(i) Given 3x+2y=k
Put x=2, y=1 then
3(2)+2(1)=k \(\Rightarrow\) k=8
(ii) Given, 2x-ky=6
Put x=2, y=1, then
2(2)-k(1)=6
\(\Rightarrow\) 4-k=6\(\Rightarrow\) k=4-6=-2
(iii) Given, \(\frac{x}{4}+\frac{y}{3}=5k\)
Put x=2, y=1, then
\(\frac{2}{4}+\frac{1}{3}=5k\)
\(\Rightarrow 5k=\frac{10}{12}=\frac{5}{6}\)
\(\Rightarrow k=\frac{1}{6}\) 

Given: The diagonals AC and BD of a quadrilateral ABCD are equal and bisect each other at right angles.
To Prove: Quadrilateral ABCD is a square.
Proof: In \(\Delta\)OAD and \(\Delta\)OCB,
OA = OC             I Given
OD = OB       I Given
\(\angle AOD=\angle COB\)    |Vertically Opposite Angles
 \(\therefore \Delta OAD\cong \Delta OCB\)    I SAS Congruence Rule

\(\therefore \)   AD = CB           I C.P.C.T.
 \(\angle ODA=\angle OBC\)  |C.P.C.T
\(\angle BDA=\angle DBC\)
.\(\therefore \)   AD II BC
Now, \(\because\) AD = CB and AD II CB
\(\therefore \)  Quadrilateral ABCD is a II gm.      I A quadrilateral is a parallelogram if a pair of opposite sides are parallel and  equal.
In \(\therefore \) \(\Delta\) AOB and \(\Delta\)AOD,
AO = AO                          I Common
OB = OD          I Given
 \(\angle \)AOB = \(\angle \)AOD               |Each=\(90°\)
\(\therefore \)  \(\Delta\)AOB \(\cong \)  \(\Delta\) AOD         I SAS Congruence Rule
AB = AD                 I C.P.C.T.
Now, \(\because\) ABCD is a parallelogram and
AB=AD
 \(\because\) ABCD is a rhombus.
Again, in \(\Delta\) ABC and \(\Delta\) BAD,
AC = BD            I Given
BC = AD            I \(\because\)  ABCD is a rhombus
AB =  BA                     I Common
\(\therefore \) \(\Delta\) ABC \(\cong \) \(\Delta\) BAD       I SSS Congruence Rule
\(\therefore \)    \(\angle \) ABC = \(\angle\) BAD    I C.P.C.T.
\(\because\)    AD II BC                 I Opp. sides of IIgm ABCD and transversal AB intersects them.
\(\therefore \) \(\angle ABC+\angle BAD=180°\)      I Sum of consecutive interior angles on the  same side of a transversal is 180°
ABC = \(\angle\)BAD = \(90°\)
Similarly,\(\angle\)BCD =\(\angle\)ADC =\(90°\)
 \(\therefore \)  ABCD is a square.

Yes! These postulates contain two undefined terms: Point and Line. Yes! These postulates are consistent because they deal with two different situations
(i) says that given two points A and B, there is a point C lying on the line in between them,
(ii) says that given A and B, we can take C not lying on the line through A and B. These 'postulates' do not follow from Euclid's postulates, however, they follow from Axiom 'Given two distinct lines, there is a unique line that passes through them.

(0,0), (1,-2), (2, -4)

(i) \(B\rightarrow (-5,2)\)
(ii) \(C\rightarrow \left( 5,-5 \right) \)
(iii) E
(iv) G
(v) 6
(vi) -3
(vii) \(L\rightarrow (0,5)\)
(viii) \(M\rightarrow (-3,0)\)

\({ \left[ \frac { 3 }{ 2 } x+1 \right] }^{ 3 }\)
\(={ \left( \frac { 3 }{ 2 } x \right) }^{ 3 }+{ (1) }^{ 3 }+3\left( \frac { 3 }{ 2 } x \right) (1)\left( \frac { 3 }{ 2 } x+1 \right) \) | Using Identity VI
\(=\frac { 27 }{ 8 } { x }^{ 3 }+1+\frac { 9 }{ 2 } x\left( \frac { 3 }{ 2 } x+1 \right) \)
\(=\frac { 27 }{ 8 } { x }^{ 3 }+1+\frac { 27 }{ 4 } { x }^{ 2 }+\frac { 9 }{ 2 } x\)
\(=\frac { 27 }{ 8 } { x }^{ 3 }+\frac { 27 }{ 4 } { x }^{ 2 }+\frac { 9 }{ 2 } x+1\)

