Category: Class 9

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.3 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 4th Lesson Lines and Angles Exercise 4.3

Question 1.
It is given that l // m; to prove ∠1 is supplement to ∠8. Write reasons for the Statements.

Solution:

Statement	Reasons
i) l //m	∠1 + ∠8 = 180° (exterior angles on the same side of the transversal)
ii) ∠1 = ∠5	corresponding angles
iii) ∠5 + ∠8 = 180°	linear pair of angles
iv) ∠1 + ∠8 = 180°	exterior angles on the same side of the transversal.
v) ∠1 is supplement is ∠8	exterior angles on the same side of the transversal.

Question 2.
In the given figure AB || CD; CD || EF and y: z = 3:7 find x.

Solution:
Given that AB//CD; CD//EF.
⇒ AB // EF
Also y : z = 3 : 7
From the figure x + y = 180 …………. (1)
[∵ interior angles on the same side of the transversal]
Also y + z = 180 ………….. (2)
Sum of the terms of the ratio y : z
= 3 + 7 = 10
∴ y = \(\frac{3}{10}\) x 180° = 54°
y = \(\frac{7}{10}\) x 180° = 126°
From (1) and (2)
x + y = y + z
⇒ x = z = 126°

Question 3.
In the given figure AB//CD; EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.

Solution:
Given that EF ⊥ CD; ∠GED = 126°
i. e., ∠FED = 90° and
∠GEF = ∠GED – ∠FED
∠GEF = 126° – 90° = 36°
In ∆GFE
∠GEF + ∠FGE + ∠EFG = 180°
36 + ∠FGE + 90° = 180°
∠FGE = 180° – 126° = 54°
∠AGE = ∠GFE + ∠GEF
(exterior angle in ∆GFE)
= 90°+ 36°= 126°

Question 4.
In the given figure PQ//ST, ∠PQR = 110° and ∠RST = 130°, find ∠QRS. [Hint : Draw a line parallel to ST through point R.]

Solution:

Given PQ // ST
Draw a lipe ‘l’ parallel to ST through R.
From the figure
a + 110° = 180° and c + 130 = 180°
[ ∵ Interior angles on the same side of the transversal]
∴ a = 180° -110° = 70°
c = 180° – 130° = 50°
Also a + b + c = 180° (angles at a point on a line)
70° + b + 50° = 180°
b = 180° – 120° = 60°
∴ ∠QRS = 60°

Question 5.
In the given figure m // n. A, B are any two points on in and n respectively. Let C be an interior point between the lines m and n. Find ∠ACB.

Solution:

Draw a line ‘l’ parallel to m and n through C.
From the figure
x = a [ ∵ alternate interior angles for l, m]
y = b [ ∵ alternate interior angles for l, n]
∴ z = a + b = x + y

Question 6.
Find the values of a and b, given that p // q and r // s.

Solution:
Given that p // q and r // s.
∴ From the figure
2a = 80° (∵ corresponding angles)
a = \(\frac { 80° }{ 2 }\) = 40°
Also 80° + b = 180° ( ∵ interior angles on the same side of the transversal)
∴ b = 180° – 80° = 100°

Question 7.
If in the figure a // b and c // d, then name the angles that are congruent to (i) ∠1 and (ii) ∠2.

Solution:
Given that a // b and c // d.
∠1 = ∠3 (∵ vertically opposite angles)
∠1 = ∠5 (∵ corresponding angles)
∠1 = ∠9 (∵ corresponding angles)
Also ∠1 = ∠3 = ∠5 = ∠7 ;
∠1 = ∠11 = ∠9 = ∠13 = ∠15
Similarly ∠2 = ∠4 = ∠6 = ∠8
Also ∠2 = ∠10 = ∠12 = ∠14 = ∠16

Question 8.
In the figure the arrow head segments are parallel, find the values of x and y.

Solution:
From the figure
y = 59° ( ∵ alternate interior angles)
x = 60° ( ∵ corresponding angles)

Question 9.
In the figure the arrow head segments are parallel then find the values of x and y.

Solution:
From the figure 35° + 105° + y = 180°
∴ y = 180° – 140°
= 40°
∴ x = 40° (∵ x, y are corresponding angles)

Question 10.
Find the values of x and y from the figure.

Solution:
From the figure 120° + x = 180°
(∵ exterior angles on the same side of the transversal)
∴ x = 180° – 120°
x = 60°
Also x = (3y + 6)
(∵ corresponding angles)
3y + 6 = 60°
3y = 60° – 6° = 54°
y = \(\frac { 54 }{ 3 }\) = 18°
∴ x = 60°; y = 18°

Question 11.
From the figure find x and y.

Solution:
From the figure
52° + 90° + (3y + 5)° = 180°
(∵ interior angles of a triangle)
∴ 3y + 147 = 180°
⇒ 3y = 33°
⇒ y = \(\frac { 33 }{ 3 }\) = 11°
Also x + 65° + 52° = 180°
(∵ interior angles on the same side of the transversal)
∴ x = 180° -117° = 63°

Question 12.
Draw figures for the following statement.
“If the two arms of one angle are respectively perpendicular to the two arms of another angle then the two angles are either equal or supplementary
Solution:

AO ⊥ PQ, OB ⊥ QR
Angles are supplementary.

AO ⊥ PQ, OB ⊥ QR
Angles are equal.

Question 13.
In the given figure, if AB // CD; ∠APQ = 50° and ∠PRD = 127°, find x and y.

Solution:
Given that AB // CD.
∠PRD = 127°
From the figure x = 50°
(∵ alternate interior angles)
Also y + 50 = 127°
(∵ alternate interior angles)
∴ y = 127-50 = 77°

Question 14.
In the given figure PQ and RS are two mirrors placed parallel to each other.
An incident ray \(\overline{\mathrm{AB}}\) strikes the mirror PQ at B, the reflected ray moves along the path \(\overline{\mathrm{BC}}\) and strikes the mirror RS at C and again reflected back along CD. Prove that AB // CD. [Hint : Perpendiculars drawn to parallel lines are also parallel]

Solution:

Draw the normals at B and C.
then ∠x = ∠y (angle of incidence angle of reflection are equal)
∠y = ∠w (alternate interior angles)
∠w = ∠z (angles of reflection and incidence)
∴ x + y = y + z (these are alternate interior angles to \(\overline{\mathrm{AB}}\), \(\overline{\mathrm{CD}}\))
Hence AB // CD.

Question 15.
In the figures given below AB // CD. EF is the transversal intersecting AB and CD at G and H respectively. Find the values of x and y. Give reasons.

Solution:
For fig(i)
3x = y (∵ alternate interior angles)
2x + y = 180° (∵ linear pair of angles)
∴ 2x + 3x = 180°
5x= 180°
x = \(\frac { 180 }{ 5 }\) = 36°
and y = 3x = 3 x 36 = 108°

For fig (ii)
2x + 15 = 3x- 20°
(∵ corresponding angles)
2x-3x = -20-15
– x = – 35
x = 35°

For fig (iii)
(4x – 23) + 3x = 180° ,
( ∵ interior angles on the same side of the transversal)
7x- 23 = 180°
7x = 203
x = \(\frac { 203 }{ 7 }\) = 29°

Question 16.
In the given figure AB // CD, ‘t’ is a transversal intersecting E and F re-spectively. If ∠2 : ∠1 = 5 : 4, find the measure of each marked angles.

Solution:
Given that AB // CD and ∠2 : ∠1 = 5 : 4 ∠1 + ∠2 = 180° (-. linear pair of angles) Sum of the terms of the ratio ∠2 :∠1 = 5 + 4 = 9
∴ ∠1 = \(\frac { 4 }{ 9 }\) x 180° = 80°
∠2= \(\frac { 5 }{9 }\) x 180° = 100°
Also ∠1, ∠3, ∠5, ∠7 are all equal to 80°. Similarly ∠2, ∠4, ∠6, ∠8 are all equal to 100°.

Question 17.
In the given figure AB//CD. Find the values of x, y and z.

Solution:
Given that AB // CD.
From the figure (2x + 3x) + 80° = 180°
(∵ interior angles on the same side of the transversal)
∴ 5x = 180° – 80°
x = \(\frac{100}{5}\) = 20°
Now 3x = y (∵ alternate interior angles)
y = 3 x 20° = 60° .
And y + z = 180°
(∵ linear pair of angles)
∴ z = 180°-60° = lg0°

Question 18.
In the given figure AB // CD. Find the values of x, y and z.

Solution:
Given that AB // CD.
From the figure x° + 70° + x° = 180°
(∵ The angles at a point on the line)
∴ 2x = 180° – 70°
x = \(\frac { 110° }{ 2 }\) = 55-
Also 90° + x° + y° = 180°
[∵ interior angles of a triangle]
90° + 55° + y = 180°
y = 180° – 145° = 35°
And x° + z° = 180°
[∵ interior angles on the same side of a transversal]
55° + z = 180°
z = 180°-55° = 125°

Question 19.
In each of the following figures AB // CD. Find the values of x in each case.

