Bhagya

Students can download Maths Chapter 3 Algebra Ex 3.11 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.11

Question 1.
Solve the following quadratic equations by completing the square method
(i) 9x² – 12x + 4 = 0
Answer:
9x² – 12x + 4 = 0
x² – \(\frac { 12x }{ 9 } \) + \(\frac { 4 }{ 9 } \) = 0 (Divided by 9)
x² – \(\frac { 4x }{ 3 } \) = \(\frac { -4 }{ 9 } \)
Add [\(\frac { 1 }{ 2 } \) (\(\frac { 4 }{ 3 } \))]² on both sides

The solution is \(\frac { 2 }{ 3 } \)

(ii) \(\frac { 5x+7 }{ x-1 } \) = 3x + 2
Answer:
(3x + 2) (x – 1) = 5x + 7
3x² – 3x + 2x – 2 = 5x + 7 ⇒ 3x² – x – 5x – 2 – 7 = 0
3x² – 6x – 9 = 0 ⇒ x² – 2x – 3 = 0 (divided by 3)
x² – 2x = 3
Adding (\(\frac { 1 }{ 2 } \) × 2)² on both sides
x² – 2x + 1 = 3 + 1
(x – 1)² = 4 ⇒ x – 1 = \(\sqrt { 4 }\)
x – 1 = ±2
x – 1 = 2 or x – 1 = -2
x = 3 or x = -1
The solution set is -1 and 3

Question 2.
Solve the following quadratic equations by formula method
(i) 2x² – 5x + 2 = 0
Answer:
a = 2, b = -5, c = 2

The solution set is \(\frac { 1 }{ 2 } \) and 2

(ii) \(\sqrt { 2 }\) f² – 6 f + 3 \(\sqrt { 2 }\) = 0
Answer:
Here a = \(\sqrt { 2 }\), b = -6 and c = 3\(\sqrt { 2 }\)

The solution set is \(\frac{3+\sqrt{3}}{\sqrt{2}}\) and \(\frac{3-\sqrt{3}}{\sqrt{2}}\)

(iii) 3y² – 20y – 23 = 0
Answer:
a = 3, b = -20, c = -23

The solution set is -1 and \(\frac { 23 }{ 3 } \)

(iv) 36y² – 12ay + (a² – b²) = 0
Answer:
Here a = 36, b = -12a, c = a² – b²

The solution set is \(\frac { (a+b) }{ 6 } \) and \(\frac { (a-b) }{ 6 } \)

Question 3.
A ball rolls down a slope and travels a distance d = t² – 0.75t feet in t seconds. Find the time when the distance travelled by the ball is 11.25 feet.
Answer:
Distance = t² – 0.75t
11.25 = t² – 0.75t
Multiply by 100
1125 = 100t² – 75t
100t² – 75t – 1125 = 0 (Divided by 25)
4t² – 3t – 45 = 0
a = 4,
b = -3,
c = -45

(time will not be negative)
The required time = \(\frac { 15 }{ 4 } \) seconds
= 3 \(\frac { 3 }{ 4 } \) second or 3.75 seconds

Students can download Maths Chapter 4 Geometry Ex 4.3 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 4 Geometry Ex 4.3

Question 1.
The diameter of the circle is 52 cm and the length of one of its chord is 20 cm. Find the distance of the chord from the centre.
Radius of a circle = \(\frac{56}{2}\) = 26 cm
Solution:

Length of the chord = 20 cm
AC = \(\frac{20}{2}\)
= 10 cm
In ΔOAC, OC² = OA² – AC²
= 26² – 10²
= (26 + 10) (26 – 10)
= 36 × 16
OC = \(\sqrt{30×16}\)
= 6 × 4 cm
= 24 cm
Distance of the chord from the centre = 24 cm.

Question 2.
The chord of length 30 cm is drawn at the distance of 8 cm from the centre of the circle. Find the radius of the circle.
Solution:

Distance AC = \(\frac{1}{2}\) × Length of chord
= \(\frac{1}{2}\) × 30
= 15 cm
Distance from the centre = 8 cm
In ΔOAC Radius (OA) = \(\sqrt{AC² + OC²}\)
= \(\sqrt{15² + 8²}\)
= \(\sqrt{224 + 64}\)
= \(\sqrt{289}\)
= 17
Radius of the circle = 17 cm.

Question 3.
Find the length of the chord AC where AB and CD are the two diameters perpendicular to each other of a circle with radius 4√2 cm and also find ∠OAC and ∠OCA.
Solution:
Radius of a circle = 4√2 cm
In the right ΔAOC,

AC² = OA² + OC²
AC² = (4√2)² + (4√2)²
= 32 + 32 = 64
AC = \(\sqrt{64}\)
= 8
Length of the chord = 8 cm,
∠OAC = ∠OCA = 45°
Since OAC is an isosceles right angle triangle.

Question 4.
A chord is 12 cm away from the centre of the circle of radius 15cm. Find the length of the chord.
Solution:
Radius of a circle (OA) = 15 cm
Distance from centre to the chord (OC) = 12 cm
In the right ΔOAC,

AC² = OA² – OC²
= 15² – 12²
= 225 – 144
= 81
AC = \(\sqrt{81}\)
= 9
Length of the chord (AB) = AC + CB
= 9 + 9 = 18 cm.

Question 5.
In a circle, AB and CD are two parallel chords with centre O and radius 10 cm such that AB = 16 cm and CD = 12 cm determine the distance between the two chords?
Solution:
Length of the chord (AB) = 16 cm
∴ AF = \(\frac{1}{2}\) × 16
= 8 cm
Length of the chord (CD) = 12 cm
∴ CE = \(\frac{1}{2}\) × 12
= 6 cm

In the right ΔOCE,
OE² = OC² – CE²
= 10² – 6²
= 100 – 36
= 64
OE = \(\sqrt{64}\)
= 8 cm
In the right ΔOAF,
OF² = OA² – AF²
= 10² – 8²
= 100 – 64
= 36
OE = \(\sqrt{36}\)
= 6 cm
Distance between the two chords = OE + OF
= 8 + 6
= 14 cm.

Question 6.
Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.
Ans

Two circules intersect at C and D
CD is the common chord.
CD = AC + AD
= 3 + 3
= 6 cm
Length of the common chord = 6 cm.

