105 x 106 = (100 + 5) x (100 + 6)
= (100)2 + (5 + 6) (100) + (5 x 6), using Identity IV
= 10000 + 1100 + 30
= 11130

Let f(x) = x³ + 3x² + 3x + 1
(i) Here, the zero of x + 1 is x=-1.  [\(\because\) x + 1 = 0 \(\Rightarrow\) x = -1]
On putting x =- 1 in f(x), we get
f (-1) = (-1)³ + 3 x (-1)² + 3 x (-1)+1
= - 1 + 3 - 3 + 1 = 0
Hence, the remainder is zero.
(ii) \(\frac {27}{8}\)
(iii) 1
(iv) \(-{ \pi }^{ 3 }+3{ \pi }^{ 3 }-3\pi +1\)
(v) \(\frac {-27}{8}\)

(i)

(ii)

By inspection, x = 2, y = 2 is a solution because for x = 2, y = 2
x + 2y = 2 + 4 = 6
Now, let us choose x = 0. With this value of x, the given equation reduces to 2y = 6 which has the unique solution y = 3. So x = 0, y = 3 is also a solution of x + 2y = 6. Similarly, taking y = 0, the given equation reduces to x = 6. So, x = 6, y = 0 is a solution of x + 2y = 6 as well. Finally, let us take y = 1. The given equation now reduces to x + 2 = 6, whose solution is given by x = 4. Therefore, (4, 1) is also a solution of the given equation. So four of the infinitely many solutions of the given equation are:
(2, 2), (0, 3), (6, 0) and (4, 1).

(i) 4,3, (4,3)
(ii) -3,4, (-3,4)
(iii) -5,-4, (-5,-4)
(iv) 3,-4, (3,-4)

Let p(x) = x² – 2x
Then p(2) = 2² – 4 = 4 – 4 = 0
and p(0) = 0 – 0 = 0
Hence, 2 and 0 are both zeroes of the polynomial x² – 2x.

\(6 \sqrt{5} \times 2 \sqrt{5}=6 \times 2 \times \sqrt{5} \times \sqrt{5}=12 \times 5=60\)

(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.

Let AB = BC = 1 unit length
Using Pythagoras theorem, we see that
OC =\(\sqrt{1^{2}+1^{2}}=\sqrt{2}\)
Construct CD=1 unit length perpendicular to OC, then using Pythagoras theorem, we see that
OD =\(\sqrt{(\sqrt{2}^{2}+1^{2}}=\sqrt{3}\)
Using a compass with centre O and radius OD, draw an arc which intersects the number line at the point Q, then Q corresponds to √3

The given equation is x-2y = 4
(4,0)
 Put x = 4 and y = 0 in (1) we get
x - 2y = 4 - 2(0) = 4
(4,0) is a solution of (1).

\(2x+3y=9.3 \overset{-}{5}\)=0
Comparing with ax+by+c=0, we get
a=2, b=3, c=-9.3\(\overset{-}5\).

(i) The point (-2,4) lies in the II quadrant.
(ii) The point (3, -1) lies in the IV quadrant.
(iii) The point (- 1,0) lies on the negative x-axis.
(iv) The point (1, 2) lies in the I quadrant.
(v) The point (- 3, - 5) lies in the Ill quadrant.

Consider the lamp as a point and table as a plane.Choose any two perpendicular edges of the table.Measure the distance of the lamp from the longer edge, suppose it is 25 cm. Again, measure the distance of the lamp from the shorter edge, and suppose it is 30 cm.You can write the position of the lamp as (30,25) or (25, 30), depending on the order you fix.

(99)³
(99)³=(100-1)³
=(100)³-(1)³-3(100)(1)(100-1) | Using Identity VII
=1000000-1-300(100-1)
=1000000-1-30000+300
=970299

\(p(x)={ x }^{ 3 }+{ 3x }^{ 2 }-3x+1,\ g(x)=x+2\)
g(x)=0
\(\Rightarrow x+2=0\ \Rightarrow \ x=-2\)
\(\therefore\) Zero of g(x) is -2.
Now, p(-2)
\(={ (-2) }^{ 3 }+{3 (-2) }^{ 2 }+3(-2)+1\)
\(=-8+12-6+1=-1\neq 0\)
\(\therefore\) By factor theorem, g(x) is not a factor of p(x)

\(4\sqrt { 5 } +5\sqrt { 2 } \)

(2, 2) and (-1, -3) both satisfy y=3x

5=2x-3  ⇒ x=4

\(x+p=0\ \Rightarrow \ x=-p\)
By factor theorem,
\((-p)^2+p(-p)+3-p=0 \ \Rightarrow \ p=3\)

\(p(-1)=2+\frac{-1}{2}+(-1)^2-\frac{(-1)^3}{3}=\frac{17}{6}\)

Evident

(i) 4x² + 9y² + 16z² + 12xy - 24yz - 16xz
= (2x)² + (3y)² + (-4z)² + 2(2x)(3y) + 2(3y)(-4z) + 2(-4z)(2x)
= (2x + 3y - 4z)² [Using Identity V]
= (2x + 3y - 4z)(2x + 3y - 4z)
(ii) 2x² + y²+ 8z² - 2 \(\sqrt{2}\)xy + 4\(\sqrt{2}\) yz - 8xz
= (-2 \(\sqrt{2}\)x)² + (y)² + (2\(\sqrt{2}\)z)² + 2 (-\(\sqrt{2}\)x)(y) + 2 (2\(\sqrt{2}\)z)(y) + 2 (2\(\sqrt{2}\)z) (-\(\sqrt{2}\)x)
= (-\(\sqrt{2}\)x + y + 2 \(\sqrt{2}\))²
= (-\(\sqrt{2}\)x + y + 2\(\sqrt{2}\)z)( -\(\sqrt{2}\)x + y + 2\(\sqrt{2}\)z)