Solution:

In each case draw a line ‘l’ parallel to AB and CD through F.
fig (i)
a + 104° = 180° ⇒ a = 180° – 104° = 76°
b+ 116° = 180° ⇒ -b = 180°- 116° = 64°
[∵ interior angles on the same side]
∴ a + b = x = 76° + 64° = 140°

fig-(ii)
a = 35°, b = 65° [∵ alt. int. angles]
x = a + b = 35° + 65° = 100°

fig- (iii)
a + 35° = 180° ⇒ a = 145°
b + 75° = 180° ⇒ b = 105°
[ ∵ interior angles on the same side]
∴ x = a + b = 145° + 105° = 250°

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.2 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 4th Lesson Lines and Angles Exercise 4.2

Question 1.
In the given figure three lines \(\overline{\mathrm{AB}}\) , \(\overline{\mathrm{CD}}\) and \(\overline{\mathrm{EF}}\) intersecting at ‘O’. Find the values of x, y and z, it is being given that x : y : z = 2 : 3 : 5

Solution:
From the figure
2x + 2y + 2z = 360°
⇒ x + y + z = 180°
x : y : z = 2 : 3 : 5
Sum of the terms of the ratio
= 2 + 3 + 5 = 10

Question 2.
Find the value of x in the following figures.
i)

Solution:
From the figure
3x + 18= 180° – 93° (∵ linear pair )
3x + 18 = 87
3x = 87- 18 = 69
∴ x = \(\frac{69}{3}\) = 23

ii)

Solution:
From the figure
(x – 24)° + 29° + 296° = 360“
(∵ complete angle)
x + 301° = 360°
∴ x = 360° – 301° = 59°

iii)

Solution:
From the figure
2 + 3x = 62
(∵ vertically opposite angles)
3x = 62 – 2
∴ 3x = 60° ⇒ x = \(\frac{60}{3}\)
∴ x = 20°

iv)

Solution:
From the figure
40 + (6x + 2) = 90°
(∵ complementary angles)
6x = 90° – 42°
6x = 48
x = \(\frac{48}{6}\) = 8°

Question 3.
In the given figure lines \(\overline{\mathrm{AB}}\) and \(\overline{\mathrm{CD}}\) intersect at ’O’. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.

Solution:
Given that ∠AOC +∠BOE = 70°
∠BOD = 40°
∠AOC = 40°
(∵ ∠AOC, ∠BOD are vertically opposite angles)
∴ 40° + ∠BOE = 70°
⇒ ∠BOE = 70° – 40° = 30°
Also ∠AOC + ∠COE +∠BOE = 180°
( ∵ AB is a line)
⇒ 40° + ∠COE + 30° = 180°
⇒ ∠COE = 180° -70° = 110°
∴ Reflex ∠COE = 110°

Question 4.
In the given figure lines \(\overline{\mathrm{XY}}\) and \(\overline{\mathrm{MN}}\) . intersect at O. If ∠POY = 90° and a : b = 2:3, find c.

Solution:
Given that XY and MN are lines.
∠POY = 90°
a : b = 2 : 3
From the figure a + b = 90°
Sum of the terms of the ratio a : b
= 2 + 3 = 5
∴ b = \(\frac{3}{5}\) x 90° = 54°
From the figure b + c = 180°
(∵ linear pair of angles)
54° + c = 180°
c = 180°-54° = 126°

Question 5.
In the given figure ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.

Solution:
Given that ∠PQR = ∠PRQ
From the figure
∠PQR + ∠PQS = 180° ………….. (1)
∠PRQ + ∠PRT = 180° …………..(2)
From (1) and (2)
∠PQR + ∠PQS = ∠PRQ + ∠PRT
But ∠PQR = ∠PRQ
So ∠PQS = ∠PRT
Hence proved.

Question 6.
In the given figure, if x + y = w + z, then prove that AOB is a line.

Solution:
Given that x + y = w + z = k say
From the figure
x + y + z + w = 360°
(∵ Angle around a point)
Also x + y = z + w
∴ x + y = z + w = \(\frac{360^{\circ}}{2}\)
∴ x + y = z + w = 180°

OR

k + k = 360°
2k = 360°
k = \(\frac{360^{\circ}}{2}\)

(i.e.) (x,y) and (z, w) are pairs of adjacent angles whose sum is 180°.
In other words (x, y) and (z, w) are linear pair of angles ⇒ AOB is a line.

Question 7.
In the given figure \(\overline{\mathrm{PQ}}\) is a line. Ray \(\overline{\mathrm{OR}}\) is perpendicular to line \(\overline{\mathrm{PQ}}\).\(\overline{\mathrm{OS}}\) os is another ray lying between rays \(\overline{\mathrm{OP}}\) and \(\overline{\mathrm{OR}}\) Prove that

Solution:
Given : OR ⊥ PQ ⇒ ∠ROQ = 90°
To prove: ∠ROS = \(\frac{1}{2}\) (∠QOS – ∠POS)
Solution:
Proof: From the figure
∠ROS = ∠QOS – ∠QOR ……………(1)
∠ROS = ∠ROP – ∠POS ……………..(2)
Adding (1) and (2)
∠ROS + ∠ROS = ∠QOS – ∠QOR +∠ROP – ∠POS [ ∵ ∠QOR = ∠ROP = 90° given]
⇒ 2∠ROS = ∠QOS – ∠POS
⇒ ∠ROS = \(\frac{1}{2}\) [∠QOS – ∠POS]
Hence proved.

Question 8.
It is given that ∠XYZ = 64° and XY is produced to point P. A ray \(\overline{\mathrm{YQ}}\) bisects ∠ZYP. Dräw a figure from the given Information. Find ∠XYQ and reflex ∠QYP.
Solution:

∠XYQ = 32°
∠QYP = 32°

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 4 Lines and Angles Ex 4.1 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 4th Lesson Lines and Angles Exercise 4.1

Question 1.
In the given figure, name:

i) Any six points
Solution:
A, B, C, D, P, Q, M, N etc.

ii) Any five line segments
Solution:
\(\overline{\mathrm{AX}}, \overline{\mathrm{XM}}, \overline{\mathrm{MP}}, \overline{\mathrm{PB}}, \overline{\mathrm{MN}}, \overline{\mathrm{PQ}}, \overline{\mathrm{AB}} \ldots \ldots\) etc.

iii) Any four rays
Solution:
\(\overline{\mathrm{MA}}, \overline{\mathrm{PA}}, \overline{\mathrm{PB}}, \overline{\mathrm{NC}}, \overline{\mathrm{QD}} \ldots \ldots\) etc.

iv) Any four lines
Solution:
\(\overline{\mathrm{MA}}, \overline{\mathrm{PA}}, \overline{\mathrm{PB}}, \overline{\mathrm{NC}}, \overline{\mathrm{QD}} \ldots \ldots\)

v) Any four collinear points
Solution:
A, X, M, P and B are collinear points on the line \(\overline{\mathrm{AB}}\).

Question 2.
Observe the following figures and identify the type of angles in them.

Solution:
∠A – reflex angle
∠B – right angle
∠C – acute angle

Question 3.
State whether the following state¬ments are true or false
i) A ray has no end point.
ii) Line \(\overline{\mathrm{AB}}\) is the same as line \(\overline{\mathrm{BA}}\)
iii) A ray \(\overline{\mathrm{AB}}\) is same as the ray \(\overline{\mathrm{BA}}\)
iv) A line has a definite length.
v) A plane, has length and breadth but no thickness.
vii) Two lines may intersect in two points.
viii) Two intersecting lines cannot both be parallel to the same line.
Solution:
i) A ray has no end point. – False
ii) Line \(\overline{\mathrm{AB}}\) is the same as line \(\overline{\mathrm{BA}}\) – True
iii) A ray \(\overline{\mathrm{AB}}\) is same as the ray \(\overline{\mathrm{BA}}\) – False
iv) A line has a definite length. – False
v) A plane, has length and breadth but no thickness. – True
vi) Two distinct points always determine a unique line. – True
vii) Two lines may intersect in two points. – False
viii) Two intersecting lines cannot both be parallel to the same line. – True

Question 4.
What is the angle between two hands of a clock when the lime in the clock
is
a) 9 ‘o clock
b) 6 ‘o clock
c) 7 ; 00 p.m.
Solution:

a) 12 hours = = 360°
1 hour = \(\frac{360^{\circ}}{12}\) = 30°
∴Angle between hands when the time is 9 o clock = 3 x 30 = 90

b)

Angle between hands = 6 x 30° = 180°

c)

Angle between hands = 7 x 30° = 210°

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 3 The Elements of Geometry Ex 3.1 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 3rd Lesson The Elements of Geometry Exercise 3.1

Question 1.
Answer the following:
i) How many dimensions a solid has ?
Solution:
A solid has three dimensions namely length, breadth and height or depth.

ii) How many books are there in Euclid’s Elements ?
Solution:
There are 13 volumes in Euclid’s elements.

iii) Write the number of faces of a cube and cuboid.
Solution:
Cube : 6 faces
Cuboid : 6 faces

iv) What is the sum of interior angles of a triangle ?
Solution:
The sum of interior angles of a triangle is 180°.

v) Write three undefined terms of geometry.
Solution:
Point, line and plane are three undefined terms in geometry.