Question 7.
Find the value of x° in the following figures:

Solution:
(i) ∠BOC = 30° + 60°
= 90°
∠BAC (x) = \(\frac{1}{2}\) ∠BOC (By theorem)
= \(\frac{1}{2}\) × 90°
x = 45°

(ii) Join OP
∠QPR = \(\frac{80}{2}\) = 40° (Angle of the circumference is \(\frac{1}{2}\) the angle at the centre)
∠RPO = 30° (OP and OR are equal sides)
∠OPQ = 40° – 30°
= 10°
∠OQP = 10° (OQ and OP are equal sides)
∴ x = 10°

(iii) ∠OPN = ∠ONP (ON and OP are the radius of the circle)
= 90°
In ΔOPN
∠PON + ∠ONP + ∠NPO = 180° (Sum of the angles of a triangle)
70° + x° + x° = 180°
2x° + 70° = 180°
2x = 180° – 70°
2x = 110°
X = \(\frac{110°}{2}\)
= 55°
The value of x = 55°

(iv) Reflex ∠YOZ = 2∠YXZ
= 2(120°)
= 240°
∠YOZ = 360° – reflex ∠YOZ
= 360° – 240°
= 120°
∴ x = 120°

(v) ∠OAC + ∠OCA = 180° – 100°
= 80°
∠OAC = \(\frac{1}{2}\) × 80° [Since OA = OC, (∴ ∠OAC = ∠OCA)]
= 40°
∠OBA + ∠OAB = 180° – 140°
= 40° [Since ∠OBA = ∠OAB, since OB = OA]
∴∠OAB = \(\frac{40°}{2}\)
= 20°
∠BAC = ∠OAB + ∠OAC
= 20° + 40°
x = 60°

Question 8.
In the given figure, ∠CAB = 25°, find ∠BDC, ∠DBA and ∠COB.

Solution:
In ΔACP,
∠ACP = 180° – (25° + 90°)
= 180° – 115°
= 65°
∠CBA = ∠CAB = 25° [Both the angles are standing in the same base]
∠DBA = 65° [∠DBA and ∠BCA standing in the same base]
∠COB = 50°

Students can download Maths Chapter 3 Algebra Ex 3.10 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.10

Question 1.
Solve the following quadratic equations by factorization method
(i) 4x² – 7x – 2 = 0
Answer:
4x² – 7x – 2 = 0
4x² – 8x + x – 2 = 0
4x(x – 2) + 1(x – 2) = 0
(x – 2) + (4x + 1) = 0

x – 2 = 0 or 4x + 1 = 0 (equate the product of factors to zero)
x = 2 or 4x = -1 ⇒ x = \(\frac { -1 }{ 4 } \)
The roots are 2; \(\frac { -1 }{ 4 } \)

(ii) 3(p² – 6) = p(p + 5)
Answer:
3p² – 18 = p² + 5p

2p² – 5p – 18 = 0
2p² – 9p + 4p – 18 = 0
p(2p – 9) + 2(2p – 9) = 0
(2p – 9)(p + 2) = 0
2p – 9 = 0 or p + 2 =
The roots are p = \(\frac { 9 }{ 2 } \), -2

(iii) \(\sqrt{a(a-7)}\) = 3 \(\sqrt{2}\)
Answer:
Squaring on both sides
a(a – 7) = (3\(\sqrt{2}\))²

a² – 7a = 18
a² – 7a – 18 = 0
(a – 9) (a + 2) = 0
a – 9 = 0 or a + 2 = 0
The roots are -2 and 9

(iv) \(\sqrt { 2 }\) x² + 7x + 5\(\sqrt { 2 }\) = 0
Answer:
\(\sqrt { 2 }\) x² + 7x + 5 \(\sqrt { 2 }\) = 0
\(\sqrt { 2 }\) x² + 2x + 5x + 5\(\sqrt { 2 }\) = 0
\(\sqrt { 2 }\) x (x + \(\sqrt { 2 }\)) + 5(x + \(\sqrt { 2 }\)) = 0

(x + \(\sqrt { 2 }\)) + (\(\sqrt { 2 }\)x + 5) = 0 (equate the product of factors to zero)
x + – \(\sqrt { 2 }\) = 0 or \(\sqrt { 2 }\)x + -5
x = \(\frac{-5}{\sqrt{2}}\)
The roots are – \(\sqrt { 2 }\), \(\frac{-5}{\sqrt{2}}\)

(v) 2x² – x + \(\frac { 1 }{ 8 } \) = 0
Answer:
2x² -x + \(\frac { 1 }{ 8 } \) = 0
16x² – 8x + 1 = (multiply by 8)

16x² – 4x – 4x + 1
4x(4x – 1) – 1 (4x – 1) = 0
(4x- 1) (4x- 1) = 0
4x = 1, 4x = 1
x = \(\frac { 1 }{ 4 } \), x = \(\frac { 1 }{ 4 } \)
The roots are \(\frac { 1 }{ 4 } \) and \(\frac { 1 }{ 4 } \)

Question 2.
The number of volleyball games that must be scheduled in a league with n teams is given by G(n) = \(\frac{n^{2}-n}{2}\) where each team plays with every other team exactly once. A league schedules 15 games. How many teams are in the league?
Solution:
G(n) = \(\frac{n^{2}-n}{2}\)
⇒ 15 = \(\frac{n^{2}-n}{2}\) ⇒ 30 = n² – n
n² – n – 30 = 0
⇒ n² – 6n + 5n – 30 = 0
n(n – 6) + 5 (n – 6) = 0
(n – 6)(n + 5) = 0 ⇒ n = 6, -5
As n cannot be (-ve), n = 6.
∴ There are 6 teams in the league.

Students can download Maths Chapter 5 Coordinate Geometry Additional Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 5 Coordinate Geometry Additional Questions

I. Multiple Choice Questions.

Question 1.
On which quadrant does the point (- 4, 3) lie?
(a) I
(b) II
(c) III
(d) IV
Solution:
(b) II

Question 2.
The point whose abscissa is 5 and lies on the x-axis is …….
(a) (-5, 0)
(b) (5, 5)
(c) (0, 5)
(d) (5, 0)
Solution:
(d) (5, 0)

Question 3.
A point which lies in the III quadrant is ……..
(a) (5, 4)
(b) (5, -4)
(c) (-5, -4)
(d) (-5, 4)
Solution:
(c) (-5, -4)

Question 4.
A point on the y-axis is ……..
(a) (1, 1)
(b) (6, 0)
(c) (0, 6)
(d) (-1, -1)
Solution:
(c) (0, 6)

Question 5.
The distance between the points (4, -1) and the origin is ……..
(a) \(\sqrt{24}\)
(b) \(\sqrt{37}\)
(c) \(\sqrt{26}\)
(d) \(\sqrt{17}\)
Solution:
(d) \(\sqrt{17}\)

Question 6.
The distance between the points (-1, 2) and (3, 2) is ……..
(a) \(\sqrt{14}\)
(b) \(\sqrt{15}\)
(c) 4
(d) 0
Solution:
(c) 4

Question 7.
The centre of a circle is (0, 0). One end point of a diameter is (5, -1), then the radius is …….
(a) \(\sqrt{24}\)
(b) \(\sqrt{37}\)
(c) \(\sqrt{26}\)
(d) \(\sqrt{17}\)
Solution:
(c) \(\sqrt{26}\)

