Category: Class 11

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 3 Trigonometry Ex 3.1 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.1

Question 1.
Identify the quadrant in which an angle of each given measure lies,
(i) 25°
(ii) 825°
(iii) – 55°
(iv) 328°
(v) – 230°
Answer:
(i) 25°

25° First quadrant

(ii) 825°

825° = 9 × 90° + 15°
825° = 2 × 360° + 105°
∴ 825° lies in the second quadrant.

iii) -55°

-55° lies in the fourth quadrant

iv) 328°

328° = 270° + 58° lies in the fourth quadrant.

v) -230°

– 230° = – 180° + (- 50°) lies in the second quadrant.

Question 2.
For each given angle, find a co-terminal angle with a measure of 9 such that 0 ≤ θ < 360°.
(i) 395°
(ii) 525°
(iii) 1150°
(iv) – 270°
(v) – 450°
Answer:
(i) 395°
395° = 360° + 35°
395° – 35° = 360°
∴ Coterminal angle for 395° is 35°

(ii) 525°
525° = 360° + 165°
360° – 165° = 360°
∴Coterminal angle for 525° is 165°

(iii) 1150°
1150° = 360° + 360° + 360° + 70°
1150° = 3 × 360° + 70°
1150° – 70° = 3 × 360°
∴ Coterminal angle for 1150° is 70°.

(iv) – 270°
– 270° = 360° + 90°
– 270° – 90° = 360°
∴ Coterminal angle for -270° is 90°

(v) – 450°
– 450° = – 720° + 270°
– 450° – 270° = – 2 × 360°
∴ Coterminal angle for – 450° is 270°

Question 3.
If a cos θ – b sin θ = c , show that a sin θ + b cos θ = ± \(\sqrt{\mathbf{a}^{2}+\mathbf{b}^{2}-\mathbf{c}^{2}}\)
Answer:
a cos θ – b sin θ = c
(a cos θ – b sin θ)² + (a sin θ + b cos θ)² = a² cos² θ – 2 ab sin θ cos θ + b² sin²θ + a² sin² θ + b² cos² θ + 2 ab sin θ cos θ
c² + (a sin 0 + b cos θ )² = a² cos² θ + a² sin² θ + b² sin² θ + b²cos²θ
= a² (cos²θ + sin²θ) + b²(sin²θ + cos²θ)
c² + (a sin θ + b cos θ )2 = a² + b²
(a sin θ + b cos θ)² = a² + b² – c²
a sin θ + b cos θ = ± \(\sqrt{\mathbf{a}^{2}+\mathbf{b}^{2}-\mathbf{c}^{2}}\)

Question 4.
If sin θ + cos θ = m , show that cos⁶ θ + sin⁶ θ = \(\frac{4-3\left(m^{2}-1\right)^{2}}{4}\) where m² ≤ 2.
Answer:
sin θ + cos θ = m
(sin θ + cos θ)² = m²

Question 5.

Answer:

Question 6.

Answer:

Question 7.

Answer:

Question 8.
If tan² θ = 1 – k², show that sec θ + tan³ θ cosec θ = ( 2 – k²)^3/2. Also, find the values of k for which this result holds.
Answer:
tan² θ = 1 – k²
1 + tan² θ = 1 + 1 – k²
sec²θ = (2 – k²)
sec²θ = (2 – k²)^1/2


tan² θ = 1 – k²
When θ = \(\frac{\pi}{2}\), tan \(\frac{\pi}{2}\) = ∞, not defined 2
When θ = 0, tan² 0 = 1 – k²
1 – k² = 0 ⇒ k² = 1 ⇒ k = ± 1
When θ = 45°, tan² 45° = 1 – k²
1 – k² = 1 ⇒ – k² = 0 ⇒ k = 0
When θ > 45°, say θ = 60°
tan² 60° = 1 – k² = (√3)² = 1 – k²
3 = 1 – k² ⇒ k² = 1 – 3 = – 2
∴ θ > 45°, k² is negative ⇒ k is imaginary
∴ k lies between -1 and 1 ⇒ k ∈ [-1 , 1]

Question 9.
If sec θ + tan θ = p, obtain the values of sec θ, tan θ and sin θ in terms of p.
Answer:
Given sec θ + tan θ = p
We have sec² θ – tan² θ = 1
(sec θ + tan θ) (sec θ – tan θ) = 1
p (sec θ – tan θ) = 1
sec θ – tan θ = \(\frac{1}{p}\)
(sec θ – tan θ) + (sec θ – tan θ) = p + \(\frac{1}{p}\)