Question 2.
State whether the following statements are true or false. Also give reasons for your answers.
a) Only one line can pass through a given point
b) All right angles are equal
c) Circles with same radii are equal
d) A finite line can be extended on its both sides endlessly to get a straight line

e) From figure AB > AC
Solution:
a) Only one line can pass through a given point – False.
Reason : (Since, infinitely many lines can pass through a given point)
b) All right angles are equal – True.
c) Circles with same radii are equal – True.
d) A finite line can be extended on its both sides endlessly to get a straight line – True.
e) From figure AB > AC – True.

Question 3.
In the figure given below, show that the length AH > AB + BC + CD.

Solution:
Given a line \(\stackrel{\leftrightarrow}{\mathrm{AH}}\)
To prove AH > AB + BC + CD
From the figure AB + BC + CD = AD
AD is a part of whole AH.
From Euclid’s axiom whole is greater than part.
∴ AH > AD
⇒ AH > AB + BC + CD

Question 4.
If a point Q lies between two points P and R such PQ = QR, prove that PQ = \(\frac{1}{2}\)PR.

Let PR be a given line.
Given that PQ = QR
i. e., Q is a point on PR.
⇒ PQ + QR = PR
⇒ PQ + PQ = PR [∵ PQ = QR]
⇒ 2PQ = PR
⇒ PQ = \(\frac{1}{2}\) PR
Hence proved.

Question 5.
Draw an equilateral triangle whose sides are 5.2 cm
Soluton:
Step – 1 : Draw a line segment AB of length 5.2 cm. *
Step – 2 : Draw an arc of radius 5.2 cm with centre A.
Step – 3 : Draw an arc of radius 5.2 cm with centre B.
Step – 4 : Two arcs intersect at C; join C to A and B.

Δ ABC is the required triangle.

Question 6.
What is a conjecture? Give an example for it.
Solution:
Mathematical statements which are neither proved nor disproved are called conjectures. Mathematical discoveries often start out as conjectures. This may be an educated guess based on observations.
Eg : Every even number greater than 4 can be written as sum of two primes. This example is called Gold Bach Conjecture

Question 7.
Mark two points P and Q. Draw a line through P and Q. Now how many lines are parallel to PQ, can you draw ?
Solution:

Infinitely many lines parallel to PQ can be drawn.

Question 8.
In the figure given below, a line n falls on lines / and m such that the sum of the interior angles 1 and 2 is less than 180°, then what can you say about lines l and m ?

Solution:
Given : l, m and n are lines, n is a transversal.
∠1 < 90°
∠2 < 90°
If the lines l and m are produced on the side where angles 1 and 2 are formed, they intersect at one point.

Question 9.
In the figure given below, if ∠1 = ∠3, ∠2 = ∠4 and ∠3 = ∠4 write the rela-tion between ∠1 and ∠2 using Euclid’s postulate.

Solution :
Given : ∠1 = ∠3
∠3 = ∠4
∠2 = ∠4
∴∠1 = ∠2

∵Both ∠1 and ∠2 are equal to ∠4. (By Euclid’s axiom things which are equal to same things are equal to one another).

Question 10.
In the figure given below, we have BX = \(\frac{1}{2}\) AB, BY= \(\frac{1}{2}\) BC and AB = BC. Show that BX = BY.

Solution:
Given : BX = \(\frac{1}{2}\) AB
BY = \(\frac{1}{2}\)BC
AB = BC
To prove : BX = BY
Proof: Given AB = BC [ ∵ By Euclid’s axiom things which are halves of the same things are equal to one another]
\(\frac{1}{2}\) AB = \(\frac{1}{2}\) BC
BX = BY
Hence proved.

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.5 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 2nd Lesson Polynomials and Factorisation Exercise 2.5

Question 1.
Use suitable identities to find the following products.
i) (x + 5) (x + 2)
Solution:
(x + 5) (x + 2)
= x² + (5 + 2)x + 5 x 2
[ ∵ (x + a) (x + b) = x² + (a + b) x + ab]
= x² + 7x + 10

ii) (x – 5) (x – 5)
Solution:
(x – 5) (x – 5)
= (x – 5)² = x² – 2(x) (5) + 5²
[ ∵(x – y)² = x² – 2xy + y²]
= x² – 10x + 25

iii) (3x + 2) (3x – 2)
Solution:
(3x + 2) (3x – 2) = (3x)² – (2)²
[∵ (x + y) (x – y) =x² – y²]
= 9x² – 4

iv) \(\left(x^{2}+\frac{1}{x^{2}}\right)\left(x^{2}-\frac{1}{x^{2}}\right)\)
Solution:

v) (1 + x) (1 + x)
Solution:
(1 + x) (1 + x)
= (1 + x)² = 1² + 2 (1) (x) + x²
[∵(x + y)² = x² + 2xy + y²]
= 1 + 2x + x²

Question 2.
Evaluate the following products with¬out actual multiplication.
i) 101 x 99
Solution:
101 x 99
= (100 + 1) (100 – 1)
= 100² – 1²
= 10000 – 1
= 9999

ii) 999 x 999
Solution:
999 x 999
= 999²
= (1000 – 1)²
= 1000² – 2 x (1000) x 1 + 1²
= 1000000-2000 + 1
= 998001

iii) \(50 \frac{1}{2} \times 49 \frac{1}{2}\)
Solution:

iv) 501 x 501
Solution:
501 x 501
= (500 + 1) (500 + 1)
= (500 + 1)²
= 500² + 2 x (500) x 1 + 1²
= 250000 + 1000 + 1 = 251001

v) 30.5 x 29.5 = (30 + 0.5) (30 – 0.5)
= 30² – (0.5)²
= 900 – 0.25
= 899.75

Question 3.
Factorise the following using appro-priate identities.
i) 16x² + 24xy + 9y²
Solution:
16x² + 24xy + 9y2
= (4x)² + 2 (4x) (3y) + (3y)²
= (4x + 3y)² = (4x + 3y) (4x + 3y)
[ ∵ (x + y)² = x² + 2xy + y²]

ii) 4y² – 4y + 1
Solution:
4y² – 4y + 1
= (2y)² – 2 (2y) (1) + (1)²
[ ∵ (x -y)² = x² – 2xy + y²]
= (2y -1)² = (2y – 1) (2y-1)

iii) \(4 x^{2}-\frac{y^{2}}{25}\)
Solution:

iv) 18a² – 50
Solution:
18a² – 50 = 2 (9a² – 25)
= 2[(3a)² – (5)²]
[ ∵ x² – y² = (x + y) (x – y)]
= 2 (3a + 5) (3a – 5)

v) x² + 5x + 6
Solution:
x² + 5x + 6 = x² + (3 + 2) x + 3 x 2
[ ∵ (x + a) (x + b) = x² + (a + b) x + a . b]
= (x + 3) (x + 2)

vi) 3p² – 24p + 36
Solution:
3p² – 24p + 36
= 3[p² – 8p + 12]
= 3[p² + (- 6 – 2)p + (- 6) (- 2)]
[ ∵ (x + a) (x + b) = x² + (a + b) x + ab]
= 3 (p – 6) (p – 2)

Question 4.
Expand each of the following, using suitable identities.
i) (x + 2y + 4z)2
(x + 2y + 4z)² = (x)² + (2y)² + (4z)² + 2(x) (2y) + 2 (2y) (4z) + 2 (4z) (x)
[ ∵ (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx]
= x² + 4y² + 16z² + 4xy + 16yz + 8zx

ii) (2a – 3b)³
Solution:
(2a – 3b)³ = (2a)³ – 3 (2a)² (3b) + 3 (2a) (3b)² – (3b)³
[ ∵ (a – b)³ = a³ – 3a²b + 3ab² – b³]
= 8a³ – 3(4a²) (3b) + 3 (2a) (9b²) – 27b³
= 8a³ – 36a²b + 54ab²-27b³
(or)
∵ (a – b)³ = a³ – b³– 3ab (a – b)]
= (2a)³ – (3b)³ – 3(2a) (3b) (2a – 3b)
= 8a³ – 27b³ – 18ab (2a – 3b)

iii) (- 2a + 5b – 3c)²
Solution:
(- 2a + 5b – 3c)²
= (- 2a)² + (5b)² + (- 3c)² + 2 (- 2a) (5b) + 2 (5b) (- 3c) + 2 (- 3c) (- 2a)
= 4a² + 25b² + 9c² – 20ab – 30bc + 12ca
[ ∵ (x + y + z)² = x² + y² + z² + 2xy +2yz +2za]

iv) \(\left[\frac{a}{4}-\frac{b}{2}+1\right]^{2}\)
Solution:

v) (p + 1)³
Solution:
(p + 1)³
= (P)³ + 3 (p)² (1) + 3 (p) (1)² + (1)³
[ ∵ (x + y)³ = x³ + 3x²y + 3xy² + y³]
= p³ + 3p² + 3p + 1

vi) \(\left(x-\frac{2}{3} y\right)^{3}\)
Solution:

Question 5.
Factorise
i) 25x² + 16y² + 4z² – 40xy + 16yz – 20xz
Solution:
25x² + 16y² + 4z² – 40xy + 16yz – 20xz
= (5x)² + (- 4y)² + (- 2z)² + 2(5x) (- 4y) + 2 (- 4y) (- 2z) + 2 (- 2z) (5x)
= (5x – 4y – 2z)² = (- 5x + 4y +, 2z)²

ii) 9a² + 4b² + 16c² + 12ab – 16bc – 24ca
Solution:
9a² + 4b² + 16c² + 12ab – 16bc -24ca
= (3a)² + (2b)² + (- 4c)²+ 2 (3a) (2b) + 2 (2b) (- 4c) + 2(- 4c) (3a)
= (3a + 2b – 4c)²

Question 6.
If a + b + c = 9 and ab + be + ca = 26, find a² + b² + c².
Solution:
Given that a + b + c = 9
Squaring on both sides,
(a + b + c)² = 9²
⇒ a²+ b² + c²+ 2 (ab + be + ca) = 81 ⇒ a² + b² + c² = 81 – 2 (ab + be + ca)
(by problem)
= 81 – 2 x 26
= 81 – 52 = 29

Question 7.
Evaluate the following by using suit¬able identities. m EachgM)
i) (99)³
Solution:
(99)² = (100 – 1)³
= 100³ – 3 (100)² (1) + 3 (100) (1)² – 1³
[ ∵ (x – y)³ = x³ – 3x²y + 3xy² + y³]
= 1000000 – 30000 + 300 – 1
= 970299

ii) (102)³
Solution:
(102)³ = (100 + 2)³
= 100³ + 3 (100)² (2) + 3 (100) (2)² + 2³
[ ∵ (x + y)³ = x³ + 3x²y + 3xy² + y³]
= 1000000 + 60000 + 1200 + 8
= 1061208

iii) (998)³
Solution:
(998)³ =(1000 – 2)³
[ ∵ (x – y)³ = x³ – 3x²y + 3xy² – y³] = 1000³– 3(1000)²(2) + 3(1000)(2)²– 2³
= 1000000000 – 6000000 + 12000 – 8
= 994011992

iv) (1001)³
Solution:
(1001)³ = (1000 + 1)³ .
[ ∵ (x + y)³ = x³ + 3x²y + 3xy² + y³] = 1000³ + 3(1000)²(1) + 3(1000) (1)² + 1³
= 1000000000 + 3000000 + 3000 + 1
= 1003003001

Question 8.
Factorise each of the following.
i) 8a³ + b³ + 12a² b + 6ab²
Solution:
8a³ + b³ + 12a² b + 6ab²
= (2a)³ + (b)³ + 3 (2a)² (b) + 3 (2a) (b)²
= (2a + b)³

ii) 8a³ – b³ – 12a² b + 6ab²
Solution:
8a³ – b³ – 12a² b + 6ab²
= (2a)³ – (b)³ – 3 (2a)² (b) + 3 (2a) (b)²
= (2a – b)³

iii) 1 – 64a³ -12a + 48a²
Solution:
1 – 64a³ – 12a + 48a²
= (1)³ – (4a)³ – 3(1)² (4a) + 3(1) (4a)²
= (1 – 4a)³

iv) \(8 p^{3}-\frac{12}{5} p^{2}+\frac{6}{25} p-\frac{1}{125}\)
Solution:

Question 9.
Verify i) x³ + y³ = (x + y) (x² – xy + y²);
ii) x³ – y³ = (x – y) (x² + xy + y²)
Using some non-zero positive integers and check by actual multiplication. Can you
call these as identities ?
i) x³ + y³ = (x + y) (x² – xy + y²)
Solution:
Given x³ + y³ = (x + y) (x² – xy + y²)
L.H.S = x³ + y³
R.H.S = (x + y) (x² – xy + y²)
= x (x² – xy + y²) + y (x² – xy + y²)
= x³ -x²y + xy² + x²y – xy² + y³
= x³ + y³
= L.H.S
∴ L.H.S = R.H.S

Take x = 3, y = 2
L.H.S = 3³ + 2³ = 27 + 8 = 35
R.H.S = (3 + 2) (3² – 3 x 2 + 2²)
= 5 x (9 – 6 + 4)
= 5 x 7 = 35
∴ L.H.S = R.H.S

ii) x³ – y³ = (x – y) (x² + xy + y²)
Solution:
Given that x³ – y³ = (x – y) (x² + xy + y²)
L.H.S = x³ – y³
R.H.S = (x – y) (x² + xy + y²)
= x (x² + xy + y²) – y (x² + xy + y²)
= x³ + x²y + xy² – x²y – xy² – y³
= x³ – y³= L.H.S

L.H.S = 3³ – 2³ = 27 – 8 = 19
R.H.S = (3 – 2) (3² + 3 x 2 + 2²)
= 1 x (9 + 6 + 4)
= 1 x 19 = 19
∴ L.H.S = R.H.S
We can call the above two expressions as identities

Question 10.
Factorise by using the above results (identities).
i) 27a³ + 64b³
Solution:
27a³+ 64b³ = (3a)³ + (4b)³
= (3a + 4b) {(3a)² – (3a) (4b) + (4b)²}
= (3a + 4b) (9a² – 12ab + 16b²)

ii) 343y³ – 1000
Solution:
343y³ – 1000 = (7y)³ – (10)³
= (7y – 10) [(7y)² + (7y) (10) + (10)²]
= (7y – 10) (49y² + 70y + 100)

Question 11.
Factorise 27x³ + y³ + z³ – 9xyz using identity.
Solution:
Given 27x³ + y³ + z³ – 9xyz
= (3x)³ + (y)³ + (z)³ – 3 (3x) (y) (z)
= (3x + y + z)
[(3x)² + y² + z² – (3x) (y) – (y) (z) – (z) (3x)]
[ ∵ (x³ + y³ + z³ – 3xyz = (x + y + z) (x² + y² + z²– xy – yz – zx)
= (3x + y + z) (9x² + y² + z² – 3xy – yz – 3xz)

Question 12.
Verify that x³+ y³ + z³ – 3xyz = 1/2 (x + y + z) [(x – y)² + (y – z)² + (z – x)² ]
(OR)
Verify that
p³ + q³ + r³ – 3pqr = 1/2 (p + q + r)
[(p – q)² + (q – r)² + (r – p)²]
Solution:
Given x³+ y³ + z³ – 3xyz = 1/2 (x + y + z) [(x – y)² + (y – z)² + (z – x)² ]
R-H.S = 1/2 (x + y + z) [(x – y)² + (y – z)²+ (z – x)²]
= 1/2 (x + y + z) [x² + y² – 2xy + y² + z² – 2yz + z² + x² – 2xz]
= 1/2 (x + y + z) [2x² + 2y² + 2z² – 2xy – 2yz – 2zx]
= 1/2 (x + y + z) (2) [x² + y² + z² – xy – yz – zx]
= (x + y + z) (x² + y² + z² – xy – yz – zx)
= L.H.S
Hence proved.

Question 13.
If x + y + z = 0, show that x³ + y³ + z³ = 3xyz
Solution:
Given x + y + z = 0
To prove x³ + y³ + z³ = 3xyz
We have an identity
(x + y + z) (x² + y² + z² – xy – yz – zx)
= x³ + y³ + z³ – 3xyz
Substituting x + y + z = 0in the above equation, we get
0 x (x² + y² + z² -xy-yz-zx)
= x³ + y³ + z³ – 3xyz
⇒ x³ + y³ + z³ – 3xyz = 0
⇒ x³ + y³ + z³ = 3xyz

Question 14.
Without actual calculating the cubes, find the value of each of the following.
i) (- 10)³ + 7³ + 3³
Solution:
Given (-10)³ + 7³ + 3³
Sum of the bases = -10 + 7 + 3 = = 0
∴ (- 10)³ + 7³ + 3³
= 3 (- 10) x (7) x 3
= -630
[ ∵ x + y + z = 0 then x³ + y³ + z³ = 3xyz]

ii) (28)³ + (- 15)³ + (- 13)³
Solution:
Given (28)³ + (- 15)³+ (- 13)³
Sum of the bases = 28 + (- 15) + (- 13) = 0
∴ (28)³ + (- 15)³ + (- 13)³
= 3 x 28 x (- 15) x (- 13)
= 16380

iii) \(\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{3}\right)^{3}-\left(\frac{5}{6}\right)^{3}\) read it as \(\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{3}\right)^{3}+\left(\frac{-5}{6}\right)^{3}\)
Solution:

iv) (0.2)³ – (0.3)³ + (0.1)³
Solution:
Given that (0.2)³ – (0.3)³ + (0.1)³
= (0.2)³ + (- 0.3)³ + (0.1)³
Sum of the bases = 0.2 – 0.3 + 0.1 = 0
∴ (0.2)³ + (-0.3)³ + (0.1)³
= 3 x (0.2) (- 0.3) (0.1)
= -0.018

Question 15.
Give possible expressions for the length and breadth of the rectangle whose area is given by
i) 4a² + 4a – 3
Given that area = 4a² + 4a – 3
= 4a² + 6a – 2a – 3
= 2a (2a + 3) – 1 (2a + 3)
= (2a – 1) (2a + 3)
∴ Length = (2a + 3); breadth = (2a – 1).

ii) 25a² – 35a + 12
Solution:
Given that area = 25a² – 35a +12
= 25a² – 20a – 15a + 12
= 5a (5a – 4) – 3 (5a – 4)
= (5a – 4) (5a – 3)
∴ (5a – 4) (5a – 3) are the length and breadth.