Question 8.
The point (0, -3) lies on
(a) + ve x-axis
(b) + ve y-axis
(c) – ve x-axis
(d) – ve y-axis
Solution:
(d) – ve y-axis

Question 9.
The point which is on y-axis with ordinate -5 is ……..
(a) (0, -5)
(b) (-5, 0)
(c) (5, 0)
(d) (0, 5)
Solution:
(a) (0, -5)

Question 10.
The diagonal of a square formed by the points (1, 0), (0, 1), (-1, 0) and (0, -1) is …….
(a) 2
(b) 4
(c) √2
(d) 8
Solution:
(a) 2

Question 11.
The distance between the points (-2, 2) and (3, 2) is ……..
(a) 10 units
(b) 5 units
(c) 5√3 units
(d) 20 units
Solution:
(b) 5 units

Question 12.
The midpoint of the line joining the points (1, -1) and (-5, 3) is ……..
(a) (2, 1)
(b) (2, -1)
(c) (-2, -1)
(d) (-2, 1)
Solution:
(d) (-2, 1)

Question 13.
If the centroid of a triangle is at (1, 3) and two of its vertices are (-7, 6) and (8, 5) then the third vertex is ……..
(a) (-2, 2)
(b) (2, -2)
(c) (-2, -2)
(d) (2, 2)
Solution:
(b) (2, -2)

Question 14.
The ratio in which the X-axis divides the line segment joining the points (6, 4) and (1, -7) is ……..
(a) 1 : 2
(b) 2 : 3
(c) 4 : 7
(d) 7 : 4
Solution:
(c) 4 : 7

Question 15.
The centroid of a triangle (3, -5), (-7, 4) and (10, -2) is …….
(a) (2, -1)
(b) (2, 1)
(c) (-2, 1)
(d) (1, -2)
Solution:
(a) (2, -1)

II. Answer the Following Questions.

Question 1.
Show that the given points (1, 1), (5, 4), (-2, 5) are the vertices of an isosceles right angled triangle.
Solution:
Let A (1, 1), B (5, 4) and G (-2, 5)

AB = 5, AC = 5
∴ ABC is an isosceles triangle …….. (1)
BC² = AB² + AC²
50 = 25 + 25 ⇒ 50 = 50
∴ ∠A = 90° ……… (2)
From (1) and (2) we get ABC is an isosceles right angle triangle.

Question 2.
Show that the point (3, -2), (3, 2), (-1, 2) and (-1, -2) taken in order are the vertices of a square.
Solution:

= \(\sqrt{16}\)
= 4
AB = BC = CD = DA = 4. All the four sides are equal.
∴ ABCD is a Rhombus ……..(1)

Diagonal AC = Diagonal BD = \(\sqrt{32}\) ……..(2)
From (1) and (2) we get ABCD is a square.

Question 3.
Show that the point A (3, 7) B (6, 5) and C (15, -1) are collinear.
Solution:

AB + BC = AC ⇒ \(\sqrt{13}\) + 3\(\sqrt{13}\) = 4\(\sqrt{13}\)
∴ The points A, B, C are collinear.

Question 4.
Find the type of triangle formed by (-1, -1), (1, 1) and (-√, √3)
Solution:
Let the point A (-1, -1), B (1, 1) and C (-√3, √3)

AB = BC = AC = √8
∴ ABC is an equilateral triangle.

Question 5.
Find x such that PQ = QR where P(6, -1) Q(1, 3) and R(x, 8) respectively.
Solution:

But PQ = QR
\(\sqrt{(x-1)^{2}+25}\) = \(\sqrt{41}\)
Squaring on both sides
(x – 1)² + 25 = 41
(x – 1)² = 41 – 25 = 16
x – 1 = \(\sqrt{16}\) = ± 4
x – 1 = 4 (or) x – 1 = – 4
x = 5 (or) x = -4 + 1 = -3
The value of x = 5 or – 3

Question 6.
Find the coordinate of the point of trisection of the line segment joining (4, -1) and
Solution:
Let A (4, -1) and B (-2, -3) are the given points
Let P (a, b) and Q (c, d) be the points of trisection of AB.
∴ AP = PQ = QB

The required coordinate P is (2, –\(\frac{5}{3}\)) and Q is (0, –\(\frac{7}{3}\))

Question 7.
Find the ratio in which the line segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6).
Solution:
Given points are A(-3, 10), B(6, -8) and P(-1, 6)
divides AB internally in the ratio m : n
By section formula.
A line divides internally in the ratio m : n the point P =

∴ \(\frac{6m-3n}{m+n}\) = -1
6m – 3n = -m – n
6m + m = 3n – n
7m = 2n ⇒ \(\frac{m}{n}\) = \(\frac{2}{7}\)
∴ m : n = 2 : 7
and
\(\frac{-8m+10n}{m+n}\) = 6
-8m + 10n = 6m + 6n
-8m – 6m = 6n – 10n
14m = 4n
∴ \(\frac{m}{n}\) = \(\frac{14}{4}\) = \(\frac{2}{7}\)
Hence P divides AB internally in the ratio 2 : 7

Question 8.
If (1, 2) (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find “x” and “y”.
Solution:
Let A(1, 2), B(4, y), C(x, 6) and D(3, 5)

Since ABCD is a parallelogram the diagonal bisect each other
Mid point of AC = Mid point of BD
(\(\frac{1+x}{2}\), 4) = (\(\frac{7}{2}\), \(\frac{y+5}{2}\))
\(\frac{1+x}{2}\) = \(\frac{7}{2}\)
1 + x = 7
x = 7 – 1
= 6
and
\(\frac{y+5}{2}\) = 4
y + 5 = 8
y = 8 – 5
= 3
∴ The value of x = 6 and y = 3

Students can download Maths Chapter 3 Algebra Ex 3.19 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.19

Question 1.
A system of three linear equations in three variables is inconsistent if their planes
(1) intersect only at a point
(2) intersect in a line
(3) coincides with each other
(4) do not intersect
Answer:
(4) do not intersect

Question 2.
The solution of the system x + y -3z = -6, -7y + 7z = 7,3z = 9 is
(1) x = 1, y = 2, z = 3
(2) x = -1, y = 2, z = 3
(3) x = -1, y = -2, z = 3
(4) x = 1, y = 2, z = 3
Solution:
(4) x = 1, y = 2, z = 3
Hint:
x + y – 3z = -6
-7y + 7z = 7
3z = 9
z = 3
-7y + 21 = 7
-7y = -14
y = 2
x + 2 -3 × 3 = -6
x + 2 – 9 = -6 .
x = -6 + 7 = 1

