A table

x	0	1	2	3
f(x)	1	3	5	7

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)}

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}

tan⁴\(\theta \) + tan²\(\theta \) = sec⁴\(\theta \) - sec²\(\theta \) 
L.H.S = tan²θ (tan²θ + 1)
= tan²θ.sec²θ 
= sec⁴θ - sec²θ
= R.H.S

2x² - 2x + 9 = 0
Here, a = 2, b = -2, c = 9
Now, Δ = b² - 4ac = (-2)² - 4(2)(9) = -68
Here, Δ = -68 < 0. So, the equation will have no real roots.

\(\frac{t_{2}}{t_{1}}=\frac{44}{4}=11 \)
\(\frac{t_{3}}{t_{2}}=\frac{444}{44}=\frac{111}{11} \)
\(\frac{t_{4}}{t_{3}}=\frac{4444}{444}=\frac{1111}{111} \)
The ratios between the successive terms are not equal. Therefore the sequence 4, 44, 444,... are not in G.P.

\(\sqrt { { 16x }^{ 4 }+{ 8x }^{ 2 }+1 } \)

\(\therefore \sqrt { { 16x }^{ 4 }+{ 8x }^{ 2 }+1 } \)|4x² + 1|

Given points (2, 3) and (- 7, - 1)
Equation of the line passing through (x₁ , y₁) and (x₂, y₂) is
\( \frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}} \)
\(\frac{y-3}{-1-3}=\frac{x-2}{-7-2} \)
9 (y - 3) - 4 (x - 2)
9y - 27 = 4 x - 8
4 x - 9y +19 = 0

Slope 'm' = 1
tanθ = 1 = tan 45⁰
θ = 45⁰

t₂ - t₁ = 5\(\sqrt { 2 } \) - 3\(\sqrt { 2 } \) = 2\(\sqrt { 2 } \)
t₃ - t₁ = 7\(\sqrt { 2 } \) - 5\(\sqrt { 2 } \) = 2\(\sqrt { 2 } \)
t₄ - t₃ =9\(\sqrt { 2 } \) - 7\(\sqrt { 2 } \) = 2\(\sqrt { 2 } \) 
Thus, the differences between consecutive terms are equal. Hence the terms of the sequence 3\(\sqrt { 2 } \), 5\(\sqrt { 2 } \), 7\(\sqrt { 2 } \), 9\(\sqrt { 2 } \) ..... are in A.P

(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

Standard deviation \(\sigma=6.5\)
Mean \(\bar{x}=12.5\)
Coefficient of variation C.V  \(=\frac{\sigma}{x} \times 100 \% \)
\(=\frac{6.5}{12.5} \times 100 \% \)
\(=\frac{65}{125} \times 100 \% \)
\(=\frac{13}{25} \times 100 \% \)
= 52 %
Co-efficient of variation is 52%

Let AD is the first building.
BC is the second building.
AD = BE = BC - CE
o
From the right triangle CED
\(\tan 45^{\circ} =\frac{C E}{D E} \)
\(1 =\frac{C E}{A B}=\frac{C E}{70 m} \)
CE = 70 m
BE = BC - EC
= 120 m - 70 m = 50 m
AD = 50 m
Height of the first building is 50 m.

Show of the straight lines x - 2y + 3 = 0 is
m₁ = \(\frac{-1}{-2}=\frac{1}{2}\)
Slope of the straight line 6x + 3y + 8 = 0 is
m₂ = \(\frac{-6}{3}\)=-2
Now, m₁ x m₂ = \(\frac12\) x (-2) = -1
Hence, the two straight lines are perpendicular

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

We have AB = 56.cm, AD = 14. cm, AC = 72. cm and AE = 18.cm.
BD = AB - AD = 5.6 –1.4 = 4.2 cm
and EC = AC – AE = 7.2–1.8 = 5.4 cm
\(\frac { AD }{ DB } =\frac { 1.4 }{ 4.2 } =\frac { 1 }{ 3 } \) and \(\frac { AE }{ EC } =\frac { 1.8 }{ 5.4 } =\frac { 1 }{ 3 } \)
\(\frac { AD }{ DB } =\frac { AE }{ EC } \)
Therefore, by converse of Basic Proportionality Theorem, we have DE is parallel to BC. Hence proved.

f(x) = x² - 2 where x \(\in \){ -2, -1, 0, 3}
(i) f( -2) = ( -2)² - 2 = 2; f( -1) = ( -1)² - 2 = -1
f(0) = (0)² - 2 = - 2 ; f(3) = (3)² - 2 = 7
Therefore, f = {(-2, 2), (-1, -1), (0, -2), (3, 7)}
(ii) We note that each element in the domain of f has a unique image. Therefore f is a function.

