(A - B)^T= A^T-B^T
L.H.S = \((A-B)=\left[ \begin{matrix} 1 & 2 \\ 1 & 3 \end{matrix} \right] -\left[ \begin{matrix} 4 & 0 \\ 1 & 5 \end{matrix} \right] =\left[ \begin{matrix} -3 & 2 \\ 0 & -2 \end{matrix} \right] ...(2)\)
\({ (A-B) }^{ T }=\left[ \begin{matrix} -3 & 0 \\ 2 & -2 \end{matrix} \right] ...(1)\)
\({ A }^{ T }=\left[ \begin{matrix} 1 & 1 \\ 2 & 3 \end{matrix} \right] ,{ B }^{ T }=\left[ \begin{matrix} 4 & 1 \\ 0 & 5 \end{matrix} \right] \)
\(=\left[\begin{array}{rr} -3 & 0 \\ 2 & -2 \end{array}\right]\)
From (1) and (2)
(A - B)^T= A^T-B^T
Hence verified.

Let AB be the lamp post and CD be the apartment given CD = 66 m = EB.
\( \angle A C E=60^{\circ} \)
\(\angle E C B=\angle C B D=30^{\circ} \)
In the right triangle \(\triangle\)BDC
\(\tan 30^{\circ}=\frac{C D}{B D}\)
\( \frac{1}{\sqrt{3}}=\frac{66}{B D} \)
\(B D=66 \sqrt{3}\)
= 66 x 1.732 = 114.312 m
The distance between the lamp post and the apartment = 114.31 m
Now BD = EC = 114.31 m
In the right triangle \(\triangle\)ACE
\( \tan 60^{\circ} =\frac{A E}{C E} \)
\(\sqrt{3} =\frac{A E}{66 \sqrt{3}} \)
\(A E =66 \sqrt{3} \times \sqrt{3}[\text { From (1)] }\)
= 66 x 3 = 198m
The distance between the lamp post and apartment = 114.31 m.

2x² - x - 1 = 0 here, a = 2, b = -1, c = -1
α + β = \(\frac {-b}{a} = \frac {-(-1)}{2} = \frac {1}{2}\), αβ = \(\frac {c}{a} = -\frac {1}{2}\)
2α + β, 2β + α
Sum of the roots 2α + β + 2β + α = 3(α + β) = \(3\left( \frac { 1 }{ 2 } \right) =\frac { 3 }{ 2 } \)
Product of the roots = (2α + β) (2β + α) = 4αβ + 2α² + 2β² + αβ
= 5αβ + 2(α² + β²) = 5αβ + 2[(α + β)² - 2αβ]
= \(5\left( -\frac { 1 }{ 2 } \right) +2\left[ \frac { 1 }{ 4 } -2\times -\frac { 1 }{ 2 } \right] \) = 0
The required equation is x² - (Sum of the roots)x + (Product of the roots) = 0
x² - \(\frac { 3 }{ 2 } x\) + 0 = 0 gives 2x² - 3x = 0

∵ m-16 = 8
m = 8 + 16 = 24
n = -32

Total number of cards remaining
= 11 + 11 + 13 + 11 = 46
n(S) = 46
(i) Let A be the event of getting a clavor card
n(A) = 13
\(\mathrm{P}(\mathrm{A})=\frac{n(A)}{n(S)}=\frac{13}{46}\)
(ii) Let B be the event of getting queen of red card
n(B) = 0 (Removed red queens)
\(\mathrm{P}(\mathrm{B})=\frac{n(B)}{n(S)}=0\)
(iii) Let C be the event of getting a king of black card
n(C) = 1
\(\mathrm{P}(\mathrm{C})=\frac{n(C)}{n(S)}=\frac{1}{46}\)

Let a, b be the intercepts
Given, Sum of the intercept a + b = 1      .........(1)
Product of the intercepts ab = -6              .........(2)
\(b=\frac{-6}{a}\)
Substituting in (1)
\(a-\frac{6}{a}=1\)
a² - a - 6 = 0
(a - 3)(a + 2) = 0
a = -2, 3
When a = -2, \(b=\frac{-6}{-2}=3\)
When a = 3 , \(b=\frac{-6}{-2}=-2\)
Equation of the line in intercepts form is
\(\frac{x}{a}+\frac{y}{b}=1\)
(i) a = -2, b = 3 \(\Rightarrow \frac{x}{-2}+\frac{y}{3}=1\)
3x - 2y + 6 = 0
(ii) a = 3,b = -2 \(\Rightarrow \frac{x}{3}+\frac{y}{-2}=1\)
-2x + 3y + 6 = 0
= 2x - 3y - 6 = 0