\(p\left( -\frac { 1 }{ 3 } \right) =3\left( -\frac { 1 }{ 3 } \right) +1=-1+1=0\)
\(\therefore -\frac { 1 }{ 3 } \) is a zero of p(x)

\(\frac { 1 }{ \sqrt { 7 } -\sqrt { 6 } } \)=\(\frac { 1 }{ \sqrt { 7 } -\sqrt { 6 } } \)\(\times \frac { \sqrt { 7 } +\sqrt { 6 } }{ \sqrt { 7 } +\sqrt { 6 } } \)
Multiplying and dividing by \(\sqrt { 7 } +\sqrt { 6 } \)
\(=\frac { \sqrt { 7 } +\sqrt { 6 } }{ \sqrt { 7 } -\sqrt { 6 } } =\sqrt { 7 } +\sqrt { 6 } \)

Let x = \(0.\overline { 6 } \) = 0.6666...
Multiplying both sides by 10, we get
10 x = 6.6666...
10x = 6+0.6666...
10x = 6+x
10x - x = 6
9x = 6
x = 6/9
x = 2/3
Thus, \(0.\overline { 6 } \)= 2/3
Here p= 2
q = 3(\(\neq 0\))

Yes! zero is a rational number.We can write zero in the form \(\frac { p }{ q } \),where p and q are integers and \(q\neq 0\)as follows:
  \(0=\frac { 0 }{ 1 } =\frac { 0 }{ 2 } =\frac { 0 }{ 3 } \)etc.

(i) Let the plot be ABCD.
Construction: Join AC. Draw BE||AC.
Proof: ar(\(\Delta\)ADE) = ar(Quad.ABCD)
Health centre can be constructed in triangular plot (\(\Delta\)AOB) and the farmer can have the triangular plot \(\Delta\)ADE.
Proof:
ar(\(\Delta\)ADE) = ar(\(\Delta\)ADC) + ar(\(\Delta\)ACE)
= ar(\(\Delta\)ADC) + ar(\(\Delta\)ABC)
= ar(Quad.ABCD)
(ii) Helpful, wise, and co-operating.
(iii) Yes, constructing a Health Centre is justified and essential also, because Health is Wealth.

Through O, draw AB || PS
Also  PA || BS
PABS is a parallelogram.
ar(POS) =\(\frac { 1 }{ 2 } ar(PABS)\) 

(Triangle and a parallelogram are on the same base and between the same parallels)
Similarly, ar(QOR )= \(\frac { 1 }{ 2 } \)ar(QABR)
\(\therefore \ ar(POS)+ar(QOR)=\frac { 1 }{ 2 } [ar(PABS)+ar(QABR)]\)
\(=\frac { 1 }{ 2 } ar(PQRS)\)

Two triangles ABC and DEF, such that

AB = DE and AB II DE
Also BC = EF and BC II EF
Proof: (i) In a quadrilateral ABED,
AB = DE and AB II DE
⇒ One pair of opposite sides are equal and parallel.
ABED is a parallelogram
⇒ AD = BE and AD II BE....(i)
(ii) In quadrilateral BCFE,
BC = EF and BC II EF
⇒ One pair of opposite sides are equal and parallel.
BCFE is a parallelogram
CF = BE and CF II BE ...(ii)
(iii) From equations (i) and (ii), we get
AD = CF and AD II CF
⇒ ACFD is a parallelogram
AC = DF andAC II DF
(iv) In ∆ABC and ∆DEF
AB = DE (Given)
BC = EF (Given)
and AC = DF (Proved above in part (c))
So by S.S.s.,ΔABC≅ΔDEF.