In Is \(\triangle\)ABC ~ \(\triangle\)PQR
\(\frac { PQ }{ AB } =\frac { 3 }{ 6 } =\frac { 1 }{ 2 } ;\frac { QR }{ BC } =\frac { 4 }{ 10 } =\frac { 2 }{ 5 } \)
Since \(\frac { 1 }{ 2 } \neq \frac { 2 }{ 5 } ,\frac { PQ }{ AB } \neq \frac { QR }{ BC } \)
The corresponding sides are not proportional.
Therefore \(\triangle\)ABC  is not similar to \(\triangle\)PQR.

(a) Capital employed - Goodwill + current liabilities

           

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.

-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.

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

Let A be the initial position of the bird.
B be the position after travelling 30 km at an angle of 35^o from A.
Let C be the position after travelling 32 km at an angle of 48^o from B.
[complementary angle]
From the right triangle AOB
\(\cos 55^{\circ}=\frac{O A}{A B} \)
\(0.5736=\frac{O A}{30} \)
OA = 30 x 0.5736 = 17.21 km
B is 17 .21 km to the West of A

City A			City B
x1	d1 = x - \({ \overline { x } }_{ 1 }\)	d12	x2	d2 = x - \({ \overline { x } }_{ 2 }\)	d22
20	0	0	10	-5	25
22	2	4	20	5	25
19	-1	1	18	3	9
23	-3	9	12	-3	9
16	-4	16	15	0	0
100	6	30	75	0	68

\(\bar { { x }_{ 1 } } =\frac { 100 }{ 5 } =20\)
\(\sigma =\sqrt { \frac { \sum { { d }^{ 2 } } }{ n } } \)
\(=\sqrt { \frac { 30 }{ 5 } } \)
\(=\sqrt { 6 } \)
= 2.44
\(CV=\frac { \sigma }{ \bar { x } } \times 100\)
\(=\frac { 2.44 }{ 20 } \times 100\)
= 12.2
\(\bar { { x }_{ 2 } } =\frac { \sum { x } }{ n } =\frac { 75 }{ 5 } =15\)
\(\sigma =\sqrt { \frac { \sum { { d }^{ 2 } } }{ n } } \)
\(=\sqrt { \frac { 68 }{ 5 } } \)
\(=\sqrt { 13.6 } \)
= 3.68
\(CV=\frac { \sigma }{ \bar { x } } \times 100\)
\(=\frac { 3.86 }{ 15 } \times 100\)
\(=\frac { 368 }{ 15 } \)
C.V. of city A < C.V of city B.
∴ City A is more consistents.

Mean \(\overline { x } \) = 62.8
Σx = 50

Class interval	Mid value of x1	Frequency f1	Σf1x1
0-20	10	5	50
20-40	30	f1	30f1
40-60	50	10	500
60-80	70	f2	70f2
80-100	90	7	630
100-120	110	8	880
		30+f1+f2	2060+30f1+70f2

\(\overline { x } =\frac { \sum { { f }_{ i }{ x }_{ i } } }{ \sum { { f }_{ i } } } =\frac { 2060+30{ f }_{ 1 }+70{ f }_{ 2 } }{ 30+{ f }_{ 1 }+{ f }_{ 2 } } \)
\(\frac { 2060+30{ f }_{ 1 }+70{ f }_{ 2 } }{ 30+{ f }_{ 1 }+{ f }_{ 2 } } \) = 62.8 ....(1)
30 + f₁ + f₂ = 50 (given)
f₁ +f₂ = 20  ...(2)
2060 + 30f₁ + 70f₂ = 3140
30f₁ + 70f₂ = 3140 - 2060
30f₁ + 70f₂ = 1080  ....(3)
Solving (2) & (3) we get,