Question 10.
If cot θ(1 + sin θ) = 4m and cot θ (1 – sin θ) = 4n then prove that (m² – n²)² = mn.
Answer:

Question 11.
If cosec θ – sin θ = a³, sec θ – cos θ = b³ then prove that a²b²(a² + b²) = 1.
Answer:

Question 12.
Eliminate θ from the equations a sec θ – c tan θ = b, b sec θ + d tan θ = c.
Answer:

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 3 Trigonometry Ex 3.3 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.3

Question 1.
Find the values of
(i) sin 480°
(ii) sin (-1110°)
(iii) cos 300°
(iv) tan (1050°)
(v) cot 660°
(vi) tan \(\left(\frac{19 \pi}{3}\right)\)
(vii) sin \(\left(\frac{-11 \pi}{3}\right)\)
Answer:
(i) sin(480°) = sin(360° + 120°) = sin 120°
= sin(90° + 30°) = cos 30° = \(\sqrt{3}\)/2

(ii) sin(-1110°) = -sin(1110°)
= – sin (360° × 3 + 30°)
= -sin 30° = -1/2

(iii) cos(300°) = cos(270° + 30°) = sin 30° = 1/2

(iv) tan (1050°)
tan (1050°) = tan(12 × 90 – 30°)
= – tan30° = – \(\frac{1}{\sqrt{3}}\)

(v) cot 660°
cot 660° = cot (7 × 90 + 30°)
= – tan 30° = – \(\frac{1}{\sqrt{3}}\)

(vi) tan \(\left(\frac{19 \pi}{3}\right)\)

(vii) sin \(\left(\frac{-11 \pi}{3}\right)\)

Question 2.
\(\left(\frac{5}{7}, \frac{2 \sqrt{6}}{7}\right)\) is a point on the terminal side of an angle θ in standard position. Determine the six trigonometric function values of angle θ.
Answer:
Given \(\left(\frac{5}{7}, \frac{2 \sqrt{6}}{7}\right)\) is a point on the terminal side of an angle θ in standard position.

Question 3.
Find the values of the other five trigonometric functions of the following
(i) cos θ = –\(\frac{1}{2}\), θ lies in the III quadrant
(ii) cos θ = \(\frac{2}{3}\), θ lies in the I quadrant
(iii) sin θ = –\(\frac{2}{3}\), θ lies in the IV quadrant
(iv) tan θ = – 2, θ lies in the II quadrant
(v) sec θ = \(\frac{13}{5}\), θ lies inthe IVquadrant
Answer:
(i) cos θ = –\(\frac{1}{2}\), θ lies in the III quadrant

(ii) cos θ = \(\frac{2}{3}\), θ lies in the I quadrant
We know that cos²θ + sin²θ = 1
\(\left(\frac{2}{3}\right)^{2}\) + sin²θ = 1
\(\frac{4}{9}\) + sin²θ = 1

Since θ lies in the I quadrant all trigonometric functions are positive.

(iii) sin θ = –\(\frac{2}{3}\), θ lies in the IV quadrant
We know that cos²θ + sin²θ = 1
cos²θ + \(\left(-\frac{2}{3}\right)^{2}\) = 1
cos²θ + \(\frac{4}{9}\) = 1

Since θ lies in the fourth quadrant cos θ is positive.

(iv) tan θ = – 2, θ lies in the II quadrant
We know that sec²θ – tan²θ = 1
sec²θ – (-2)² = 1
sec²θ – 4 = 1
sec²θ = 1 + 4 = 5
sec θ = ± √5
Since θ lies in the second quadrant sec θ is negative.