Let p(x) = x³ – 23x² + 142x – 120
We shall now look for all the factors of –120. Some of these are ±1, ±2, ±3, ±4, ±5, ±6, ±8, ±10, ±12, ±15, ±20, ±24, ±30, ±60.
By trial, we find that p(1) = 0. So x – 1 is a factor of p(x).
Now we see that x³ – 23x² + 142x – 120 = x³ – x² – 22x² + 22x + 120x – 120
= x²(x –1) – 22x(x – 1) + 120(x – 1) 
= (x – 1) (x² – 22x + 120) [Taking (x – 1) common]
We could have also got this by dividing p(x) by x – 1.
Now x² – 22x + 120 can be factorised either by splitting the middle term or by using the Factor theorem. By splitting the middle term, we have:
x² – 22x + 120 = x² – 12x – 10x + 120
= x(x – 12) – 10(x – 12)
= (x – 12) (x – 10)
So, x³ – 23x² – 142x – 120 = (x – 1)(x – 10)(x – 12)

Let the cost of a notebook = Rs x
and the cost of a pen = Rs y
According to the question,
Cost of a notebook = 2 (Cost of a pen)
\(\Rightarrow\)x = 2y
\(\Rightarrow\) x-2y = 0
which is the required linear equation in two variables.

Let EW and NS be two main roads such that the road EW is along the East-West direction and road NS is along the North-South direction. Then, the angle between the two roads is 90°, i.e. the roads EW and NS are perpendicular to each other.
Let us consider EW along X-axis and NS along Y-axis and let the center of ciry is 0
Here, the distance be in two consecutive streets in same direction is 200 m and all the streets are parallel to the main

The street plan is shown in the above figure:
(i) Because of the two reference lines that we have used for locating them, there is only one cross-street which can be referred to as (4, 3).
(ii) Similarly, there is only one cross-street which can be referred to as (3, 4).

(i) We have, \(\frac { 1 }{ \sqrt { 7 } } \)
On multiplying both numerator and denominator by \(\sqrt { 7 } \) , we get
\(\frac { 1 }{ \sqrt { 7 } } \times \frac { \sqrt { 7 } }{ \sqrt { 7 } } =\frac { \sqrt { 7 } }{ 7 } \)
(ii) We have,\(\frac { 1 }{ \sqrt { 7 } -\sqrt { 6 } } \)
On multiplying both numerator and denominator by \(\sqrt { 7 } +\sqrt { 6 } ,\) we get
\(\frac { 1 }{ \sqrt { 7 } -\sqrt { 6 } } \times \frac { \left( \sqrt { 7 } +\sqrt { 6 } \right) }{ \left( \sqrt { 7 } +\sqrt { 6 } \right) } \)
\(=\frac { \sqrt { 7 } +\sqrt { 6 } }{ { \left( \sqrt { 7 } \right) }^{ 2 }-{ \left( \sqrt { 6 } \right) }^{ 2 } } [\because (a-b)(a+b)={ a }^{ 2 }-{ b }^{ 2 }]\)
\(=\frac { \sqrt { 7 } +\sqrt { 6 } }{ 7-6 } =\frac { \sqrt { 7 } +\sqrt { 6 } }{ 1 } \)
\(=\sqrt { 7 } +\sqrt { 6 } \)
(iii) \(\left[ \frac { \sqrt { 5 } -\sqrt { 2 } }{ 3 } \right] \)
(iv) \(\left[ \frac { \sqrt { 7 } +1 }{ 3 } \right] \)

First, we draw AB = 9.3 units. Now, from B, mark a distance of 1 unit. Let this point be C. Let D be the mid-point of AC. Now, draw a semi-circle with centre D and radius DA. Let us draw a line perpendicular to AC passing through point B and intersecting the semi-circle at point D.
\(\therefore\)   Distance, BD=\(\sqrt { 9.3 } \)
Draw an arc with centre B and radius BD, which intersects the number line at point E. So point E represents \(\sqrt { 9.3 } \) .

(a) (iii) 2x + 3y = 60
Let the number of children be x and the number of adults be y then the linear equation in two variable for the given situation is 
2x + 3y = 60.
(b) (i) 10
2x + 3y =60 ⇒ 2(15) + 3y = 60
⇒ 3y = 60 - 30 = 30 
⇒ y = 10
(c) (i) 12
2x + 3y = 60 ⇒ 2x + 3(12) = 60
⇒ 2x 60 - 36 = 24
⇒ x = 12
(d) (iii) 15
On putting x = 5 and y = 0 in the equation 3x + 5y = b, we have 
3 x 5 + 5 x 0 = b
⇒ 15 + 0 = b
⇒ b = 15
(e) (ii) 1.x - 1.y + 5 = 0
y - x = 5 ⇒ y = x + 5
⇒ x - y + 5 = 0
⇒ 1.x - 1.y + 5 = 0

(i) (b) (2, 3)
(ii) (a) (2, -1)
(iii) (a) 4
(iv) (d) (6, 2)
(v) (b) II