Question 16.
What are the possible polynomial expressions for the dimensions of the cuboids whose volumes are given below ?
i) 3x³ – 12x
Solution:
Volume = 3x³ – 12x
= 3x (x² – 4)
= 3x (x + 2) (x – 2) are the dimensions.

ii) 12y² + 8y – 20
Solution:
Given that volume = 12y² + 8y – 20
= 4 (3y² + 2y – 5)
= 4 [3y² + 5y – 3y – 5]
= 4 [y (3y + 5) – 1 (3y + 5)]
= 4 (3y + 5) (y – 1)
Hence 4, (3y + 5) and (y – 1) are the dimensions.

Question 17.
Show that if 2 (a² + b² ) = (a + b)² then a = b
Solution:
Given that 2 (a² + b² ) = (a + b)²
To prove a = b
As 2 (a² + b² ) = (a + b)²
We have
2a² + 2b² = a² + 2ab + b²
2a² – a² + 2b² – b² = 2ab
a² + b² = 2ab
This is possible only when a = b
∴ a = b

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.4 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 2nd Lesson Polynomials and Factorisation Exercise 2.4

Question 1.
Determine which of the following polynomials has (x + 1) as a factor.
i) x³ – x² – x + 1
Solution:
f(- 1) = (- 1)³ – (- 1)² – (- 1) + 1
= -1 – 1 + 1 + 1 = 0
∴ (x + 1) is a factor.

ii) x⁴ -x³ +x² – x + 1
Solution:
f(- 1) = (- 1)⁴ – (- 1)³ + (- 1)² – (- 1) + 1
= 1 + 1 + 1 + 1 + 1= 5
∴ (x + 1) is not a factor.

iii) x⁴ + 2x³ + 2x² + x + 1
Solution:
f(- 1) = (-1)⁴ + 2 (- 1)³ + 2 (- 1)² + (-1) + 1
= 1 – 2 + 2 – 1 + 1 = 1
∴ (x + 1) is not a factor.

iv) x³ – x² – (3 – √3)x + √3
Solution:
f(- 1) = (- 1)³ – (- 1)² – (3 – √3)(-1) + √3
= – 1 – 1 + 3 – √3 + √3 = 1
∴ (x + 1) is not a factor.

Question 2.
Use the factor theorem to determine whether g(x) is a factor of f(x) in each of the following cases:
i) f(x) = 5x³ + x² – 5x – 1; g(x) = x + 1
[Factor theorem : If f(x) is a polynomial; f(a) = 0 then (x – a) is a factor of f(x); a ∈ R]
Solution:
g(x) = x+ 1 = x- a say
∴ a = – 1
f(a) = f(- 1) = 5 (- 1)³ + (- 1)² – 5 (- 1) – 1
= -5 + 1 + 5 – 1 = 0
∴ x + 1 is a factor of f(x).

ii) f(x) = x³ + 3x² + 3x + 1; g(x) = x + 1
Solution:
g(x) = x + 1 = x – a
∴ a = – 1
f(a) = f(- 1) = (- 1)³ + 3 (- 1)² + 3(-1) + 1
= -1 + 3 – 3 + 1 =0
∴ f(x) is a factor of g(x).

iii) f(x) = x³ – 4x² + x + 6;
g(x) = x – 2
Solution:
g(x) = x- 2 = x- a
∴ a = 2
f(a) = f(2) = 2³ – 4(2)² + 2 + 6
= 8 – 16 + 2 + 6 = 0
∴ g(x) is a factor of f(x).

iv) f(x) = 3x³+ x² – 20x +12; g(x) = 3x – 2
Solution:
g(x) = 3x – 2 = \(x-\frac{2}{3}\) = x – a
∴ a = 2/3

v) f(x) = 4x³+ 20x²+ 33x + 18; g(x) = 2x + 3
Solution:
g(x) = 2x + 3 = x + \(\frac{3}{2}=\) = x – a
∴ a = -3/2

∴ g(x) is a factor of f(x).

Question 3.
Show that (x – 2), (x + 3) and (x – 4) are factors of x³ – 3x² – 10x + 24.
Solution:
Given f(x) = x³ – 3x² – 10x + 24
To check whether (x – 2), (x + 3) and (x – 4) are factors of f(x), let f(2), f(- 3) and f(4)
f(2) = 2³ – 3(2)² – 10(2) + 24
= 8- 12-20 + 24 = 0
∴ (x – 2) is a factor of f(x).

f(- 3) = (- 3)³ – 3(- 3)²– 10(- 3) + 24
= – 27 – 27 + 30 + 24 = 0
∴ (x + 3) is a factor of f(x).

f(4) = (4)³ – 3 (4)² – 10 (4) + 24
= 64 – 48 – 40 + 24
= 88 – 88
= 0
∴ (x – 4) is a factor of f(x).

Question 4.
Show that (x + 4), (x – 3) and (x – 7) are factors of x³ – 6x² – 19x + 84.
Solution:
Let f(x) = x³ – 6x² – 19x + 84
To verify whether (x + 4), (x – 3) and (x – 7) are factors of f(x) we use factor theorem.

Let f(- 4), f(3) and f(7)
f(- 4) = (- 4)³ – 6 (- 4)² – 19 (- 4) + 84
= -64 – 96 + 76 + 84
= 0 .
∴ (x + 4) is a factor of f(x).

f(3) = 3³ – 6(3)² – 19(3) + 84
= 27 – 54 – 57 + 84
= 0
∴ (x – 3) is a factor of f(x).

f(7) = 7³ – 6(7)² – 19(7) + 84
= 343 – 294 – 133 + 84
= 427 – 427
= 0
∴ (x – 7) is a factor of f(x).

Question 5.
If both (x – 2) and \(\left(x-\frac{1}{2}\right)\) of px² + 5x + r, show that p = r.
Solution:
Let f(x) = px²+ 5x + r
As (x – 2) and \(\left(x-\frac{1}{2}\right)\) are factor of f(x), we have f(2) = 0 and f(1/2) = 0
∴ f(2) = p(2)² + 5(2) + r
= 4p + 10 + r = 0
= 4p + r
= – 10 ………………(1)

⇒ p + 10 + 4r = 0
⇒ p + 4r = – 10 ………………. (2)
From (1) and (2);
4p + r = p + 4r
4p – p = 4r – r
3p = 3r
∴ P = r

Question 6.
If (x² – 1) is a factor of ax⁴ + bx³ + cx² + dx + e, show that a + c + e = b + d = 0.
Solution:
Let f(x) = ax⁴ + bx³ + cx² + dx + e
As (x – 1) is a factor of f(x) we have
x² – 1 = (x + 1) (x – 1) hence f(1) = 0 and f(-1) = 0
f(1) = a + b + c + d + e = 0 ……………. (1)
and f(-1) = a- b + c- d + e = 0
⇒ a + c + e = b + d
Substitute this value in equation (1)
a + c + e + b + d=0
b + d + b + d=0
2 (b + d) = 0
⇒ b + d = 0
∴ a + c + e = b + d = 0

Question 7.
Factorise
i) x³ – 2x² – x + 2
Solution:
Let f(x) = x³ – 2x² – x + 2
By trial, we find f(l) = 1³ – 2(1)² – 1 + 2
= 1 – 2 – 1 + 2
= 0 .
∴ (x – 1) is a factor of f(x).
[by factor theorem]
Now dividing f(x) by (x – 1).

f(x) = (x – 1) (x² – x – 2)
= (x – 1) [x² – 2x + x- 2]
= (x – 1) [x (x – 2) + 1 (x – 2)]
= (x – 1) (x – 2) (x + 1)

ii) x³ – 3x² – 9x – 5
Solution:
Let f(x) = x³ – 3x² – 9x – 5By trial,
f(- 1) = (- 1)³ – 3(- 1)² – 9(- 1) – 5
=-1 – 3 + 9 – 5
=0
∴ (x + 1) is a factor of f(x).
[ ∵ by factor theorem]
Now dividing f(x) by (x + 1).

f(x)=(x + 1)(x² – 4x – 5)
But x²– 4x – 5 = x² – 5x + x – 5
= x (x – 5) + 1 (x – 5)
=(x – 5)(x + 1)
∴ f(x)=(x + 1)(x + 1)(x – 5)

iii) x³ + 13x² + 32x + 20
Solution:
Let f(x) = x³ + 13x² + 32x + 20
Let f(- 1)
= (- 1)³ + 13 (- 1)² + 32 (- 1) + 20
= – 1 + 13 – 32 + 20 = 33 – 33 = 0
∴ (x + 1) is a factor of f(x).
[ ∵ by factor theorem] Now dividing f(x) by (x + 1).

iv) y³ + y² – y – 1
Let f(y) = y³ + y² – y – 1
f(1) = 1³+ 1²– 1 – 1 = 0
(y – 1) is a factor of f(y).
Now dividing f(y) by (y – 1).