Question 3.
If (x – 6) is the HCF of x² – 2x – 24 and x² – kx – 6 then the value of k is ……………..
(1) 3
(2) 5
(3) 6
(4) 8
Answer:
(2) 5
Hint.
HCF = x – 6
p(x) = x² – 2x – 24
= (x – 6) (x + 4)
g(x) = x² – kx – 6
∴ x – 6 is the common factor.
g (6) = 6² – k(6) – 6
= 36 – 6k – 6
= 30 – 6k
g(6) = 0
30 – 6k = 0
30 = 6k ⇒ k = \(\frac { 30 }{ 6 } \) = 5
The value of k = 5

Question 4.
\(\frac { 3y-3 }{ y } \) + \(\frac { 7y-7 }{ 3y2 } \) is …………

Answer:
(1) \(\frac{9 y}{7}\)

Question 5.
y² + \(\frac{1}{y^{2}}\) is not equal to ………….
(1) \(\frac{y^{4}+1}{y^{2}}\)
(2) (y + \(\frac { 1 }{ y } \))²
(3) (y – \(\frac { 1 }{ y } \))²
(4) (y + \(\frac { 1 }{ y } \))² – 2
Answer:
(2) (y + \(\frac { 1 }{ y } \))²
Hint.

Question 6.
\(\frac{x}{x^{2}-25}-\frac{8}{x^{2}+6 x+5}\) gives …………..

Answer:

Hint.

Question 7.
The square root of \(\frac{256 x^{8} y^{4} z^{10}}{25 x^{6} y^{6} z^{6}}\) is equal to ………….

Answer:

Hint.

Question 8.
Which of the following should be added to make x⁴ + 64 a perfect square
(1) 4x²
(2) 16x²
(3) 8x²
(4) -8x²
Solution:
(2) 16x²
Hint:
x⁴ + 64 = (x²)² + 8² + 2 × 8x²
= (x² + 8)²

Question 9.
The solution of (2x – 1)² = 9 is equal to ………..
(1) -1
(2) 2
(3) -1, 2
(4) None of these
Answer:
(3) -1, 2
Hint.
(2x – 1)² = 9 ⇒ (2x – 1) = \(\sqrt { 9 }\)
2x – 1 = ± 3 ⇒ 2x – 1 = 3 or 2x – 1 = 3
2x – 1 = 3
2x = 4 ⇒ x = \(\frac { 4 }{ 2 } \) = 2
2x – 1 = -3 ⇒ 2x = -3 + 1
x = -1

Question 10.
The values of a and b if 4x⁴ – 24x³ + 76x² + ax + b is a perfect square are ………..
(1) 100,120
(2) 10,12
(3) -120,100
(4) 12,10
Answer:
(3) -120,100
Hint.
Since it is a perfect square
a + 120 = 0
a = -120
b – 100 = 0
b = 100

Question 11.
If the roots of the equation q²x² + p²x + r² = 0 are the squares of the roots of the equation qx² + px + r = 0 , then q,p,r are in ……………..
(1) A.P.
(2) G.P.
(3) Both A.P and G.P
(4) none of these
Answer:
(2) G.P.
Hint.
q²x² + p²x + r² = 0
Let the roots be α² + β²
α² + β² = \(\frac{-p^{2}}{q^{2}}\)
α²β² = \(\frac{r^{2}}{q^{2}}\)
qx² + px + r = 0
Let the root be α and β

Question 12.
Graph of a linear polynomial is a
(1) straight line
(2) circle
(3) parabola
(4) hyperbola
Solution:
(1) straight line

Question 13.
The number of points of intersection of the quadratic polynomial x2 + 4x + 4 with the X axis is …………….
(1) 0
(2) 1
(3) 0 or 1
(4) 2
Answer:
(2) 1
Hint:
(x + 2)² = 0 ⇒ (x + 2) (x + 2) = 0
x + 2 = 0 or x + 2 = 0 ⇒ x = -2 or x = -2
Number of points of intersection is 1 (both the values are same)

Question 14.
For the given matrix

the order of the matrix A^T is ……………
(1) 2 × 3
(2) 3 × 2
(3) 3 × 4
(4) 4 × 3
Answer:
(3) 3 × 4

Question 15.
If A is a 2 × 3 matrix and B is a 3 × 4 matrix, how many columns does AB have ………..
(1) 3
(2) 4
(3) 2
(4) 5
Answer:
(2) 4
Hint.
The order of AB is 2 × 4
∴ Number of columns is 4.

Question 16.
If the number of columns and rows are not equal in a matrix then it is said to be a
(1) diagonal matrix
(2) rectangular matrix
(3) square matrix
(4) identity matrix
Solution:
(2) rectangular matrix

Question 17.
Transpose of a column matrix is …………..
(1) unit matrix
(2) diagonal matrix
(3) column matrix
(4) row matrix
Answer:
(4) row matrix

Question 18.
Find the matrix X if


Answer:

Hint.

Question 19.
Which of the following can be calculated from the given matrices?

(i) A²
(ii) B²
(iii) AB
(iv) BA
(1) (i) and (ii) only
(2) (ii) and (iii) only
(3) (ii) and (iv) only
(4) all of these
Answer:
(3) (ii) and (iv) only
Hint: (i) A² is possible to find
(ii) B² is also possible to find
(iii) not possible: order of A= (3 × 2) order of B is (3 × 3). AB is not possible number of column of matrix A ≠ number of rows of the matrix B.
(iv) Possible number column of the matrix is equal to the number of the matrix A.

Question 20.

Which of the following statements are correct?

(1) (i) and (ii) only
(2) (ii) and (iii) only
(3) (iii) and (iv)
(4) all of these
Answer:
(1) (i) and (ii) only
Hint.
(i) AB + C : order of A = 2 × 3
order of B = 3 × 2
order of AB = 2 × 2
order of C = 2 × 2
It is possible to add AB + C

(ii) BC :
order of B = 3 × 2
order of C = 2 × 2
It is possible to find BC

(iii) BA is not possible
order of B = 3 × 2
order of A = 3 × 2
BA does not exist.
BA + C is not a correct statement

(iv) ABC is not possible
order of A = 2 × 3 order of
B = 3 × 2
order of AB = 2 × 2
order of C = 3 × 2
It is not possible to multiply AB and C.
∴ The statement ABC is not correct.