\(x=\frac { \sqrt { 5 } +1 }{ \sqrt { 5 } -1 } \)
\({ x }^{ 2 }={ \left[ \frac { \sqrt { 5 } +1 }{ \sqrt { 5 } -1 } \right] }^{ 2 }=\frac { 6+2\sqrt { 5 } }{ 6-2\sqrt { 5 } } =\frac { 3+\sqrt { 5 } }{ 3-\sqrt { 5 } } \)
\(y=\frac { \sqrt { 5 } -1 }{ \sqrt { 5 } +1 } \)
\({ y }^{ 2 }={ \left[ \frac { \sqrt { 5 } -1 }{ \sqrt { 5 } +1 } \right] }^{ 2 }=\frac { 6-2\sqrt { 5 } }{ 6+2\sqrt { 5 } } =\frac { 3-\sqrt { 5 } }{ 3+\sqrt { 5 } } \)
\({ x }^{ 2 }+{ y }^{ 2 }=\frac { { \left( 3+\sqrt { 5 } \right) }^{ 2 }+{ \left( 3-\sqrt { 5 } \right) }^{ 2 } }{ \left( 3-\sqrt { 5 } \right) \left( 3+\sqrt { 5 } \right) } \)
\(=\frac { 9+5+6\sqrt { 5 } +9+5-6\sqrt { 5 } }{ 9-5 } \)
\(=\frac { 28 }{ 4 } \)
\({ x }^{ 2 }+{ y }^{ 2 }=7\)

Let p(x)=2x³+ax²+7x-6
p(x) is divisble by 2x-1
\(\Rightarrow p\left( \frac { 1 }{ 2 } \right) =0\)
\(p\left( \frac { 1 }{ 2 } \right) ={ p\left( \frac { 1 }{ 2 } \right) }^{ 3 }+a{ \left( \frac { 1 }{ 2 } \right) }^{ 2 }+7\left( \frac { 1 }{ 2 } \right) -6=0\)
\(\Rightarrow 2\times \left( \frac { 1 }{ 8 } \right) +a\left( \frac { 1 }{ 4 } \right) +7\left( \frac { 1 }{ 2 } \right) -6=0\)
\(\Rightarrow \frac { 1 }{ 4 } +\frac { a }{ 4 } +\frac { 7 }{ 2 } -6=0\)
\(\Rightarrow \ \frac { a }{ 4 } -\frac { 9 }{ 4 } =0\)
\(\Rightarrow \ a=9\)
\(\therefore\)  p(x)=2x³+ax²+7x-6

\(\therefore\) x²+5x+6=(x+2)(x+3)
Hence p(x)=(2x-1)(x+2)(x+3)

Equation of p is x=-1
Equation of r is y=-2
Yes, the equation of q is 
x+y=1
Area \(=\frac{1}{2}\times4\times 4=8\) sq.units

Let us recall what a rectangle is.
A rectangle is a parallelogram in which one angle is a right angle.
 
Let ABCD be a rectangle in which \(\angle\) A = 90°.
We have to show that \(\angle\) B = Ð C = \(\angle\) D = 90°
We have, AD || BC and AB is a transversal
(see Fig.).
So, \(\angle\) A + \(\angle\) B = 180° (Interior angles on the same
side of the transversal)
But, \(\angle\) A = 90°
So, \(\angle\) B = 180° – \(\angle\) A = 180° – 90° = 90°
Now, \(\angle\) C = Ð A and \(\angle\) D = \(\angle\) B
(Opposite angles of the parallellogram)
So, \(\angle\) C = 90° and \(\angle\) D = 90°.
Therefore, each of the angles of a rectangle is a right angle.

\(\because\)  C is the midpoint of AB
\(\therefore \)  AC = CB
AC + AC = CB + AC
| If equals are added to equals, then   the wholes are equal (Euclid's Axiom (ii))]
\(\Rightarrow \)  2AC =  AB      I CB + AC coincides with AB 
\(\Rightarrow \)  \(1\over2\)(2AC) = \(1\over2\) AB
| Things which are halves of the same thing are equal (Euclid's Axiom (vii»]
\(\Rightarrow \)  AC =\(1\over2\)AB
\(\Rightarrow \) \(1\over2\)AC  = \(1\over2\)(\(1\over2\)AB)
| Things which are halves of the same thing are equal to one another (Euclid's Axiom (vii))]
\(1\over2\)AC = \(1\over2\)AB
AD = \(1\over4\)AB
\(\because\) D is the mid-point of AC
\(\therefore \)AD = DC =\(1\over2\)AC (as above)

The figure obtained is a right angled triangle.