∴ f(x) = (x + 1)(x² + 12x + 20)
But (x² + 12x + 20) = x²+ 10x + 2x + 20
=x(x + 10)+2(x + 10)
=(x + 10)(x + 2)
∴f(x) = (x + 1)(x + 2)(x + 10)

Question 8.
If ax² + bx + c and bx² + ax + c have a common factor x + 1 then show that c = 0 and a = b.
Solution:
Let f(x) = ax² + bx + c and g(x) = bx² + ax + c given that (x + 1) is a common factor for both f(x) and g(x).
∴ f(-1) = g(- 1)
⇒a(- 1)² + b(- 1) + c
= b(- 1)² + a (- 1) + c
⇒ a – b + c = b – a + c
⇒ a + a = b + b
⇒ 2a = 2b
⇒ a = b
Also f(- 1) = a – b + c = 0
⇒ b – b + c = 0
⇒ c = 0

Question 9.
If x² – x – 6 and x² + 3x – 18 have a common factor x – a then find the value of a.
Solution:
Let f(x) = x² – x – 6 and
g(x) = x² + 3x – 18
Given that (x – a) is a factor of both f(x) and g(x).
f(a) = g(a) = 0
⇒ a² – a – 6 = a² + 3a – 18
⇒ – 4a = – 18 + 6
⇒ – 4a = – 12
∴ a = 3

Question 10.
If (y – 3) is a factor of y³– 2y²– 9y + 18, then find the other two factors.
Solution:
Let f(y) = y³– 2y² – 9y + 18
Given that (y – 3) is a factor of f(y).
Dividing f(y) by (y – 3)

∴ f(y) = (y – 3) (y + y – 6)
But y² + y – 6
= y² + 3y – 2y – 6
= y (y + 3) – 2 (y + 3)
= (y + 3) (y – 2)
∴ f(y) = (y – 2)(y – 3)(y + 3)
The other two factors are (y – 2) and (y + 3).

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.3 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 2nd Lesson Polynomials and Factorisation Exercise 2.3

Question 1.
Find the remainder when
x³ + 3x² + 3x + 1 is divided by the following Linear polynomials i) x + 1 Each
Solution:
Let f(x) = x³ + 3x² + 3x + 1
By remainder theorem, the remainder is f (- 1)
f (- 1) = (- 1)³ + 3(- 1)² + 3(- 1) + 1
= – 1 + 3 – 3 + 1 = 0

ii) \(x-\frac{1}{2}\)
f(x) = x³ + 3x² + 3x + 1
By remainder theorem, the remainder is \(\mathrm{f}\left(\frac{1}{2}\right)\)

iii) x
Solution:
f(x) = x³ + 3x² + 3x + 1
The remainder is f(0)
∴ f(0) = 0³ + 3(0)² + 3(0) + 1 = 1

iv) x + π
Solution:
f(x) = x³ + 3x² + 3x + 1
By remainder theorem, the remainder is f(- π)
f(- π) = (- π)³ + 3(-π)² + 3 (-π) + 1 .
= – π³ + 3π² – 3π + 1

v) 5 + 2x
f(x) = x³ + 3x² + 3x + 1
The remainder is \(\mathrm{f}\left(\frac{-5}{2}\right)\)

Question 2.
Find the remainder when x³ – px² + 6x – p is divided by x – p.
Solution:
Let f(x) = x³ – px² + 6x – p
(x – a) = x – p)
By Remainder theorem, the remainder is f(p)
∴ f(P) = P³ – P(P)² + 6p – p
= p³ – p³ + 5p = 5p

Question 3.
Find the remainder when 2x² – 3x + 5 is divided by 2x – 3. Does it exactly divide the polynomial ? State reason.
Solution:
Let f(x) = 2x² – 3x + 5 and
x – a = 2x – 3 = x – \(\frac{3}{2}\)
By Remainder theorem f(x) when divided by (x – \(\frac{3}{2}\) ) leaves a remainder \(\mathrm{f}\left(\frac{3}{2}\right)\)

As the remainder is 5 we say that (2x – 3) is not a factor of f(x).

Question 4.
Find the remainder when 9x³ – 3x² + x – 5 is divided by x – \(\frac{2}{3}\)
Solution:
Let f(x) = 9x³ – 3x² + x – 5
x-a = x – \(\frac{2}{3}\)
Remainder theorem the remainder is \(\mathrm{f}\left(\frac{2}{3}\right)\)

Question 5.
If the polynomials 2x³ + ax² + 3x – 5 and x³ + x² – 4x + a leave the same remainder, when divided by x – 2, find the value of a.
Solution:
Let f(x) = 2x³ + ax² + 3x – 5
g(x) = x³ + x² – 4x + a
Given that f(x) and g(x) divided x – 2
give same remainder.
i e., f(2) = g(2)
By Remainder theorem.
But f(2) = 2(2)³ + a(2)² + 3(2) – 5
= 2 x 8 + 4a + 6 – 5
= 17 +4a
g(2) = 2³ + 2² – 4(2) + a .
= 8 + 4 – 8 + a = 4 + a
i.e., 4 + a = 17 + 4a
∴ a – 4a = 17 – 4
– 3a = 13
a = -13/3

Question 6.
If the polynomials x³ + ax² + 5 and x³ – 2x² + a are divided by (x + 2) leave the same remainder, find the value of a.
Solution:
Let f(x) = x³ + ax² + 5
g(x) = x³ – 2x² + a
Given that when f(x) and g(x) divided by (x + 2) leaves the same remainder.
i.e.,f(-2) = g(-2)
By Remainder theorem
f(- 2) = (- 2)³ + a(- 2)² + 5
= -8 + 4a + 5 = 4a – 3
g(- 2) = (- 2)³ – 2(- 2)² + a
= -8 – 8 + a = a – 16
By problem,
4a – 3 = a – 16
4a – a = – 16 + 3
⇒ 3a = – 13 ⇒ a = -13/3

Question 7.
Find the remainder when f(x) = x⁴ – 3x² + 4 is divided by g(x) = x – 2 and verify the result by actual division.
Solution:
Given f(x) = x⁴ – 3x² + 4
g(x) = x – 2
The remainder when f(x) is divided by g(x) is f(2).
f(2) = 2⁴ – 3(2)² + 4
= 16 – 12 + 4
= 8
Actual division

∴ The remainder either by Remainder theorem or by actual division is the same.

Question 8.
Find the remainder when p(x) = x³ – 6x² + 14x – 3 is divided by g(x) = 1 – 2x and verify the result by long division method.
Solution:
Given p(x) = x³ – 6x² + 14x – 3
g(x) = 1 – 2x
By Remainder theorem when p(x) is divided by g(x) is p(1/2).

Question 9.
When a polynomial 2x³ + 3x² + ax + b is divided by (x – 2) leaves remainder 2, and (x + 2) leaves remainder – 2. Find a and b.
Solution:Let f(x) = 2x³ + 3x² + ax + b
The remainder when f(x) is divided by (x – 2) is 2.
i.e., f(2) = 2
⇒ 2(2)³ + 3(2)²+ a(2) + b = 2
⇒ 16 + 12 + 2a +b = 2
⇒ 2a + b = – 26 …………………..(1)

Also the remainder when f(x) is divided by (x + 2) is – 2.
i.e., f(- 2) = – 2
⇒ 2(- 2)³ + 3(- 2)² + a (- 2) + b = – 2
⇒ -16 + 12 – 2a + b = – 2
– 2a + b = 2 ………………..(2)
Solving (1) and (2),

b = – 12
and 2a – 12 = – 26
2a = -26+ 12
a = -14/2 = -7,
a = -7, b = – 12

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.2 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 2nd Lesson Polynomials and Factorisation Exercise 2.2

Question 1.
Find the value of the polynomial 4x² – 5x + 3, when
(i) x = 0
Solution:
The value at x = 0 is
4(0)² – 5(0) + 3
= 3

(ii) x = – 1
Solution:
The value at x = – 1 is
4 (- 1)² – 5 (- 1) + 3
= 4 + 5 + 3
= 12

iii) x = 2
Solution:
The value at x = 2 is
4(2)² – 5(2) + 3
= 16 -10 + 3
= 9

iv) x = \(\frac{1}{2}\)
Solution:

Question 2.
Find p(0), p(1) and p(2) for each of the following polynomials.
i) p(x) = x² – x + 1
Solution:
p(0) = 0² – 0 + 1 = 1
p(1) = 1² – 1 + 1 = 1
p(2) = 2² – 2 + 1 = 3

ii) P(y) = 2 + y + 2y² – y³
Solution:
p(0) = 2 + 0 + 2(0)² – 0³ = 2
p(1) = 2+ 1 + 2(1)² – 1³ = 4
p(2) = 2 + 2 + 2(2)² -2³ = 4 + 8- 8 = 4

iii) P(z) = z³
Solution:
p(0) = 0³ = 0
p(1) = 1³ = 1
p(2) = 2³ = 8

iv) p(t) = (t – 1)(t + 1) = t² – 1
Solution:
p(0) = (0 – 1) (0 + 1) = – 1
p(1) = t² – 1 = 1² – 1 = 0
p(2) = 2² – 1 = 4 – 1 = 3

v) p(x) = x² – 3x + 2
Solution:
p(0) = 0² – 3(0) + 2 = 2
p(1) = 1² – 3(1) + 2 = 1 – 3 + 2 = 0
p(2) = 2² – 3(2) + 2 = 4- 6 + 2 = 0

Question 3.
Verify whether the values of x given in each case are the zeroes of the polynomial or not ?
i) p(x) = 2x + 1; x = \(\frac{-1}{2}\)
Solution:
The value of p(x) at x = \(\frac{-1}{2}\) is
\(\mathrm{p}\left(\frac{-1}{2}\right)=2\left(\frac{-1}{2}\right)+1\)
= -1 + 1 = 0
∴ x = \(\frac{-1}{2}\) is a zero of p(x).

(ii) p(x) = 5x – π ; x = \(\frac{-3}{2}\)
Solution:
The value of p(x) at x = \(\frac{-3}{2}\) is
\(\mathrm{p}\left(\frac{-3}{2}\right)=5\left(\frac{-3}{2}\right)-\pi=\frac{-15}{2}-\pi \neq 0\)
∴ x = \(\frac{-3}{2}\) is not a zero of p(x).

iii) p(x) = x² – 1; x = ±1
Solution:
The value of p(x) at x = 1 and – 1 is
p(1) = 1² – 1 = 0
p(-1) = (-1)² -1 = 0
∴ x = ±1 is a zero of p(x).

iv) p(x) = (x – 1) (x + 2); x = – 1, – 2
Solution:
The value of p(x) at x = – 1 is
p(-1) = (-1 – 1) (-1 + 2)
=-2 x 1 =-2 ≠ 0
Hence x = – 1 is not a zero of p(x).
And the value of p(x) at x = – 2 is
p (- 2) = (- 2 – 1) (- 2 + 2) = – 3 x 0 = 0
Hence, x = – 2 is a zero of p(x).

v) p(y) = y²; y = o
Solution:
The value of p(y) at y = 0 is p(0) = 0² = 0
Hence y = 0 is a zero of p(y).

vi) p(x) = ax + b ; x = \(\frac{-\mathbf{b}}{\mathbf{a}}\)
Solution:
The value of p(x) at x = \(\frac{-\mathbf{b}}{\mathbf{a}}\) is
\(\mathrm{p}\left(\frac{-\mathrm{b}}{\mathrm{a}}\right)=\mathrm{a}\left(\frac{-\mathrm{b}}{\mathrm{a}}\right)+\mathrm{b}\)
= -b + b = 0
∴ x = \(\frac{-\mathbf{b}}{\mathbf{a}}\) is a zero of p(x).

vii) f(x) = 3x² – 1; x = \(\frac{-1}{\sqrt{3}}, \frac{2}{\sqrt{3}}\)
Solution:

viii) f(x) = 2x – 1; x = \(\frac{1}{2} ;-\frac{1}{2}\)
Solution:

Question 4.
Find the zero of the polynomial in each of the following cases.
i) f(x) = x + 2
Solution:
x + 2 = 0
x = – 2

ii) f(x) = x – 2
Solution:
x – 2 = 0
x = 2

iii) f(x) = 2x + 3
Solution:
2x + 3 = 0
2x = – 3
x = \(\frac{-3}{2}\)

iv) f(x) = 2x – 3
Solution:
2x – 3 = 0
2x = 3
x = \(\frac{3}{2}\)

v) f(x) = x²
Solution:
x² = 0
x = 0

vi) f(x) = px, p ≠ 0
Solutin:
px = 0
x = 0

vii) f(x) = px + q; p ≠ 0; p, q are real numbers.
Solution:
px + q = 0
px = -q
x = \(\frac{-\mathrm{q}}{\mathrm{p}}\)

Question 5.
If 2 is a zero of the polynomial p(x) = 2x² – 3x + 7a, find the value of
a.
Solution:
Given that 2 is a zero of p(x) = 2x² – 3x + 7a
(i.e.) p(2) = 0
⇒ 2(2)² – 3(2) + 7a = 0
⇒ 8 – 6 + 7a = 0
⇒ 2 + 7a = 0
⇒ 7a = – 2
⇒ a = \(\frac{-2}{7}\)

Question 6.
If 0 and 1 are the zeroes of the polynomial f(x) = 2x³ – 3x² + ax + b, find the values of a and b.
Solution:
Given that f(0) = 0; f(1) = 0 and
f(x) = 2x³ – 3x² + ax + b
∴ f(0) = 2(0)³ – 3(0)² + a(0) + b
⇒ 0 = b
Also f(1) = 0
⇒ 2(1)³ – 3(1)² + a(1) + 0 = 0
⇒ 2 – 3 + a = 0 .
⇒ a = 1
Hence a = 1; b = 0

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.1 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 2nd Lesson Polynomials and Factorisation Exercise 2.1

Question 1.
Find the degree of each of the polynomials given below,
i) x⁵ – x⁴ + 3
Solution:
Degree is 5.

ii) x² + x – 5
Solution:
Degree is 2.

iii) 5
Solution:
Degree is 0.

iv) 3x⁶ + 6y³ – 7
Solution:
Degree is 6.

v) 4 – y²
Solution:
Degree is 2.

vi) 5t – √3
Solution:
Degree is 1.

Question 2.
Which of the following expressions are polynomials in one variable and which are not ? Give reasons for your answer.
i) 3x² – 2x + 5
Solution:
Given expression is a polynomial in one variable.

ii) x² + √2
Solution:
Given expression is a polynomial in one variable.

iii) p² – 3p + q
Solution:
Given expression is not a polynomial in one variable. It involves two variables p and q.

iv) y + \(\frac{2}{\mathbf{y}}\)
Solution:
Given expression is not a polynomial. Since the second term contains the variable in its denominator.

v) \(5 \sqrt{x}+x \sqrt{5}\)
Solution:
Given expression is not a polynomial. Since the first term’s exponent is not an integer.

vi) x¹⁰⁰ + y¹⁰⁰
Solution:
Given expression has two variables. So it is not a polynomial in one variable.

Question 3.
Write the coefficient of x³ in each of the following.
i) x³ + x + 1
ii) 2 – x³+ x²
iii) \(\sqrt{2} x^{3}+5\)
iv) 2x³ + 5
v) \(\frac{\pi}{2} x^{3}+x\)
vi) \(-\frac{2}{3} x^{3}\)
vii) 2x² + 5
viii) 4
Solution:
i) x³ + x + 1 : co-efficient of x³ is 1.
ii) 2 – x³+ x² : co-efficient of x³ is – 1.
iii) \(\sqrt{2} x^{3}+5\) co-efficient of x³ is √2
iv) 2x³ + 5 : co-efficient of x³ is 2.
v) \(\frac{\pi}{2} x^{3}+x\) co-efficient of x³ is \(\frac{\pi}{2}\)
vi) \(-\frac{2}{3} x^{3}\) co-efficient of x³ is \(-\frac{2}{3}\)
vii) 2x² + 5 : co-efficient of x³ is ‘0’.
viii) 4 : co-efficient of x³ is ‘0’.

Question 4.
Classify the following as linear, quadratic and cubic polynomials.
i) 5x²+ x – 7 : degree 2 hence quadratic polynomial.
ii) x – x³ , : degree 3 hence cubic polynomial.
iii) x² + x + 4 : degree 2 hence quadratic polynomial.
iv) x – 1 : degree 1 hence linear polynomial.
v) 3p : degree 1 hence linear polynomial.
vi) πr² : degree 2 hence quadratic polynomial.

Question 5.
Write whether the following statements are True or False. Justify your answer.
i) A binomial can have at the most two terms
ii) Every polynomial is a binomial
iii) A binomial may have degree 3
iv) Degree of zero polynomial is zero
v) The degree of x² + 2xy + y² is 2
vi) πr² is monomial
Solution :
i) A binomial can have at the most two terms -True
ii) Every polynomial is a binomial – False
[∵ A polynomial can have more than two terms]
iii) A binomial may have degree 3 – True
iv) Degree of zero polynomial is zero – False
v) The degree of x² + 2xy + y² is 2 – True
vi) πr² is monomial – True

Question 6.
Give one example each of a monomial and trinomial of degree 10.
Solution :
– 7x¹⁰ is a monomial of degree 10.
3x²y⁸ + 7xy – 8 is a trinomial of degree 10.