Sub.f₂ = 12 in (2), we get
f₂ +12 = 20 ⇒ f₁ = 8
f₁ = 8, f₂ = 12

3A + 2B - C = \(3\left[ \begin{matrix} 1 & 8 & 3 \\ 3 & 5 & 0 \\ 8 & 7 & 6 \end{matrix} \right] +2\left[ \begin{matrix} 8 & -6 & -4 \\ 2 & 11 & -3 \\ 0 & 1 & 5 \end{matrix} \right] -\left[ \begin{matrix} 5 & 3 & 0 \\ -1 & -7 & 2 \\ 1 & 4 & 3 \end{matrix} \right] \)
= \(\left[ \begin{matrix} 3 & 24 & 9 \\ 9 & 15 & 0 \\ 24 & 21 & 18 \end{matrix} \right] +\left[ \begin{matrix} 16 & -12 & -8 \\ 4 & 22 & -6 \\ 0 & 2 & 10 \end{matrix} \right] +\left[ \begin{matrix} -5 & -3 & 0 \\ 1 & 7 & -2 \\ -1 & -4 & -3 \end{matrix} \right] \)
= \(\left[ \begin{matrix} 14 & 9 & 1 \\ 14 & 44 & -8 \\ 23 & 19 & 25 \end{matrix} \right] \)

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.

Let us solve 8x + 3y = 18 and 4x + 5y = 9
8x + 3y = 18         ....(1)
4x + 5y = 9         ....(2)  

Substitute in (1)
8x + 3(0) = 18

The point of intersection of (1) and (2) is \(\left( \frac { 9 }{ 4 } ,0 \right) \)
Mid point of the line segment joining (5, -4) and (-7,6) is
\(\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)=\left(\frac{5-7}{2}, \frac{-4+6}{2}\right)=(-1,1)\)
Equation of the straight line joining \(\left( \frac { 9 }{ 4 } ,0 \right) \) and (-1, 1) is
\(\frac{y-y_{1}}{y_{2}-y_{1}} =\frac{x-x_{1}}{x_{2}-x_{1}} \)
\(\frac{y-0}{1-0} =\frac{x-\frac{9}{4}}{-1-\frac{9}{4}} \)
\(\frac{y}{1} =\frac{4 x-9}{-13} \)
-13y = 4x - 9
4x + 13y - 9 = 0

Coin is in the form of a cylinder
Diameter of the coin = 1.5 cm
Radius of the coin = \(\frac{1.5}{2}\)
Thickness = height = 2 mm = \(\frac{2}{10}=0.2 \mathrm{~cm}\)
Volume of coin (cylinder) = \(\pi r^{2} h\)
\(=\pi\left(\frac{1.5}{2}\right)^{2}(0.2) \)
\(=0.1125 \pi \mathrm{cm}^{3} \)
Diameter of cylinder = 4.5 cm
radius = \(\frac{4.5}{2}=2.25 \mathrm{~cm}\)
height = 10 cm
volume = \(\pi r^{2} h\ sq. units \)
= \(\pi(2.25)^{2}(10) \)
= \(50.625 \pi \)
No.of coins \(=\frac{\text { Volume of cylinder }}{\text { Volume of Coin }} \)
\(=\frac{50.625 \pi}{0.1125 \pi}=450 \text { coins. } \)

The mean of marks is 35.9 which is not an integer. Hence we take assumed mean, A=35,n=10.

xi	di = xi - A di = x - 35	di2
25	-10	100
29	-6	36
30	-5	25
33	-2	4
35	0	0
37	2	4
38	3	9
40	5	25
44	9	81
48	13	169
	\({ \Sigma d }_{ i }\) = 9	\(\frac { { \Sigma d }_{ i }^{ 2 } }{ n } \) = 453