(a) The points whose ordinates are O and Q and N.
(b) The points whose abscissae are 0 are T and P.
(c) The points whose abscissae are -3 are Q and R.
(d) The points whose ordinates are 4 are S and L.

L.H.S \(=xy[(x+y)\frac{1}{x}+\frac{1}{y}-4]\)
\(=xy[(x+y)(\frac{y+x}{xy})-4]\)
\(=xy[\frac{(x+y)^2}{xy}-4]\)
\(=(x+y)^2-4xy\)
\(=(x^2+y^2+2xy)-4xy\)  | Using Identity I
\(=x^2+y^2-2xy=(x-y)^2\)  | Using Identity II

\((3x-5y-4)(9x^2+25y^2+15xy+12x-20y+16)\)
\(=\left\{ (3x)+(-5y)+(-4)(3x)^{ 2 }+(-5y)^{ 2 }+(-4)^{ 2 }-(3x)(-5y)-(-5y)(-4)-(-4)(3x) \right\} \)
\(=(3x)^2+(-5y)^3+(-4)^3-3(3x)(-5y)(-4)\)
\(=27x^3-125y^3-64-180xy\)

\({ 5 }^{ p-3 }\times { 3 }^{ 2p-8 }=225\)
\(\\ { 5 }^{ p-3 }\times { 3 }^{ 2p-8 }={ 5 }^{ 2 }\times { 3 }^{ 2 }\)
\(\\ p-3=2\)
\(\\ 2p-8=2\)
\(\\ p=5\)

Infinitely many irrational numbers lie between \(\sqrt { 2 } \) and \(\sqrt { 3 } \)
One irrational number between \(\sqrt { 2 } \) and \(\sqrt { 3 } \) is \(\sqrt { \sqrt { 2 } \sqrt { 3 } } =\sqrt { \sqrt { 6 } } ={ 6 }^{ \frac { 1 }{ 4 } }\)
Another irrational number between \(\sqrt { 2 } \)and \(\sqrt { 3 } \) is \(\sqrt { \sqrt { 2 } { 6 }^{ \frac { 1 }{ 4 } } } \)
\(={ 2 }^{ \frac { 1 }{ 4 } }.{ 6 }^{ \frac { 1 }{ 8 } }={ 2 }^{ \frac { 1 }{ 4 } }.{ 2 }^{ \frac { 1 }{ 8 } }.{ 3 }^{ \frac { 1 }{ 8 } }\)
\(\\ ={ 2 }^{ \frac { 1 }{ 4 } +\frac { 1 }{ 8 } }.{ 3 }^{ \frac { 1 }{ 8 } }={ 2 }^{ \frac { 3 }{ 8 } }.{ 3 }^{ \frac { 1 }{ 8 } }\)
Third irrational number between \(\sqrt { 2 } \) and \(\sqrt { 3 } \) is
\(\sqrt { \sqrt { 2 } .{ 2 }^{ \frac { 3 }{ 8 } }.{ 3 }^{ \frac { 1 }{ 8 } } } ={ 2 }^{ \frac { 1 }{ 4 } }.{ 2 }^{ \frac { 3 }{ 16 } }.{ 3 }^{ \frac { 1 }{ 16 } }={ 2 }^{ \frac { 7 }{ 16 } }.{ 3 }^{ \frac { 1 }{ 16 } }\)

\(\frac { 3 }{ 7 } =\frac { 30 }{ 70 } \)
\(\\ \frac { 5 }{ 7 } =\frac { 50 }{ 70 } \)
30<31<32<33<34<35
\(\frac { 30 }{ 70 } <\frac { 31 }{ 70 } <\frac { 32 }{ 70 } <\frac { 33 }{ 70 } <\frac { 34 }{ 70 } <\frac { 50 }{ 70 } \)
So four rational numbers between \(\frac { 3 }{ 7 } \) and \(\frac { 5 }{ 7 } \)
\(\frac { 31 }{ 70 } ,\frac { 32 }{ 70 } ,\frac { 33 }{ 70 } \) and \(\frac { 34 }{ 70 } \)
\(\frac { 31 }{ 70 } ,\frac { 16 }{ 35 } ,\frac { 33 }{ 70 } \)and \(\frac { 17 }{ 35 } \)

f(x) = x³+3x²-kx-3
(x+3) is a factor of f(x) = x³+3x²-kx-3
\(\Rightarrow\) f(-3) = 0
\(\Rightarrow\) (-3)³+3(-3)²-k(-3)-3=0
\(\Rightarrow\) -27 + 27 + 3k -3=0
\(\Rightarrow\) 3k-3 = 0
\(\Rightarrow\) k = 1

a = 2 and b = 3
a^b + b^a=2³ + 3²
= 8 + 9
= 17

6 Units.

IV

-2

\(\frac { 65 }{ 16 } \)