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Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 6 Trigonometry Ex 6.3

Question 1.
Find the value of the following:

(iii) tan 15° tan 30° tan 45° tan 60° tan 75°

Solution:
(i) cos 45° = \(\frac{1}{√2}\)

= 1² + 1² – 2(\(\frac{1}{√2}\))²
= 1 + 1 – 2(\(\frac{1}{2}\))
= 2 – 1
= 1

(ii) cos 60° = \(\frac{1}{2}\)

= 1 + 1 + 1 – 8(\(\frac{1}{2}\))²
= 3 – 8 × \(\frac{1}{4}\)
= 3 – 2
= 1

(iii) tan 30° = \(\frac{1}{√3}\), tan 45° = 1, tan 60° = √3
tan 15° . tan 30°. tan 45° . tan 60°. tan 75° = tan 15° . \(\frac{1}{√3}\) . 1 . √3 tan 75°
= tan 15° × tan 75° × \(\frac{1}{√3}\) × 1 × √3
= tan(90° – 75°) × \(\frac{1}{cot 75°}\) × 1 [tan 90° – θ = cot θ]
= cot 75° × \(\frac{1}{cot 75°}\) × 1
= 1

= 1 + 1
= 2

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Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 6 Trigonometry Ex 6.2

Question 1.
Verify the following equalities:
(i) sin² 60° + cos² 60° = 1
Solution:
sin 60° = \(\frac{√3}{2}\); cos 60° = \(\frac{1}{2}\)
L.H.S = sin² 60° + cos² 60°

L.H.S = R.H.S
Hence it is proved.

(ii) 1 + tan² 30° = sec² 30°
Solution:
tan 30° = \(\frac{1}{√3}\); sec 30° = \(\frac{2}{√3}\)
L.H.S = 1 + tan² 30°

∴ L.H.S = R.H.S
Hence it is proved.

(iii) cos 90° = 1 – 2sin² 45° = 2cos² 45° – 1
Solution:
cos 90° = 0, sin 45° = \(\frac{1}{√2}\), cos 45° = \(\frac{1}{√2}\)
cos 90° = 0 ……. (1)
1 – 2 sin² 45° = 1 – 2 (\(\frac{1}{√2}\))²
= 1 – 2 × \(\frac{1}{2}\)
= 1 – 1 = 0 → (2)
2 cos² 45° – 1 = 2(\(\frac{1}{√2}\))² – 1
= \(\frac{2}{2}\) – 1
= \(\frac{2 – 2}{2}\) = 0 → (3)
From (1), (2) and (3) we get
cos 90° = 1 – 2 sin² 45° = 2 cos² 45° – 1
Hence it is proved.

(iv) sin 30° cos 60° + cos 30° sin 60° = sin 90°
Solution:
sin 30° = \(\frac{1}{2}\); cos 60° = \(\frac{1}{2}\); cos 30° = \(\frac{√3}{2}\); sin 60° = \(\frac{√3}{2}\); sin 90° = 1
L.H.S = sin 30° cos 60° + cos 30° sin 60°

= 1
R.H.S = sin 90° = 1
L.H.S = R.H.S
Hence it is proved.

Question 2.
Find the value of the following:
(i) \(\frac{tan 45°}{cosec 30°}\) + \(\frac{sec 60°}{cot 45°}\) – \(\frac{5 sin 90°}{2 cos 0°}\)
(ii) (sin 90° + cos 60° + cos 45°) × (sin 30° + cos 0° – cos 45°)
(iii) sin²30° – 2 cos³ 60° + 3 tan⁴ 45°
Solution:
(i) tan 45° = 1, cosec 30° = 2; sec 60° = 2; cot 45° = 1; tan 45°, sin 90° = 1; cos 0° = 1

= 0

(ii) sin 90° = 1; cos 60° = \(\frac{1}{2}\); cos 45° = \(\frac{1}{√2}\); sin 30° = \(\frac{1}{2}\); cos 0° = 1
(sin 90° + cos 60° + cos 45°) × (sin 30° + cos 0° – cos 45°)

= \(\frac{7}{4}\)

(iii) sin 30° = \(\frac{1}{2}\); cos 60° = \(\frac{1}{2}\); tan 45° = 1
sin² 30° – 2 cos³ 60° + 3 tan⁴ 45°

= 3

Question 3.
Verify cos 3 A = 4 cos³ A – 3 cos A, when A = 30°
Solution:
L.H.S = cos 3 A
= cos 3 (30°)
= cos 90°
= 0
R.H.S = 4 cos³ A – 3 cos A
= 4 cos³ 30° – 3 cos 30°

= 0
∴ L.H.S = R.H.S
Hence it is proved.

Question 4.
Find the value of 8 sin2x, cos 4x, sin 6x, when x = 15°.
Solution:
8 sin 2x cos 4x sin 6x = 8 sin 2 (15°) × cos 4 (15°) × sin (6 × 15°)
= 8 sin 30° × cos 60° × sin 90°
= 8 × \(\frac{1}{2}\) × \(\frac{1}{2}\) × 1
= 2

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Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 5 Coordinate Geometry Ex 5.5

Question 1.
Find the centroid of the triangle whose vertices are
(i) (2, -4), (-3, -7) and (7, 2)
Solution:
Let the vertices of a triangle be A (2, -4), B (-3, -7) and C (7, 2) Centroid

Centroid is (2, -3)

(ii) (-5, -5), (1, -4) and (-4, -2)
Solution:
Let the vertices of a triangle be A (-5, -5), B (1, -4) and C (-4, -2)

Question 2.
If the centroid of a triangle is at (4, -2) and two of its vertices are (3, -2) and (5, 2) then find the third vertex of the triangle.
Solution:
Let the vertices of a triangle be A (3, -2), B (5, 2) and C (x₃, y₃)
Centroid of a triangle is (4, -2)

∴ \(\frac{8+x_{3}}{3}\) = 4
8 + x₃ = 12
x₃ = 12 – 8
= 4
and
\(\frac{y_{3}}{3}\) = -2
y₃ = -6
∴ The third vertex is (4, -6)

Question 3.
Find the length of median through A of a triangle whose vertices are A(-1, 3), B (1, -1) and C (5, 1).
Solution:
AD is the median of the ΔABC
D is the mid-point of BC

Length of the median AD is 5 units.

Question 4.
The vertices of a triangle are (1, 2), (h, -3) and (-4, k). If the centroid of the triangle is at the point (5, -1) then find the value of \(\sqrt{(h+k)^{2}+(h+3k)^{2}}\)
Solution:
Let the vertices A (1, 2), B (h, -3) and C (-4, k)


\(\frac{-3+h}{3}\) = 5
-3 + h = 15
h = 15 + 3 = 18
and
\(\frac{-1+k}{3}\) = -1
-1 + k = -3
k = -3 + 1
k = -2

Question 5.
Orthocentre and centroid of a triangle are A(-3, 5) and B(3, 3) respectively. If C is the circumcentre and AC is the diameter of this circle, then find the radius of the circle.
Solution:
Let PQR be any triangle orthocentre, centroid and circumcentre.