Standard deviation
σ =\(\sqrt { \frac { { \Sigma d }_{ i }^{ 2 } }{ n } -\left( \frac { { \Sigma d }_{ i } }{ n } \right) ^{ 2 } } \)
= \(\sqrt { \frac { 453 }{ 10 } -\left( \frac { 9 }{ 10 } \right) ^{ 2 } } \)
= \(\sqrt { 45.3-0.81 } \)
= \(\sqrt { 44.49 } \)
σ ≃ 6.67
(ii) Step deviation method
Let x₁, x₂, x_n,... be the given data. Let A be the assumed mean.
Let c be the common divisor of x_i-A
Let d_i = \(\frac { x_{ i }-A }{ c } \)
Then x_i = d_ic + A             ....(1)
Σx_i = Σ(d_ic + A) = cΣd_i + A x n
\(\frac { \Sigma { x }_{ i } }{ n } =c\frac { \Sigma d_{ i } }{ n } +A\)
\(\bar { x } \) = c \(\bar { d } \)+A            ..(2)
\({ x }_{ i }-\bar { x } =c{ d }_{ i }+A-c\bar { d } -A=c(\bar { d_{ i } } -\bar { d } )\)(using (1) and (2))
σ = \(\sqrt { \frac { \Sigma ({ x }_{ i }-\bar { x } )^{ 2 } }{ n } } =\sqrt { \frac { \Sigma (c({ d }_{ i }-\bar { d } )^{ 2 } }{ n } } =\sqrt { \frac { { c }^{ 2 }\Sigma ({ d }_{ i }-\bar { d } )^{ 2 } }{ n } } \)
σ = \(c\times \sqrt { \frac { { \Sigma d }_{ i }^{ 2 } }{ n } -\left( \frac { { \Sigma d }_{ i } }{ n } \right) ^{ 2 } } \\ \)

Let TR = y. Since, OT is perpendicular bisector of PQ
PR = QR = 4 cm
In\(\triangle\)ORP, OP² = OR² + PR²
OR² = OP² - PR²
OR² = 5² - 4² = 25 - 16 = 9 \(\Rightarrow\) OR = 3cm
OT = OR + RT = 3 + y          ..(1)
In \(\triangle\)PRT, TP² + TR² + PR²      ..(2)
and \(\triangle\)OPT we have, OT² = TP² + OP²
OT² = (TR² + PR²) + OP² (substitute for TP² from (2))
(3 + y)² = y² + 4² + 5² (substitute for OT from (1))
9 + 6y² + 16 + 25 therefore \(y=TR=\frac { 16 }{ 3 } \)
6y = 41 - 9 we get  \(y=\frac { 16 }{ 3 } \)
From (2), TP² = TR² + PR²
\(TP2=\left( \frac { 16 }{ 3 } \right) +4^{ 2 }=\frac { 256 }{ 9 } +16=\frac { 400 }{ 9 } \)so, \(TP=\frac { 20 }{ 3 } \)

Diameter of Glass = 20 cm
radius = 10 cm
water upto height = 9 cm
radius of cylindrical metal = 5 cm
height of cylindrical metal = 4 cm
Volume of water displaced = Volume of cylindrical metal
\(\pi r_{1}^{2} h_{1}=\pi r_{2}^{2} h_{2} \)
\((10)^{2} h_{1}=(5)^{2}(4) \)
\(h_{1}=\frac{100}{100}=1 \mathrm{~cm} \)
Hence, the increase in water level is 1 cm.

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.

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

Given: A quadrilateral ABCD in which AB = AD and the bisectors of \(\angle\)BAC and \(\angle\)CAD meet the sides BC and CD at E and F respectively.
To prove: EF || BD
Construction: Join AC, BD and EF.
Proof: In \(\triangle\)CAB, AE is the bisector of \(\angle\)BAC
\(\therefore \frac{A C}{A B}=\frac{C E}{B E}\)            ......(1)
In \(\triangle\)ACD , AF is the bisector of \(\angle\)CAD
\( \frac{A C}{A D} =\frac{C F}{D F} \)
\(\Rightarrow \ \frac{A C}{A B} =\frac{C F}{D F}\)     ......(2)
From (1) and (2) we get
\( \frac{C E}{B E}=\frac{C F}{D F} \)
\(\frac{C E}{E B}=\frac{C F}{F D}\)
Thus in \(\triangle\)CBD, E and F divide the sides CB and CD respectively in the same ratio.
By the converse of Thales theorem, we have EF || BD.

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.