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 1 Real Numbers Ex 1.3 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 1st Lesson Real Numbers Exercise 1.4

Question 1.
Simple the following expressions.
i) (5 + √7) (2 + √5)
Solution:
(5 + √7) (2 + √5)
= 10 + 5√5 + 2√7 + √35

ii) (5 + √5) (5 – √5)
Solution:
(5 + √5) (5 – √5)
= 5² + (√5)²
= 25 – 5 = 20

(iii) (√3 + √7)²
Solution:
(√3 + √7)²
= (√3)² + (√7)² + 2(√3)(√7)
= 3 + 7 + 2√21
= 10 + 2√21

iv) (√11 – √7) (√11 + √7)
= (√11)² – (√7)²
= 11 – 7 = 4

Question 2.
Classify the following numbers as rational or irrational.
i) 5 – √3
ii) √3 + √2
iii) (√2 – 2)²
iv) \(\frac{2 \sqrt{7}}{7 \sqrt{7}}\)
v) 2π
vii) (2 +√2) (2 – √2)
Solution:
i) 5 – √3 – irrational
ii) √3 + √2 – irrational
iii) (√2 – 2)² – irrational
iv) \(\frac{2 \sqrt{7}}{7 \sqrt{7}}\) – rational
v) 2π – Transcendental number. (not irrational)
vi) \(\frac{1}{\sqrt{3}}\)– irrational
vii) (2 +√2) (2 – √2) – rational

Question 3.
In the following equations, find whether variables x, y, z etc., represents rational or irrational numbers.
i) x² = 7
ii) y² = 16
iii) z² = 0.02
iv) u² = \(\frac{17}{4}\)
v) w² = 27
vi) t⁴ = 256
Solution:
i) x² = 7
⇒ x = √7 is an irrational number.
ii) y² = 16 ⇒ y = 4 is a rational number.
iii) z² = 0.02 ⇒ z = \(\sqrt{0.02}\) is an irrational number.
iv) u² = \(\frac{17}{4}\) ⇒ x = \(\frac{\sqrt{17}}{2}\) is an irrational number.
v) w² = 27 ⇒ w = \(3 \sqrt{3}\) an irrational number.
vi) t⁴ = 256 ⇒ t² = \(\sqrt{256}\) = 16
⇒ t = \(\sqrt{16}\) = 4 is a rational number

Qeustion 4.
The ratio of circumference to the diameter of a circle c/d is represented by π. But we say that π is an irrational number. Why?

Question 5.
Rationalise the denominators of the following.
i) \(\frac{1}{3+\sqrt{2}}\)
Solution:

ii) \(\frac{1}{\sqrt{7}-\sqrt{6}}\)
Solution:

iii) \(\frac{1}{\sqrt{7}}\)
Solution:

iv) \(\frac{\sqrt{6}}{\sqrt{3}-\sqrt{2}}\)
Solution:

Question 6.
Simplify each of the following by rationalising the denominator.
i) \(\frac{6-4 \sqrt{2}}{6+4 \sqrt{2}}\)
Solution:

ii) \(\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}\)
Solution:

iii) \(\frac{1}{3 \sqrt{2}-2 \sqrt{3}}\)
Solution:

iv) \(\frac{3 \sqrt{5}-\sqrt{7}}{3 \sqrt{3}+\sqrt{2}}\)
Solution:

Question 7.
Find the value of \(\frac{\sqrt{10}-\sqrt{5}}{2 \sqrt{2}}\) upto three decimal places. (take \(\sqrt{2}\) = 1.414, \(\sqrt{3}\) = 1.732 and \(\sqrt{5}\) = 2.236).
Solution:

Question 8.
Find
i) 64^1/6
Solution:
= (2⁶)^1/6
= 6

ii) 32^1/5
Solution:
32^1/5
= (2⁵)^1/5
= 2

iii) 625^1/4
625^1/5
= (5⁴)^1/4
= 5

iv) 16^3/2
Solution:
16^3/2
= (4²)^3/2

v) 243^2/5
Solution:
243^2/5
= (3⁵)^2/5

vi) (46656)^-1/6
Solution:

Question 9.
Simplify \(\sqrt[4]{81}-8 \sqrt[3]{343}+15 \sqrt[5]{32}+\sqrt{225}\)
Solution:

Question 10.
If ‘a’ and ‘b’ are rational numbers, find the values of a and b in each of the following equations.
i) \(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}=\mathbf{a}+\mathbf{b} \sqrt{6}\)
Solution:
Given that \(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}=\mathbf{a}+\mathbf{b} \sqrt{6}\)
Rationalising the denominator we get

Comparing 5 + 2√6 with a + b√6
We have a = 5 and b = 2

ii) \(\frac{\sqrt{5}+\sqrt{3}}{2 \sqrt{5}-3 \sqrt{3}}=a-b \sqrt{15}\)
Solution:
Given that \(\frac{\sqrt{5}+\sqrt{3}}{2 \sqrt{5}-3 \sqrt{3}}=a-b \sqrt{15}\)
Rationalising the denominator we get

AP Board 9th Class Maths Solutions

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 1 Real Numbers Ex 1.3 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 1st Lesson Real Numbers Exercise 1.3

Question 1.
Visualise \(2.8 \overline{74}\) on the number line, using successive magnification.
Solution:

Question 2.
Visualise \(5 . \overline{28}\) on the number line, upto 3 decimal places.
Solution:

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 1 Real Numbers Ex 1.2 Textbook Questions and Answers.

AP State Syllabus 9th Class Maths Solutions 1st Lesson Real Numbers Exercise 1.2

Question 1.
Classify the following numbers as rational or irrational.
i) \(\sqrt{27}\)
ii) \(\sqrt{441}\)
iii) 30.232342345
iv) 7.484848
v) 11.2132435465
vi) 0.3030030003
Solution:
i) \(\sqrt{27}\) – irrational number
ii) \(\sqrt{441}\) = 21 – rational
iii) 30.232342345 – irrational number
iv) 7.484848 – rational number
v) 11.2132435465 – irrational number
vi) 0.3030030003 – irrational number

Question 2.
Explain with an example how irrational numbers differ from rational numbers ?
Solution:
Irrational numbers can’t be expressed in \(\frac { p }{ q }\) form where p and q are integers and q ≠ 0.
E.g.\(\sqrt{2}, \sqrt{3} ; \sqrt{5}, \sqrt{7}\) etc.
Where as a rational can be expressed in \(\frac { p }{ q }\) form
E.g. :- -3 = \(\frac { -3 }{ 1 }\) and \(\frac { 5 }{ 4 }\) etc.

Question 3.
Find an irrational number between \(\frac { 5 }{ 7 }\) and \(\frac { 7 }{ 9 }\). How many more there may be ?
Solution :
The decimal forms of \(\frac { 5 }{ 7 }\) and \(\frac { 7 }{ 9 }\) are
\(\frac{5}{7}=0 . \overline{714285} \ldots ., \frac{7}{9}=0.7777 \ldots \ldots=0 . \overline{7}\)
∴ An irrational between \(\frac { 5 }{ 7 }\) and \(\frac { 7 }{ 9 }\) is 0.727543…………
There are infinitely many irrational numbers between \(\frac { 5 }{ 7 }\) and \(\frac { 7 }{ 9 }\).

Question 4.
Find two irrational numbers between 0.7 and 0.77.
Solution:
Two irrational numbers between 0.7 and 0.77 can take the form
0.70101100111000111…………. and 0.70200200022……………

Question 5.
Find the value of √5 uPto 3 decimal places.
Solution:

[√5 is not exactly equal to 2.2350679………….. as shown ¡n calculators]

Question 6.
Find the value of √7 upto six decimal places by long division method.
Solution:

Question 7.
Locate \(\sqrt{\mathrm{10}}\) on number line.

Step – 1 : Draw a number line.
Step – 2 : Draw a rectangle OABC at zero with measures 3 x 1. i.e., length 3 units and breadth 1 unit.
Step – 3 : Draw the diagonal OB.
Step – 4 : Draw an arc with centre ‘O’ and radius OB which cuts the number line at D.
Step – 5 : ‘D’ represents \(\sqrt{\mathrm{10}}[latex] on the number line.

Question 8.
Find atleast two irrational numbers between 2 and 3.
Solution:
An irrational number between a and b is Tab [latex]\sqrt{\mathrm{ab}}\) unless ab is a perfect square.
∴ Irrational number between 2 and 3 is √6

∴ Required irrational numbers are 6^1/2, 24^1/4

Method – II:
Irrational numbers between 2 and 3 are of the form 2.12111231234………….. and 3.13113111311113…….

Question 9.
State whether the following statements are true or false. Justify your answers.
Solution:

1. Every irrational number is a real number – True (since real numbers consist of rational numbers and irrational numbers)
2. Every rational number is a real number – True (same as above)
3. Every rational number need not be a rational number – False (since all rational numbers are real numbers).
4. \(\sqrt{n}\) is not irrational if n is a perfect square – True. (since by definition of an irrational number).
5. \(\sqrt{n}\) is irrational if n is not a perfect square – True. (same as above)
6. All real numbers are irrational – False (since real numbers consist of rational