(v) sec θ = \(\frac{13}{5}\), θ lies inthe IVquadrant
We know that sec²θ – tan²θ = 1
\(\left(\frac{13}{5}\right)^{2}\) – tan²θ = 1
\(\frac{169}{25}\) – 1 = tan²θ

Question 4.
Prove that

Answer:

Question 5.
Find all the angles between 0° and 360° which satisfy the equation sin²θ = \(\frac{3}{4}\)
Answer:
sin²θ = \(\frac{3}{4}\) ⇒ sin θ = ± \(\frac{\sqrt{3}}{2}\)
sin 60° = \(\frac{\sqrt{3}}{2}\)
sin 120° = sin (180° – 60°)
= sin 60° = \(\frac{\sqrt{3}}{2}\)
∴ θ = 60° and 120°

Question 6.
Show that

Answer:

= sin² 10° + sin² 20° + [cos 20°]² + [cos 10°]²
= sin² 10° + sin² 20° + cos² 20° + cos² 10°
= sin² 10° + cos² 10° + sin² 20° + cos² 20°
= 1 + 1 = 2

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 3 Trigonometry Ex 3.2 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.2

Question 1.
Express each of the following in radian measure.
(i) 30°
(ii) 135°
(iii) -205°
(iv) 150°
(v) 330°
Answer:
(i) 30°

(ii) 135°

(iii) – 205°

(iv) 150°

(v) 330°

Question 2.
Find the degree measure corresponding to the following radian measures.
(i) \(\frac{\pi}{3}\)
(ii) \(\frac{\pi}{9}\)
(iii) \(\frac{2 \pi}{5}\)
(iv) \(\frac{7 \pi}{3}\)
(v) \(\frac{10 \pi}{9}\)
Answer:
(i) \(\frac{\pi}{3}\) radians

(ii) \(\frac{\pi}{9}\) radians

(iii) \(\frac{2 \pi}{5}\) radians

(iv) \(\frac{7 \pi}{3}\) radians

(v) \(\frac{10 \pi}{9}\) radians

Question 3.
What must be the radius of a circular running path, around which an athlete must run 5 times in order to describe 1 km?
Answer:
Let the radius of the circular path be = r m.
Length of the circular path s = 1 k. m
s = 1000 m.
Athlete runs 5 times around the path to cover 1 k. m distance
∴ θ = 360° × 5
θ = 360° × 5 × \(\frac{\pi}{180}\) radians
θ = 10 π radians
s = r θ
1000 = r 10 π
r = \(\frac{1000}{10 \pi}\)
r = \(\frac{1000 \times 7}{10 \times 22}=\frac{350}{11}\)
r = 31 .818 meters
Radius of the circular path = 31 .82 meters

Question 4.
In a circle of diameter 40 cm a chord is of length 20 cm. Find the length of the minor arc of the chord.
Answer:

Given Diameter AB = 40 cm
∴ Radius r = 20 cm
Chord CD = 20 cm
O – Centre of the circle
OC = OD = radius = 20 cm.
∴ Triangle OCD is an equilateral triangle.

To find the length of the minor arc CD.
Let s = minor arc CD.
The arc CD subtends 60° at the centre.
θ = 60°
θ = 60° × \(\frac{\pi}{180}\) radians.
θ = \(\frac{\pi}{3}\) radians
We have s = rθ

Question 5.
Find the degree measure of the angle subtended at the centre of the circle of radius 100 cm by an arc of length 22 cm.
Answer:
Given radius r = 100 cm.
Length of arc s = 22 cm.
Angle subtended by the arc at the centre = θ radians

Question 6.
What is the length of the arc intercepted by a central angle of measure 41° in a circle of radius of 10 feet?
Answer:
Central angle subtended by the arc θ = 41°
θ = 41 × \(\frac{\pi}{180}\) Radians
The radius of the circle r = 10 feet
Length of the arc = s
s = rθ

Question 7.
If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their radii.
Answer:
Let r₁ and r₂ be the radii of the two circles and l be the length of the arc.
Given central angle θ₁ = 60°

Question 8.
The perimeter of a certain sector of a circle is equal to the length of the arc of a semi-circle having the same radius. Express the angle of the sector in degrees, minutes, and seconds.
Answer:
Let OAB be the sector of a circle of radius r.
The angle of the sector is θ.

Perimeter of the sector = OA + arc AB + OB arc AB = rθ
∴ Perimeter of the sector = r + r θ + r
= 2r + rθ
= r(2 + θ) ———- (1)
Length of the arc of the semi – circle of radius
l = nπ ——– (2)
Given that perimeter the circular sector = Length of the arc of the semi circle of radius r
From equations (1) and (2), we have
r(2 + θ) = πr
2 + θ = π
θ = π – 2