Given lines are 5x - 6y - 2 and 3x + 2y = 10.
Let us solve the equations to get the point of intersection

\(x=\frac{32}{14}=\frac{16}{7} \)
Substituting \(x=\frac{16}{7} \text { in }(2) \)
\(3\left(\frac{16}{7}\right)+2 y =10 \)
\(2 y =10-\frac{48}{7}=\frac{22}{7} \)
\(y =\frac{11}{7} \)
The point of intersection is \(\left(\frac{16}{7}, \frac{11}{7}\right)\)
Slope of the line 4x - 7y + 13 = 0 is
\(-\frac{a}{b}=\frac{-4}{-7}=\frac{4}{7}\)
Slope of the perpendicular line is \(-\frac{7}{4}\)
Now, equation of the line passing through \(\left(\frac{16}{7}, \frac{11}{7}\right)\) and having slope \(m=-\frac{7}{4}\) is
\(y-y_{1} =m\left(x-x_{1}\right) \)
\(y-\frac{11}{7} =\frac{-7}{4}\left(x-\frac{16}{7}\right) \)
\(\frac{7 y-11}{7} =\frac{-7 x}{4}+4 \)
28y - 44 = -49x + 112
49x + 28y - 156 = 0

Given that AB = 13, AC = 14 and BC = 15
Let BD = x and DC = y
Using Ceva’s theorem, we have, \(\frac { BD }{ DC } \times \frac { CE }{ EA } \times \frac { AF }{ FB } =1\)
Substitute the values of  \(\frac { AF }{ FB } \ and \ \frac { CE }{ EA } \) in (1)
we have \(\frac { BD }{ DC } \times \frac { 5 }{ 8 } \times \frac { 2 }{ 5 } =1\)
\(\frac { x }{ y } \times \frac { 10 }{ 40 } =1\) we get \(\frac { x }{ y } \times \frac { 1 }{ 4 } \), Hence x = 4y  ...(2)
BC = BD + DC = 15 so, x + y = 15   ..(3)
From (2), using x = 4y in (3) we get, 4y + y = 15 gives 5y = 15 then y = 3
Substitute y = 3 in (3) we get, x = 12. Hence BD = 12, DC = 3.

Let h and r be the height and radius of the cylinder respectively.
Given that, h = 15 cm, r = 6 cm
Volume of the container V = \(\pi\)r²h cubic units.
Let, r₁ = 3 cm, h₁ = 9 cm be the radius and height of the cone.
Also, r₁ = 3 cm is the radius of the hemispherical cap.
Volume of one ice cream cone = (Volume of the cone + Volume of the hemispherical cap)
\(=\frac { 1 }{ 3 } \pi { r }_{ 1 }^{ 2 }{ h }_{ 1 }+\frac { 2 }{ 3 } \pi { r }_{ 1 }^{ 3 }\)
\(=\frac { 1 }{ 3 } \times \frac { 22 }{ 7 } \times 3\times 3\times 9+\frac { 2 }{ 3 } \times \frac { 22 }{ 7 } \times 3\times 3\times 3\)
\(=\frac { 22 }{ 7 } \times 9(3+2)=\frac { 22 }{ 7 } \times 45\)
\(Number\ of\ cones=\frac { volume\ of\ the\ cylinder }{ volume\ of\ one\ ice\ cream\ cone } \)
Number of ice cream cones needed \(=\frac { \frac { 22 }{ 7 } \times 6\times 6\times 15 }{ \frac { 22 }{ 7 } \times 45 } =12\)
Thus 12 ice cream cones are required to empty the cylindrical container.

Let the number of small spheres obtained be n.
Let r be the radius of each small sphere and R be the radius of metallic sphere.
Here, R = 16 cm, r = 2 cm
Now, n\(\times\)(Volume of a small sphere) = Volume of big metallic sphere
\(n\left( \frac { 4 }{ 3 } { \pi r }^{ 2 } \right) =\frac { 4 }{ 3 } \pi { R }^{ 3 }\)
\(n\left( \frac { 4 }{ 3 } \pi \times { 2 }^{ 3 } \right) =\frac { 4 }{ 3 } \pi \times { 16 }^{ 3 }\)
8n = 4096 gives n = 512
Therefore, there will be 512 small spheres.