A orthocentre is (-3, 5)
B centroid is (3, 3)
C orthocentre is (a, 6)
Also \(\frac{AB}{BC}\) = \(\frac{2}{1}\)
B divides AC in the ratio 2 : 1
A line divides internally in the ratio point P is (\(\frac{mx_{2}+nx_{1}}{m+n}\), \(\frac{my_{2}+ny_{1}}{m+n}\))
m = 2
x₁ = 3
y₁ = 5
amd
n = 1
x₂ = a
y₂ = b
∴ The point B (\(\frac{2a-3}{2+1}\), \(\frac{2b+5}{2+1}\))
(3, 3) = (\(\frac{2a-3}{3}\), \(\frac{2b+5}{3}\))
\(\frac{2a-3}{3}\) = 3
2a – 3 = 9
2a = 9 + 3
2a = 12
a = \(\frac{12}{2}\) = 6
and
\(\frac{2b+5}{3}\)
2b + 5 = 9
2b = 9 – 5
2b = 4
b = \(\frac{4}{2}\) = 2
∴ Orthocentre C is (6, 2)

Question 6.
ABC is a triangle whose vertices are A (3, 4), B (-2, -1) and C (5, 3). If G is the centroid and BDCG is a parallelogram then find the coordinates of the vertex D.
Solution:
The vertices of a triangle are A (3, 4), B (-2, -1) and C (5, 3)

The point G is (2, 2)
Let the vertices D be (a, b)
Since BDCG is a parallelogram
Mid-point of BC = Mid-point of DG

\(\frac{2+a}{2}\) = \(\frac{3}{2}\)
4 + 2a = 6
2a = 6 – 4
2a = 2
a = 1
and
\(\frac{2+b}{2}\) = 1
2 + b = 2
b = 2 – 2 = 0
The vertices D is (1, 0).

Question 7.

Solution:
In ΔABC, Let A (x₁, y₁), B (x₂, y₁) and C (x₃, y₃)

x₁ + x₂ = 3 → (1)
y₁ + y₂ = 10 → (2)

x₂ + x₃ = 14 → (3)
y₂ + y₃ = 10 → (4)

x₁ + x₃ = 3 → (5)
y₁ + y₃ = 10 → (6)
Add (1) + (3) + (5) We get
2x₁ + 2x₂ + 2x₃ = 30
2(x₁ + x₂ + x₃) = 30
x₁ + x₂ + x₃ = 15
From (1), x₁ + x₂ = 3
∴ x₃ = 12
From (3), x₂ + x₃ = 14
∴ x₁ = 1
From (5), x₁ + x₃ = 13
∴ x₂ = 2
Add (2), (4) and (6) we get
2y₁ + 2y₂ + 2y₃ = -12
2(y₁ + y₂ + y₃) = – 12
∴ y₁ + y₂ + y₃ = -6
From (2), y₁ + y₂ = 10
∴ y₃ = -16
From (4) y₂ + y₃ = -9
∴ y₁ = 3
From (6) y₁ + y₃ = -13
∴ y₂ = 7
The vertices of the A are A (1, 3), B (2, 7) and C (12, -16)

The Centroid of Δ is (5, -2)

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Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 6 Trigonometry Ex 6.1

Question 1.
From the given figure, find all the trigonometric ratios of angle B.

Solution:

Question 2.
From the given figure, find the values of

(i) sin B
(ii) sec B
(iii) cot B
(v) tan C
(vi) cosec C
Solution:
In the right ΔABD,
AD² = AB² – BD²
= 13² – 5²
= 169 – 25
= 144
AD = \(\sqrt{144}\)
= 12
In the right ΔADC,
AC² = AD² + DC²
= 12² + 16²
= 144 + 256
= 400
AC = \(\sqrt{400}\)
= 20

Question 3.
If 2 cos θ = √3, then find all the trigonometric ratios of angle θ.
Solution:

2 cos θ = √3 ⇒ cos θ = \(\frac{√3}{2}\)
AB² = AC² – BC²
= 2² – (√3)² ⇒ = 4 – 3 = 1
AB = √1 = 1

Question 4.
If cos A =\(\frac{3}{5}\), then find the value of \(\frac{sin A-cos A}{2 tan A}\)
Solution:

cos A = \(\frac{3}{5}\)
ΔABC
BC² = AC² – AB²
= 5² – 3²
= 25 – 9
= 16
BC = \(\sqrt{16}\) = 4
sin A = \(\frac{4}{5}\); tan A = \(\frac{4}{3}\)

∴ The value of \(\frac{sin A-cos A}{2 tan A}\) = \(\frac{3}{40}\)

Question 5.
If cos A = \(\frac{2x}{1+x_{2}}\) then find the values of sin A and tan A in terms of x.
Solution:
cos A = \(\frac{2x}{1+x_{2}}\)
In the triangle ABC

BC² = AC² – AB²
= (1 + x²)² – (2x)²
= 1 + x⁴ + 2x² – 4x²
= x⁴ – 2x² + 1
= (x² – 1)² (or) (1 – x²)² (using (a – b)²)
BC = \(\sqrt{(x^{2}-1)^{2}}\) (or) \(\sqrt{(1-x^{2})^{2}}\)
BC = x² – 1
The value of sin A = \(\frac{BC}{AC}\) = \(\frac{x²-1}{x²+1}\)
tan A = \(\frac{BC}{AB}\) = \(\frac{x²-1}{2x}\)
and
BC = 1 – x²
The value of sin A = \(\frac{1-x²}{1+x²}\)
tan A = \(\frac{1-x²}{2x}\)

Question 6.
If sin θ = \(\frac{a}{\sqrt{a²+b²}}\) then show that b sin θ = a cos θ.
Solution:

sin θ = \(\frac{a}{\sqrt{a²+b²}}\)
In the triangle ΔABC
BC² = AC² – AB²

L. H. S = R. H. S

Question 7.
If 3 cot A = 2, then find the value of \(\frac{4sin A-3cos A}{2 sin A+3 Cos A}\)
Solution:

3 cot A = 2 ⇒ cot A = \(\frac{2}{3}\)
AC² = AB² + BC²
= 3² + 2²
= 9 + 4
AC = \(\sqrt{13}\)
cos A = \(\frac{AB}{AC}\) = \(\frac{3}{\sqrt{13}}\)
sin A = \(\frac{BC}{AC}\) = \(\frac{2}{\sqrt{13}}\)
The value of \(\frac{4sin A-3cos A}{2 sin A+3 Cos A}\)

The value is (\(\frac{-1}{13}\))

Question 8.
If cos θ : sin θ = 1 : 2, then find the value of \(\frac{8cos θ-2cos θ}{4 cos θ+2 sin θ}\)
Solution:
cos θ : sin θ = 1 : 2

Aliter:
cos θ : sin θ = 1 : 2
2 cos θ = sin θ ⇒ 2 = \(\frac{sin θ}{cos θ}\) ⇒ 2 = tan θ

∴ The value is \(\frac{1}{2}\)

Question 9.
From the given figure, prove that θ + ∅ = 90°. Also prove that there are two other right angled triangles. Find sin α, cos β and tan ∅.