Question 9.
An airplane propeller rotates 1000 times per minute. Find the number of degrees that a point on the edge of the propeller will rotate in 1 second.
Answer:
Given An airplane, the propeller rotates 1000 times per minute.
∴ A point on the edge of the propeller also rotates 1000 times in 1 minute.
∴ In 1 minute the point describes 1000 × 2π radians angle at the centre.
In 60 seconds the point describes 1000 × 2π radians angle.
∴ In 1 second the angle described = \(\frac{1000 \times 2 \pi}{60}\) radians

Question 10.
A train is moving on a circular track of a 1500 m radius at the rate of 66 km/hr. What angle will it turn in 20 seconds?
Answer:
Radius of the circular track r = 1500 m.
Speed of the train = 66 km/hr
Let θ be the angle made by the path of train at the centre in 20 seconds.
In 1 hr distance moved by train along the circular path = 66 km
In 60 × 60 seconds distance moved = 66 km
∴ In 20 seconds distance moved s = \(\frac{66}{60 \times 60}\) × 20

Question 11.
A circular metallic plate of radius 8 cm and thickness 6 nuns is melted and molded into a pie (a sector of the circle with thickness) of radius 16 cm and thickness 4 mm. Find the angle of the sector.
Answer:
Radius of the circular metallic plate r = 8 cm
Thickness of the plate h = 6 mm = \(\frac{6}{10}\)
Radius of the Pie l = 16 cm
Thickness of the Pie ( h) = 4mm = \(\frac{4}{10}\) cm
Given volume of the cylinder = Volume of the sector

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 2 Basic Algebra Ex 2.12 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.12

Question 1.
Let b > 0 and b ≠ 1. Express y = b^x in logarithmic form. Also, state the domain and range of the logarithmic function.
Answer:
Given y = b^x ⇒ log_by = x, x ∈ R with range (0, ∞) (-∞, ∞)

Question 2.
Compute log₉ 27 – log₂₇ 9
Answer:
log₉27 – log₂₇9 = log₉ 3³ – log₂₇ 3²
= 3 log₉ 3 – 2 log₂₇ 3 —— (1)
By change of base rule [log_b a = \(\frac{1}{\log _{a} b}\)]

Question 3.
Solve log_ax + log₄x + log₂x = 11
Answer:

Question 4.
Solve log ₄ 2 ^8x = 2 ^log₂8
Answer:
Given log ₄ 2^8x = 2 ^log₂8
log ₄ 2^8x = 2^log₂23
log ₄ 2^8x = 2^{3 log₂2}
log ₄ 2^8x = 2³ = 8
2^8x = 4⁸
(2²)^4x = 4⁸
⇒ (4)^4x = 4⁸
⇒ 4x = 8
⇒ x = \(\frac { 8 }{ 4 }\) = 2

Question 5.
If a² + b² = 7ab, show that
log \(\left(\frac{a+b}{3}\right)\) = \(\frac{1}{2}\) (log a + log b)
Answer:
Given
a² + b² = 7ab
Adding both sides 2ab we get
a² + b² + 2ab = 7ab + 2ab
(a + b)² = 9ab

Taking square root on both sides

Taking logarithm on both sides

Question 6.
Prove that log \(\frac{\mathbf{a}^{2}}{\mathbf{b c}}\) + log \(\frac{\mathbf{b}^{2}}{\mathbf{c a}}\) + log \(\frac{c^{2}}{a b}\) = 0
Answer:

Question 7.
Prove that

Answer:

Question 8.
Prove that log_a² a + log_b² b + log_c² c = \(\frac{1}{8}\)
Answer:

Question 9.
Prove log a + log a² + log a³ + ……… + log aⁿ = \(\frac{n(n+1)}{2}\) log a
Answer:
log a + log a² + log a³ + ……… + log aⁿ
= log a + 2 log a + 3 log a + ………….. + n log a
= log a (1 + 2 + 3 + ………….. + n)
= log a × \(\frac{n(n+1)}{2}\)
= \(\frac{n(n+1)}{2}\) log a

Question 10.
If , then prove that xyz = 1
Answer:
Let \(\frac{\log x}{y-z}\) = k
log x = k(y – z)
log x = ky – kz ——— (1)
Similarly log y = k(z – x) = kz – kx ——(2)
log z = k(x – y) = kx – ky ——- (3)
Adding (1), (2) and (3)
log x + log y + log z = ky – kz + kz – kx + kx – ky
log (xyz) = 0 = log 1
xyz = 1

Question 11.
Solve log₂x – 3 log_1/2x = 6
Answer:

Question 12.
Solve log_{5 – x} (x² – 6x + 65) = 2
Answer:
log_{5 – x}(x² – 6x + 65) = 2
⇒ x² – 6x + 65 = (5 – x)²
⇒ x² – 6x + 65 = 25 + x² – 10x
⇒ x² – 6x + 65 – 25 – x² + 10x = 0
⇒ 4x + 40 = 0
⇒ 4x = -40
⇒ x = -10

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 2 Basic Algebra Ex 2.13 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.13

Question 1.
If |x + 2| ≤ 9 then x belongs to
(1) (- ∞,- 7)
(2) [- 11, 7]
(3) (-∞, – 7) ∪ [11, ∞]
(4) (-11, 7)
Answer:
(2) [- 11, 7]

Explanation:
-x – 2 ≤ 9 x + 2 ≤ 9
-x < 9 + 2 = 11 x ≤ 9 – 2 = 7
⇒ x ≥ -11
so x ∈ [-11, 7]

Question 2.
Given that x, y and b are real numbers x < y, b > 0 then
(1) xb < yb
(2) xb > yb
(3) xb < yb
(4) \(\frac{x}{b}\) ≥ \(\frac{y}{b}\)
Answer:
(1) xb < yb

Explanation:
Given x, y and b are real numbers and b ≥ 0. x > b
Multiplying by positive real number the inequality is not affected.
∴ xb < yb

Question 3.
If \(\frac{|x-2|}{x-2}\) ≥ 0 , then x belongs to
(1) [2, ∞)
(2) (2, ∞)
(3) (-∞, 2)
(4) (-2, ∞)
Answer:
(1) [2, ∞)

Explanation:
Given \(\frac{|x-2|}{x-2}\) ≥ 0
By the definition of mod function
|x – 2 | = – (x – 2) if x – 2 < 0 | x – 2 | = x – 2 if x – 2 > 0
Suppose x – 2 < 0 then

Question 4.
The solution of 5x – 1 < 24 and 5x + 1 > – 24 is
(1) (4, 5)
(2) (- 5, – 4)
(3) (- 5, 5)
(4) (- 5, 4)
Answer:
(3) (- 5, 5)

Explanation:
The given inequalities are
5x – 1 < 24 ——— (1)
5x + 1 > – 24 ——– (2)
(1) ⇒ 5x – 1 < 24
⇒ 5x < 24 + 1
⇒ 5x < 25 ⇒ x < 5 ——— (3)
(2) ⇒ 5x + 1 > – 24
⇒ 5x > – 24 – 1
⇒ 5x > – 25
⇒ x > – 5 ——– (4)
Combining (3) and (4), we have
-5 < x < 5
∴ x ∈ (- 5, 5)

Question 5.
The solution set of the following inequality
|x – 1| ≥ |x – 3| is
(1) [0, 2]
(2) [2, ∞)
(3) (0, 2)
(4) (-∞, 2)
Answer:
(2) [2, ∞)

Explanation:
The given inequality is |x – 1| ≥ |x – 3|
(x – 1)² ≥ (x – 3)²
x² – 2x + 1 ≥ x² – 6x + 9
– 2x + 1 ≥ – 6x + 9
6x – 2x + 1 – 9 ≥ 0
4x – 8 ≥ 0 ⇒ 4x ≥ 8
⇒ x ≥ 2
∴ The solution set of the given inequality lies in the interval (2, ∞)

Question 6.
The value of \(\log _{\sqrt{2}} 512\) is
(1) 16
(2) 18
(3) 9
(4) 12
Answer:
(2) 18

Explanation:

Question 7.
The value of \(\log _{3} \frac{1}{81}\) is
(1) – 2
(2) – 8
(3) – 4
(4) – 9
Answer:
(3) – 4

Explanation:
\(\log _{3} \frac{1}{81}\) = log₃1 – log₃81
= o – log₃ 3⁴
= – 4 log₃3
= – 4 × 1
= – 4

Question 8.
If log_√x0.25 = 4, then the value of x is
(1) 0.5
(2) 2.5
(3) 1.5
(4) 1.25
Answer:
(1) 0.5

Explanation:
log_√x0.25 = 4
By the definition of logarithm
0.25 = \((\sqrt{x})^{4}\)
(0.5 )2 = \(\left(x^{\frac{1}{2}}\right)^{4}\)
(0.5 )2 = x2
x = 0.5