f(x) = x², g(x) = 3x, h(x) = x - 2
f o g = f [g(x)] = f (3x)
= (3x)² - 9x²
(f o g) o h = (f o g) [h (x)] = (f  o g) [x - 2]
= 9 ( x - 2)²
g o h = g [h (x)] = g[x - 2] = 3 (x - 2)
f o (g o h) = f [g (h(x))]
= f [3(x - 2)] = [3(x - 2))]² = 9 (x - 2)²
(f o g) o h = f o( g o h)
Hence proved.

f(x) = \(\sqrt { 1+\sqrt { 1-\sqrt { 1-{ x }^{ 2 } } } } \)
\( f(x)=\sqrt{1-t} \)
\(where\ t=\sqrt{1-\sqrt{1-x^{2}}}\)
\(1-t \geq 0 \)
\(t \leq 1 \)
\(\sqrt{1-\sqrt{1-x^{2}}} \leq 1 \)
Squaring \(\sqrt{1-\sqrt{1-x^{2}}} \leq 1 \)
\(-\sqrt{1-x^{2}} \leq 0 \)
\(\sqrt{1-x^{2}} \geq 0 \)
\(1-x^{2} \geq 0 \)
\(x^{2} \leq 1 \)
= x [-1, 1] i.e., {- 1, 0, 1}

Let r₁, r₂ be the radii of two cones,
Given r₁ = r₁ and let h₁, h₂ be the heights of two
V₁ = Volumes of I cone = \(\frac{1}{3} \pi r_{1}^{2} h_{1}=\frac{\pi}{3} r_{1}^{2} h\)
= 3600 cm³
V₂ = Volume of II cone = \(\frac{1}{3} \pi r_{2}^{2} h_{2}=5040 \mathrm{~cm}^{3}\)
\(Now, \frac{V_{1}}{V_{2}}=\frac{3600}{5040} \Rightarrow \frac{\frac{\pi}{3} r_{1}^{2} h_{1}}{\frac{\pi}{3} r_{2}^{2} h_{2}}=\frac{3600}{5040}\left[\because r_{1}=r_{2}\right]. \)
\(\frac{h_{1}}{h_{2}}=\frac{5}{7}=5: 7\)

Pari: time required to complete the work = 4 hrs.
∴ In 1 hr. he will complete =\(\frac{1}{4}\) of the work.
=\(\frac{1}{4}\)w
Yuvan: Time required to complete the work = 6 hrs.
∴ In 1 hr. he will complete the \(\frac{1}{6}\) of the work
=\(\frac{1}{6}\)w
Working together, in 1 hr. they will complete \(\frac { w }{ 4 } +\frac { w }{ 6 } \) of the work.
\(=\frac { 6w+4w }{ 24 } =\frac { 5 }{ 12 } \)
∴ To complete the total work tame taken
\(=\frac { w }{ \frac { 5 }{ 12 } w } =\frac { 12 }{ 5 } =2.5\)hrs
= 2 hrs 24 minutes

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

Since, \(\triangle\)QAQC is a right triangle at C, AQ² = AC² + QC²  ...(1)
Also, \(\triangle\)BPC is a right triangle at C, BP² = BC²+ CP²        ...(2)
\(\triangle\) ABCC is a right triangle at C, AB² = AC² + BC²              ....(3)
From (1) and (2), AQ²+  BP² = AC²+ QC² + BC² + CP²
4(AQ² + BP²) = 4AC² + 4QC² + 4BC² + 4CP²
= 4AC² + (2QC)²+ ABC² + (2CP)²
=  4AC² + BC² + 4BC² + AC² (Since P and Q are mid points)
= 5(AC² + BC²) (From equation (3))
4(AQ² + BP²) = 5AB²

\(\frac { 4{ x }^{ 2 }y }{ 2{ x }^{ 2 } } \times \frac { 6x{ z }^{ 3 } }{ 20{ y }^{ 4 } } =\frac { { 3x }^{ 3 }z }{ 5{ y }^{ 3 } } \)

Let us take the four terms in the form (a - 3d), (a -d), (a + d) and (a + 3d).
Since sum of the four terms is 28,
a - 3d  + a - d + a + d + a + 3d = 28
4a = 28 gives a = 7
Similarly, since sum of their squares is 276,
(a - 3d)² + (a - d)² + (a + d)² + (a + 3d)² = 276
a² - 6ad + 9d² + a² - 2ad + d² + a² + 2ad  + d² + a² + 6ad  + 9d² = 276
4a² + 20d² = 276 \(\Rightarrow\) 4(7)² + 20d² = 276
d² = 4 gives d = \(\pm\)2
If d = 2 then the four numbers are 7 - 3(2), 7 - 2, 7 + 2, 7 + 3(2)
That is the four numbers are 1,5,9 and 13.
If a = 7, d = -2 then the four numbers are 13,9, 5 and 1
Therefore, the four consecutive terms of the  A.P are 1, 5, 9 and 13