(cosec\(\theta \) - sin\(\theta \)) (sec\(\theta \) - cos\(\theta \)) (tan\(\theta \) + cot\(\theta \))
\(\left( \frac { 1 }{ sin\theta } -sin\theta \right) \quad \left( \frac { 1 }{ cos\theta } -cos\theta \right) \left( \frac { sin\theta }{ cos\theta } +\frac { cos\theta }{ sin\theta } \right) \)
\(=\frac { 1-si{ n }^{ 2 } }{ sin\theta } \times \frac { 1-co{ s }^{ 2 } }{ cos\theta } \times \frac { si{ n }^{ 2 }\theta +co{ s }^{ 2 }\theta }{ sin\theta cos\theta } \)
\(=\frac { co{ s }^{ 2 }\theta si{ n }^{ 2 }\theta \times 1 }{ si{ n }^{ 2 }\theta co{ s }^{ 2 }\theta } =1\)

\(A=\left[ \begin{matrix} 3 & 0 \\ 4 & 5 \end{matrix} \right] ,B=\left[ \begin{matrix} 6 & 3 \\ 8 & 5 \end{matrix} \right] ,C=\left[ \begin{matrix} 3 & 6 \\ 1 & 1 \end{matrix} \right] \)
CD - AB = 0 ⇒ CD = AB
\(AB=\left[ \begin{matrix} 3 & 0 \\ 4 & 5 \end{matrix} \right] \left[ \begin{matrix} 6 & 3 \\ 8 & 5 \end{matrix} \right] =\left[ \begin{matrix} (18+0) & (9+0) \\ (24+40) & (12+25) \end{matrix} \right] \)
\(CD=\left[ \begin{matrix} 18 & 9 \\ 64 & 37 \end{matrix} \right] \)
\(Let\quad D=\left[ \begin{matrix} x & y \\ z & w \end{matrix} \right] \)
\(\left[ \begin{matrix} 3 & 6 \\ 1 & 1 \end{matrix} \right] \left[ \begin{matrix} x & y \\ z & w \end{matrix} \right] =\left[ \begin{matrix} 18 & 9 \\ 64 & 37 \end{matrix} \right] \)
\(\left[ \begin{matrix} 3x+6z & 3y+6w \\ x+z & y+w \end{matrix} \right] =\left[ \begin{matrix} 18 & 9 \\ 64 & 37 \end{matrix} \right] \)
3x + 6z = 18      ...(1)
x + z = 64    ...(2)

Sub. x=122 in(2)
122+z=64
z=64-122=-58
3y+6w=9  ....(3)
 y+w=37   ....(4)

Sub. w = -34 in (4)
y-34 = 37
y = 37 + 34 = 71
∴ Solutions: x = 122
y = 71
z = -58
w = -34
\(\therefore D=\left[ \begin{matrix} 122 & 71 \\ -58 & -34 \end{matrix} \right] \)

\(60-t=\frac { { t }^{ 2 } }{ 4 } -3\)
⇒ t²-12 = 240-4t
⇒ t²+4t-252 = 0
⇒ t²+18t-14t-252 = 0
⇒ t(t +18)-14(t +18) = 0
⇒ (t +18)(t-14) = 0
∴ t = 14 or t = -18 is not possible

Diameter = 6 cm
Radius =\(\frac { 6 }{ 2 } =3cm\) 

Length of the tangents PA = PB = 4 cm
Construction:
Steps:
(1) With centre O, draw a circle of radius 3cm.
(2) Draw a line OP = 5 cm
(3) Draw a bisector of OP, which cuts OP and 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 = 4 cm.

Construction:
Steps (1) Draw QR = 6.5 cm.
Steps (2) Draw \(\angle RQE={ 60 }^{ 0 }\)
Steps (3) Draw \(\angle FQE={ 90 }^{ 0 }\)
Steps (4) Drawn the perpendicular bisector XY to ER which intersects QF at O and ER at G.
Steps (5)  With O as center and OQ as radius drawn a circle
Steps (6) XY intersects QR at G. On XY, from G marked an arc at M, such that GM = 4.5 cm
Steps (7) Drawn AB through M which is parallel to QR
Steps (8)  AB meets the circle at P and S
Steps (9) Joined QP and RP Then \(\triangle\)PQR is the required triangle.
Steps (10) Here \(\triangle\)SQR is also another required triangle.