Solution:
In the ΔABC,
AB = 9 + 16 = 25
AC = 15; BC = 20
AB² = 25²
= 625 ……. (1)
AC² + BC² = 15² + 20²
= 225 + 400
= 625 …….. (2)
From (1) and (2) we get
AB² = AC² + BC²
ABC is a right angle triangle at C (Pythagoras theorem)
∴ ∠C = 90°
θ + ∅ = 90°
Also ADC is a right angle triangle ∠ADC = 90° (Given)
BDC is also a right angle triangle ∠BDC = 90° (since ADB is a straight line sum of the two angle is 180°)
From the given diagram

Question 10.
A boy standing at a point O finds his kite flying at a point P with distance OP = 25 m. It is at a height of 5 m from the ground. When the thread is extended by 10 m from P, it reaches a point Q. What will be the height QN of the kite from the ground? (use trigonometric ratios).

Solution:
Let the angle O be “θ”
In ΔONQ

In ΔOMP
sin θ = \(\frac{PM}{OP}\) ⇒ sin θ = \(\frac{5}{25}\)
sin θ = \(\frac{1}{5}\) ……… (2)
From (1) and (2) we get
\(\frac{h}{35}\) = \(\frac{1}{5}\)
5 h = 35 ⇒ h = \(\frac{35}{5}\) = 7
The height of the kite from the ground is 7m.

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Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Unit Exercise 3

Question 1.
Solve
\(\frac { 1 }{ 3 } \) (x + y – 5) = y – z = 2x – 11 = 9 – (x + 2z)
Answer:
\(\frac { 1 }{ 3 } \) (x + y – 5) = y – z
x + y – 5 = 3y – 3z
x + y – 3y + 3z = 5
x – 2y + 3z = 5 ….(1)
y – z – 2x – 11
-2 x + y – z = -11
2x – y + z = 11 …..(2)
2x – 11 = 9-(x + 2 z)
2x – 11 = 9 – x – 2z
2x + x + 2z = 9 + 11
3x + 2z = 20 ….(3)

3x – z = 17 …. (5)


Substitute the value of z = 1 in (3)
3x + 2(1) = 20
3x = 20 – 2
3x = 18
x = \(\frac { 18 }{ 3 } \) = 6
substitute the value of x = 6, z = 1 in (2)
2(6) – y + 1 = 11
12 – y + 1 = 11
13 – y = 11
-y = 11 – 13
-y = -2
y = 2
∴ The value of x = 6, y = 2 and z = 1

Question 2.
One hundred and fifty students are admitted to a school. They are distrbuted over three sections A, B and C. If 6 students are shifted from section A to section C, the sections will have equal number of students. If 4 times of students of section C exceeds the number of students of section A by the number of students in section B, find the number of students in the three sections.
Answer:
Let the number of students in section A be “x”
Let the number of students in section B be “y”
Let the number of students in section C be “z”
By the given first condition
x + y + z = 150 ……(1)
again by the second condition
x – 6 = z + 6
x – z = 6 + 6
x – z = 12 ….(2)
again by the third condition
x + y = 4z
x + y – 4z = 0
x + y – 4z = 0 ….(3)
Subtracting (1) and (3)

Substitute the value of z = 30 in (2)
x – 30 = 12
x = 12 + 30
= 42
Substitute the value of x = 42 and z = 30 in (1)
42 + y + 30 = 150
y + 72 = 150
y = 150 – 72
= 78
Number of students in section A, B and C are = 42, 78 and 30.

Question 3.
In a three-digit number, when the tens and the hundreds digit are interchanged the new number is 54 more than three times the original number. If 198 is added to the number, the digits are reversed. The tens digit exceeds the hundreds digit by twice as that of the tens digit exceeds the unit digit. Find the original number.
Answer:
Let the hundred digit be x
the tens digit be y and the unit digit be z
∴ The number is 100x + 10y + z
By the given first condition
100y + 10x + z = 54 + 3 (100x + 10y + z)
100y + 10x + z = 54 + 300x + 30y + 3z
-290x + 70y – 2z = 54 (÷ -2)
145x-35y + z = -27 ….(1)
Again by the second condition
198 + 100x + 10y + z = 100z + 10y + x
99x – 99z = -198 (÷ 99)
x – z = -2 ….(2)
Again by the third condition
y – x = 2(y – z)
y – x = 2y – 2z
– x – y + 2z = 0
x + y – 2z = 0 ….(3)

substitute the value of x = 1 in …….(2)
1 – z = -2
3 = z
∴ z = 3
substitute the value of x = 1 and z = 3 in …….(3)
1 – y – 6 = 0
y – 5 = 0
y = 5
∴ The number is 153

Question 4.
Find the least common multiple of xy(k² +1) + k(x² + y²) and xy(k² – 1) + k(x² – y²).
Solution:
xy (k² + 1) + k (x² + y²) …………… (1)
xy(k² – 1) + k(x² – y²) …………… (2)
(1) ⇒ xyk² + xy + kx² + ky²
(2) ⇒ xyk² – xy + kx² – ky²
(1) ⇒ yk (xk + y) + x (xk + y)
= (xk + y) (x + yk)
(2) ⇒ yk (xk – y) + x (xk – y)
= (x + yk) (xk – y)
∴ L.C.M. : (x + yk) (xk + y) (xk – y)
= (x + yk) (x²k² – y²)

Question 5.
Find the GCD of the following by division algorithm
2x⁴ + 13x³ + 27x² + 23x + 7,
x³ + 3x² + 3x + 1, x² + 2x + 1
Answer:
p(x) = 2x⁴ + 13x³ + 27x² + 23x + 7
g(x) = x³ + 3x² + 3x + 1
r(x) = x² + 2x + 1
(i) Find the G.C.D. of p(x) and g(x)

(ii) Find the G.C.D. of r(x) and the G.C.D. of p(x) and g(x)

∴ G.C.D.= x² + 2x + 1
∴ G.C.D. of the three
polynomials = x² + 2x + 1

Question 6.
Reduce the given Rational expressions to its lowest form
(i) \(\frac{x^{3 a}-8}{x^{2 a}+2 x^{a}+4}\)
Answer:
x^3a – 8 = (x^a)³ – 2³
(using the formula a³ – b³ = (a – b)(a² + ab + b²)
= (x^a – 2)[(x^a)² + x^a × 2 + 2²]
= (x^a – 2) (x^2a + 2x^a + 4)

(ii) \(\frac{10 x^{3}-25 x^{2}+4 x-10}{-4-10 x^{2}}\)
Answer:
10x³ – 25x² + 4x – 10 = 5x²(2x – 5) + 2 (2x – 5)
= (2x – 5) (5x² + 2)
– 4 – 10x² = -2 (2 + 5x²)
= -2(5x² + 2)

Question 7.
Simplify

Answer:

Question 8.
Arul, Ravi and Ram working together can clean a store in 6 hours. Working alone, Ravi takes twice as long to clean the store as Arul does. Ram needs three times as long as Arul does. How long would it take each if they are working alone?
Answer:
Let the time taken by Arul be “x” hours
Let the time taken by Ravi be “y” hours
Let the time taken by Ram be “z” hours
By the given first condition
\(\frac { 1 }{ x } \) + \(\frac { 1 }{ y } \) + \(\frac { 1 }{ z } \) = \(\frac { 1 }{ 6 } \)
Again by the given second condition
\(\frac { 1 }{ x } \) = 2 × \(\frac { 1 }{ y } \)
\(\frac { 1 }{ x } \) – \(\frac { 2 }{ y } \) = 0
By the given third condition
3 × \(\frac { 1 }{ z } \) = \(\frac { 1 }{ x } \)
– \(\frac { 1 }{ x } \) + \(\frac { 3 }{ z } \) = 0
Let \(\frac { 1 }{ x } \) = a, \(\frac { 1 }{ y } \) = b, \(\frac { 1 }{ z } \) = c
a + b + c = \(\frac { 1 }{ 6 } \)
6a + 6b + 6c = 1 …….(1)
a – 2b = 0 ……….(2)
-a + 3c = 0 …………(3)

Arul take 11 hours, Ravi take 22 hours and Ram takes 33 hours.

Question 9.
Find the square root of 289x⁴ – 612x³ + 970x² – 684x + 361.
Answer:

Question 10.
Solve \(\sqrt { y+1 }\) + \(\sqrt { 2y-5 }\) = 3
Answer:
\(\sqrt { y+1 }\) + \(\sqrt { 2y-5 }\) = 3
(squaring on bothsides)

8y² – 9y² – 12y + 78y – 20 – 169 = 0
-y² – 66y – 189 = 0
y² – 66y + 189 = 0
(y – 3) (y – 63) = 0

y – 3 or y = 63
The value of y is 3 and 63

Question 11.
A boat takes 1.6 hours longer to go 36 kins up a river than down the river. If the speed of the water current is 4 km per hr, what is the speed of the boat in still water?
Answer:
Let the speed of the boat in still water be “x”
Time taken to go for up of a river = \(\frac { 36 }{ x+4 } \)
By the given condition
\(\frac { 36 }{ x-4 } \) – \(\frac { 36 }{ x+4 } \) = 1.6

The speed of the boat in still water = 14 km/hr

Question 12.
Is it possible to design a rectangular park of perimeter 320 m and area 4800 m2? If so find its length and breadth.
Answer:
Let the length of the rectangular park be “l”
and the breadth of the rectangular park be “b”
Perimeter of the park = 320 m
2 (l + b) = 320
l + b = 160
l = 160 – b ……….(1)
Area of the park = 4800 m²
l × b = 4800 ….(2)
substitute the value of l = 160 – b in (2)
(160 – b)b = 4800
160b – b² = 4800
b² – 160b + 4800 = 0
(b – 120) (b – 40) = 0

b = -120 = 0 or b – 40 = 0
b = 120 or b = 40
If breadth is 120 length is 40
If breadth is 40 length is 120
Length of the park = 120 m
Breadth of the park = 40 m

Question 13.
At t minutes past 2 pm, the time needed to 3 pm is 3 minutes less than \(\frac{t^{2}}{4}\) Find t.
Answer:
Time needed by the minutes hand to show
3 pm = (60 – 1) minutes
By the given condition


∴ The value of t = 14 minutes

Question 14.
The number of seats in a row is equal to the total number of rows in a hall. The total number of seats in the hall will increase by 375 if the number of rows is doubled and the number of seats in each row is reduced by 5. Find the number of rows in the hall at the beginning.
Answer:
Let the number of rows in the hall be “x”
∴ Total number of rows = x
Total number of seats in the hall is “x2”
By the given condition
x² + 375 = 2x (x – 5)
x² + 375 = 2x²– 10x
x² – 2x² + 10x + 375 = 0
– x² + 10x + 375 = 0
– x² – 10x – 375 = 0
(x – 25) (x + 15)
x – 25 = 0 or x + 15 = 0
x = 25 or x = – 15

Number of rows in the hall = 25

Question 15.
If α and β are the roots of the polynomial f(x) – x² – 2x + 3, find the polynomial whose roots are
(i) α + 2, β + 2
Answer:
α and β are the roots of the polynomial
x² – 2x + 3 = 0
α + β = 2; αβ = 3
(i) Sum of the roots = α + 2 + β + 2
= α + β + 4
= 2 + 4
= 6
Product of the roots = (α + 2) (β + 2)
= αβ + 2α + 2β + 4
= αβ + 2(α + β) + 4
= 3 + 4 + 4
= 11
The quadratic polynomial
x² – (sum of the roots) x + product of the roots = 0
x² – (6) x + 11 = 0
x² – 6x + 11 = 0

(ii) \(\frac{\alpha-1}{\alpha+1}, \frac{\beta-1}{\beta+1}\)
Answer:
Sum of the roots


The quadratic polynomial is
x² – (sum of the roots) + products of the roots = 0
x² – (\(\frac { 2 }{ 3 } \)) x + \(\frac { 1 }{ 3 } \) = 0
3x² – 2x + 1 = 0

Question 16.
If -4 is a root of the equation x² + px – 4 = 0 and if the equation x² + px + q = 0 has equal roots, find the values of p and q.
Solution:
f(x) = x² + px – 4 = 0
If -4 is a root, then
f(-4) = (-4)² + P(-4) – 4 = 16 – 4p – 4 = 0
12 – 4p = 0
-4p = -12
p = 3
x² + 3x + q =0 has equal roots,
∆ = b² – 4ac = 0
3² – 4 × 1 × q = 0
9 – 4q = 0
-4 q = -9
q = \(\frac{9}{4}\)
p = 3, q = \(\frac{9}{4}\)

Question 17.
Two farmers Senthil and Ravi cultivates three varieties of grains namely rice, wheat and ragi. If the sale (in ₹) of three varieties of grains by both the farmers in the month of April is given by the matrix.

and the May month sale (in ₹) is exactly twice as that of the April month sale for each variety.
(i) What is the average sales of the months April and May.
Answer:
(i) Let A represent the sale on April

Let B represent the sale on May

Average sale of the month April and May

(ii) If the sales continue to increase in the same way in the successive months, what will be sales in the month of August?
Answer:
If it increasing in the successive months of
May sale is 2 (April sale)
June sale is 4 (April sale)
July sale is 8 (April sale)
August sale is 16 (April sale)
Sales in the month of August

Question 18.
If cos = I₂, find x.
Answer:

Question 19.
Given

and if BA = C², find p and q
Answer:

∴The value of p = 8 and q = 4

Question 20.

find the matrix D, such that CD – AB = 0
Answer:
Given