Before determining the area of quadrilateral, plot the vertices in a graph.
Let the vertices be A(8, 6), B(5, 11), C(-5, 12) and D(-4, 3).
Therefore, area of the quadrilateral ABCD
=\(\frac{1}{2}\) { (x₁y₂ + x₂y₃ + x₃y₁) - (x₂y₁ + x₃y₂ + x₁y₃) }
=\(\frac{1}{2}\) { (80 + 60  - 15 - 24) - (30 - 55 - 48 + 24)}
=\(\frac{1}{2}\) {109 + 49 }
=\(\frac{1}{2}\) { 158 } = 79 sq. units

Given: A is the set of all natural numbers less than 8 A = {1,2,3,4,5,6,7}
B is the set of all prime numbers less than 8 B = {2,3,5,7}
'C' is the set of all even prime number
C = {2}
verify
(A ∩ B) x C = (A x C) ∩ (B x C)
\(A\cap B\) = {2,3,5,7}
\((A \cap B) \times C=\{2,3,5,7\} \times\{2\}\)
= {(2,2),(3,2),(5,2),(7,2}}    ......(1)
A X C = {1,2,3,4,5,6,7} x {2}
= {(1,2),(2,2),(3,2),(4,2),(5,2),(6,2),(7,2)}
B x C = {2,3,5,7} x {2}
= {(2,2),(3,2),(5,2),(7,2)}
\((A\times C)\cap (B\times C)\) = {(2,2) ,(3,2),(5,2),(7,2)}  ...(2)
From (1) and (2) it is clear that
(A ∩ B) x C = (A x C) ∩ (B x C)
Hence verified.

\(\left[ \begin{matrix} 12 & 3 \\ x & \frac { 3 }{ 2 } \end{matrix} \right] =\left[ \begin{matrix} y & z \\ 3 & 5 \end{matrix} \right] \)
x = 3
y = 12
z = 3

Given \(P\left( A \right) :P\left( \bar { A } \right) =17:15\)
(i) \(\frac{P(A)}{P(\bar{A})} =\frac{17}{15} \)
\(\frac{P(A)}{1-P(A)} =\frac{17}{15} \quad[\therefore P(\bar{A})=1-P(A)] \)
15 P(A) = 17 [1- P(A)]
15 P(A) = 17 - 17 P(A)
\(32 \mathrm{P}(\mathrm{A})=17 ; \quad \mathrm{P}(\mathrm{A})=\frac{17}{32} \)
\(P(\bar{A})=1-\mathrm{P}(\mathrm{A})=1-\frac{17}{32}=\frac{32-17}{32}=\frac{15}{32} \)
(ii) \(\mathrm{P}(\mathrm{A})=\frac{n(A)}{n(S)}=\frac{17}{32}\)
n(A) = 17

Ascending order : 24, 26, 29, 31, 33, 37
Standard deviation \(\sigma=\sqrt{\frac{\sum d_{i}^{2}}{N}-\left(\frac{\sum d_{i}}{N}\right)^{2}}\)
Let the assumed mean A = 31

xi	\(\mathrm{d}_{\mathrm{i}}=\mathrm{x}_{\mathrm{i}}-\mathrm{A} \) \(\mathrm{d}_{\mathrm{i}}=\mathrm{x}_{\mathrm{i}}-31 \)	\(\mathrm{d}_{\mathrm{i}}^{2}\)
24	-7	49
26	-5	25
29	-2	4
31	0	0
33	2	4
37	6	36
	\(\Sigma d_{i}\) = -6	\(\Sigma d_{i}^{2}\) = 118

\(\sigma=\sqrt{\frac{118}{6}-\left(\frac{-6}{6}\right)^{2}} \)
\(\sigma=\sqrt{19.67-(-1)^{2}}=\sqrt{19.7-1} \)
 \(\sigma=\sqrt{18.67}=4.3 \)
\(\text { Mean } \ \bar{x} =\frac{\sum x_{i}}{N} \)
\(\text { Mean } \ \bar{x} =\frac{24+26+29+31+33+37}{6} \)
\(\bar{x} =\frac{180}{6}=30 \)

 