\(\frac { 68 }{ 7 } -\frac { 5 }{ 7 } \sqrt { 15 } \)

0

(i) False, because zero is a whole number but not a natural number.
(ii) True, because every integer m can be expressed in the form m/1, and so it is a rational number.
(iii) False because 3/5 is a rational number but not an integer.

\(\angle ABC=180°-\angle ADC=180°-110°=70°\) 
\(\angle ACB=90°\) 
\(\therefore \angle BAC=20°\)
\(\angle ACD=\angle BAC\)   | Alternate interior angles

By definition

(i) Since sum of adjacent angles of a parallelogram is 180°
ஃ We have ㄥA+ㄥB=180
⇒ 5x + 7 + 3x - 3 = 180
⇒ 8x + 4 = 180
⇒ 8x = 176
⇒ x=\(\frac{176}{8}\)=22

ㄥA=(5x+7)°=(5x22+7)=117°
ㄥB=(3x-3)°=(3x22-3)=63°
ㄥC=ㄥA=117°
and ㄥD=ㄥA=63°
(ii) Properties of parallelogram.
(iii) Energy conservation is necessary for a happy and prosperous future.

\(\sqrt { \frac { \left( \sqrt { 2 } -1 \right) }{ \left( \sqrt { 2 } +1 \right) } } \)is an irrational number.
\(\sqrt { \frac { \left( \sqrt { 2 } -1 \right) }{ \left( \sqrt { 2 } +1 \right) } } =\sqrt { \frac { \left( \sqrt { 2 } -1 \right) }{ \left( \sqrt { 2 } +1 \right) } \times \frac { \left( \sqrt { 2 } -1 \right) }{ \left( \sqrt { 2 } -1 \right) } } \)
\(=\sqrt { \frac { { \left( \sqrt { 2 } -1 \right) }^{ 2 } }{ 2-1 } } \)
\(=\sqrt { \frac { { \left( \sqrt { 2 } -1 \right) }^{ 2 } }{ 1 } } =\sqrt { 2 } -1\)
which is an irrational number.
Let, there is a number x such that x³ is an irrational number but x⁵ is a rational number.
Let, x =\(\sqrt[5]{7}\) be the number.
⇒ x³ = (5√7)³ = (7)^3/5
is an irrational number.
But x⁵ = (\(\sqrt[5]{7}\))⁵=(7)^5/5  = 7
=7 is a rational number.
(ii) Accepting own mistakes gracefully, co-operative learning among the classmates.

Justify, \(\frac { \sqrt { 2 } }{ \sqrt { 5 } +\sqrt { 3 } } =\frac { \sqrt { 2 } }{ \sqrt { 5 } +\sqrt { 3 } } \times \frac { \left( \sqrt { 5 } -\sqrt { 3 } \right) }{ \left( \sqrt { 5 } -\sqrt { 3 } \right) } \)
\(=\frac { \sqrt { 10 } -\sqrt { 6 } }{ { \left( \sqrt { 5 } \right) }^{ 2 }-{ \left( \sqrt { 3 } \right) }^{ 2 } } =\frac { \sqrt { 10 } -\sqrt { 6 } }{ 5-3 } \)
\(=\frac { \sqrt { 10 } -\sqrt { 6 } }{ 2 } \)
Again, \(\sqrt { 28 } +\sqrt { 98 } +\sqrt { 147 } \)
\(=\sqrt { 2\times 2\times 7 } +\sqrt { 2\times 7\times 7 } +\sqrt { 3\times 7\times 7 } \)
\(=2\sqrt { 7 } +7\sqrt { 2 } +7\sqrt { 3 } \)