac { 2 \ sin \ A \ cos \ A }{ [1+(sin \ A- \ cos \ A)][1-(sin \ A-cos \ A)] } =\frac { 2sinAcosA }{ 1-(sin \ A- \ cos \ A{ ) }^{ 2 } } \)
=\(\frac { 2sinAcosA }{ 1-(si{ n }^{ 2 }A+co{ s }^{ 2 }A-2sinAcosA) } =\frac { 2sinAcosA }{ 1-(1-2sinAcosA) } \)
=\(\frac { 2sinAcosA }{ 1-1+2sinAcosA } =\frac { 2sinAcosA }{ 2sinAcosA } =1.\)

Given ABC - DEF
then we have
\(\frac{\operatorname{Area}(\Delta \mathrm{ABC})}{\text { Area }(\Delta \mathrm{DEF})}=\frac{A B^{2}}{D E^{2}}=\frac{B C^{2}}{E F^{2}}=\frac{A C^{2}}{D F^{2}} \)
\(\frac{9}{16}=\frac{B C^{2}}{E F^{2}} \)
\(\frac{9}{16} =\frac{2.1 \times 2.1}{E F^{2}} \)
\(\mathrm{EF}^{2} =\left(\frac{2.1 \times 4}{3}\right)^{2} \)
\(\mathrm{EF} =\frac{2.1 \times 4}{3}=2.8 \)
EF = 2.8 cm

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

Since the three points P(-1, -4), Q(b, c) and R(5, -1) are collinear,
Area of triangle PQR = 0
\(\frac{1}{2}\) { (x₁y₂ + x₂y₃ + x₃y₁) - (x₂y₁ + x₃y₂ + x₁y₃) } = 0
\(\frac{1}{2}\) { (- c - b - 20) - (- 4b + 5c + 1) } = 0
- c - b - 20 + 4b - 5c - 1 = 0
b - 2c = 7 ...(1)
Also, 2b + c = 4 .....(2) (from given information)
Solving (1) and (2) we get b = 3, c = -2

kx² - (8k + 4) + 81 = 0
Since the equation has real and equal roots, Δ = 0
That is, b² - 4ac = 0
Here, a = k, b = -(8k + 4), c = 81
That is, [-(8k + 4)]² - 4(k)(81) = 0
64k² + 64k + 16 - 324k = 0
64k² - 260k + 16 = 0
dividing by 4 we get 16k² - 65k + 4 = 0
(16k - 1)(k - 4) = 0 then, k = \(\frac {1}{16}\) or k = 4

We know that the tangents drawn from are external point to a circle are equal.
Therefore AD AF = x 
BD = BE = y
and CE = CF = z 
Now, AB = 12 cm, BC = 8 cm, and CA = 10 cm.
x + y = 12,y + z = 8 and z + x = 10
(x + y) + (y + z) + (z + x) = 12 + 8 + 10
2(x + y + z) = 30
x+  y + z = 15
Now, x + y = 12 and x + y + z = 15
12 + z  = 15
Z = 3
y + z = 8 and x + y + z = 15
\(x+8=15\Rightarrow x=7\)
and z + x  = 10 and x + y + z = 15
\(10+y=15\Rightarrow y=5\)
Hence, AD = x = 7cm,
BE = y = 5 cm and
CF = z = 3 cm

The function f : N ⟶ N is defined by f(x) = 3x + 2
(i) If x = 1, f(1) = 3(1) + 2 = 5
If x = 2, f(2) = 3(2) + 2 = 8
If x = 3, f(3) = 3(3) + 2 = 11
The images of 1, 2, 3 are 5, 8, 11 respectively.
(ii) If x is the pre-image of 29, then f(x) = 29, Hence 3x + 2 = 29
3x = 27 ⇒ x = 9
Similarly, if x is the pre-image of 53, then f(x) = 53. Hence 3x + 2 = 53
3x = 51 ⇒ x = 17
Thus the pre-images of 29 and 53 are 9 and 17 respectively.
(iii) Since different elements of N have different images in the co-domain, the function f is a one-one function.
The co-domain of f is N.
But the range of f = {5, 8, 11, 14, 17,....} is a proper subset of N.
Therefore f is not an onto function. That is, f is an into function.
Thus f is a one-one and into function.