 Now co- efficient of variation C.V \(=\frac{\sigma}{x} \times 100 \%\)
\(\mathrm{C.V}=\frac{4.3}{30} \times 100 \%=0.1433 \times 100 \%\)
= 14.33 %
Co-efficient of variation of the given data
= 14.33 %

 

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%

f(x) = x - 6, g(x) = x²
fog(x) = f(g(x)) = f(x²) = x² - 6 ...(1)
gof(x) = g(f(x)) = g(f(x)) = g(x - 6) = (x - 6)²
= x² - 12x + 36
fog(x) ≠ gof

Let the radius of the sphere be 'r' cm
Surface area = \(4 \pi r^{2}\)
when radius is increased by 25% ,then new diameter = r + 25% + r
\(=r+\frac{25 r}{100}=\frac{5 r}{4}\)
Surface area of new sphere
\(=4 \pi\left(\frac{5 r}{4}\right)^{2} \)
\(=4 \pi\left(\frac{25 r^{2}}{16}\right) \)
\(=\frac{25 \pi r^{2}}{4} \)
Increase in surface area = \(\frac{25 \pi r^{2}}{4}-4 \pi r^{2}\)
\(=\frac{25 \pi r^{2}-16 \pi r^{2}}{4} \)
\(=\frac{9 \pi r^{2}}{4} \)
Percentage increase in surface area
\(=\frac{9 \pi r^{2} / 4}{4 \pi r^{2}} \times 100 \% \)
\(=\frac{900}{16} \%=56.25 \% \)

In \(\triangle\)ABC, AD is the bisector of \(\angle\)A
Therefore by Angle Bisector of \(\angle\)A
\(\frac { AB }{ AC } =\frac { BD }{ DC } \)
\(\frac{4}{3}=\frac{6}{A C}\) gives 4AC = 18. Hence, AC \(=\frac{9}{2}=4.5 \mathrm{~cm}\)

\(\sqrt { \frac { 1+cos\theta }{ 1-cos\theta } } \)=\(\sqrt { \frac { 1+cos\theta }{ 1-cos\theta } \times \frac { 1+cos\theta }{ 1+cos\theta } } \) [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]
=\(\frac { 1+cos\theta }{ sin\theta } =cosec\theta +cot\theta \)

Step 1 : Draw the graph of y = x² - 4x + 3 by preparing the table of values as below

x	-2	-1	0	1	2	3	4
y	15	8	3	0	-1	0	3

Step 2 : To solve x² - 6x + 9 = 0, subtract x² - 6x + 9 = 0 from y  = x² - 4x + 3

The equation y = 2x - 6 represent a straight line. Draw the graph of y = 2x - 6 forming the table of values as below.

x	0	1	2	3	4	5
y	-6	-4	-2	0	2	4

The line y = 2x - 6 intersect y = x² - 4x + 3 only at one point.
Step 3 : Mark the point of intersection of the curve y = x² - 4x + 3 and y = 2x - 6 that is (3,0). 
Therefore, the x coordinate 3 is the only solution for the equation x² - 6x + 9 = 0

Step 1 : Draw the graph of y = x² + x - 2 by preparing the table of values as below

x	-3	-2	-1	0	1	2
y	4	0	-2	-2	0	4

Step 2 : To solve x² + x - 2 = 0 subtract x² + x - 2 = 0 from y = x² + x - 2

The equation y = 0 represents the X axis.
Step 3 : Mark the point of intersection of the curve x² + x - 2 with the X axis. That is (–2,0) and (1,0)
Step 4 : The x coordinates of the respective points form the solution set {−2,1} for x² + x - 2 = 0

The distance between the point from the centre is 10 cm.

Length of the tangents PA - PB = 8.7 cm
Construction:
Steps:
(1) With O as centre, draw a circle of radius 5cm.
(2) Draw a line OP = 10 cm.
(3) Draw a perpendicular bisector of OP which cuts OP at 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 and PB = 8.7 cm

Construction:
Step (1) Draw a line segment PQ = 4.5 cm
Step (2) At P, draw PE such that \(\angle QPE={ 35 }^{ 0 }\)
Step (3) At P, draw PF such that \(\angle EPF={ 90 }^{ 0 }\)
Step (4) Draw \(\bot \) bisector to PQ which intersects PF at O.
Step (5) With O centre OP as radius draw a circle.
Step (6) From G, marked arcs of radius 6 cm on the circle marked them as R and S.
Step (7) Joined PR and RQ. Then \(\triangle\)PQR is the required triangle
Step (8) \(\triangle\)PQS is the required triangle