Question 9.
The value of log_ab . log_bc . log_ca is
(1) 2
(2) 1
(3) 3
(4) 4
Answer:
(2) 1

Explanation:
log_ab . log_bc . log_ca = log_ac . log_ca
= log_aa = 1

Question 10.
If 3 is the logarithm of 343 then, the base is
(1) 5
(2) 7
(3) 6
(4) 9
Answer:
(2) 7

Explanation:
⇒ log_x343 = 3 ⇒ 343 = x³
(.i.e.,) 7³ = x³ ⇒ x = 7
⇒ x = 7

Question 11.
Find a so that the sum and product of the roots of the equation 2x² + (a – 3 ) x + 3a – 5 = 0 are equal is
(1) 1
(2) 2
(3) 0
(4) 4
Answer:
(2) 2

Explanation:
Given quadratic equation is
2x² + (a – 3) x + 3a – 5 = 0
Let the roots be α, β
Sum of the roots α + β = \(-\frac{(a-3)}{2}\)
Product of the roots α β = \(\frac{3 a-5}{2}\)
Given α + β = α β
∴ \(-\frac{(a-3)}{2}=\frac{3 a-5}{2}\)
– a + 3 = 3a – 5
3a + a = 5 + 3
4a = 8 ⇒ a = 2

Question 12.
If a and b are the roots of the equation x² – kx +16 = 0 and satisfy a² + b² = 3², then the value of k is
(1) 10
(2) – 8
(3) – 8, 8
(4) 6
Answer:
(3) – 8, 8

Explanation:
a + b = k ….(1) ab = 16 ….(2)
a² + b² = (a + b)² – 2ab = 32 .
k² – 32 = 32 ⇒ k² = 64 ⇒ k = ±8

Question 13.
The number of solutions of x² + |x – 1| = 1 is
(1) 1
(2) 0
(3) 2
(4) 3
Answer:
(3) 2

Explanation:
The given quadratic equatiuon is
x² + |x – 1| = 1 ——– (1)

Case(i)
By the definition of mod function if x – 1 ≥ 0, then
|x – 1| = x – 1
∴ (1) ⇒ x² + x – 1 = 1
x² + x – 2 = 0
x² + 2x – x – 2 = 0
x(x + 2) – 1 (x + 2) = 0
(x – 1)(x + 2) = 0
x – 1 = 0 or x + 2 = 0
x = 1 or x = – 2
Since x – 1 ≥ 0
x = – 2 is not possible. ∴ x = 1

Case (ii)
Again by the definition of mod function if x – 1 < 0 then
|x – 1| = – (x – 1)
∴ (1) ⇒ x² – (x – 1) = 1
x² – x + 1 = 1
x² – x = 0
x (x – 1 ) = 0
x = 0 or x – 1 = 0
x = 0 or x = 1
Since x < 1, x = 1 is not possible
∴ x = 0
∴ The required solution set is {0, 1}
Number of solutions = 2

Question 14.
The equation whose roots are numerically equal but opposite in sign to the roots of 3x² – 5x – 7 = 0 is
(1) 3x² – 5x – 7 = 0
(2) 3x² + 5x – 7 = 0
(3) 3x² – 5x + 7 = 0
(4) 3x² + x – 7 = 0
Answer:
(2) 3x² + 5x – 7 = 0

Explanation:
The given quadratic equation is
3x² – 5x – 7 = 0 ——— (1)
Let α and β be the roots of eqn (1)
Sum of the roots α + β = \(-\left(\frac{-5}{3}\right)\)
α + β = \(\frac{5}{3}\)
Product of the roots α β = – \(\frac{7}{3}\)
The quadratic equation whose roots are – α and – β is
x² – (sum of the roots) x + Product of the roots = 0
x² – (- α – β)x + (- α)(- β) = 0
x² + (α + β)x + αβ = 0
x² + \(\frac{5}{3}\) x – \(\frac{7}{3}\) = 0
3x² + 5x – 7 = 0 is the required equation.

Question 15.
If 8 and 2 are the roots of x² + ax + c = 0 and 3, 3 are the roots of x² + dx + b = 0 , then the roots of the equation x² + ax + b = 0 are
(1) 1, 2
(2) -1, 1
(3) 9, 1
(4) -1, 2
Answer:
(3) 9, 1

Explanation:
Given that 8 and 2 are the roots of the equation
x² + ax + c = 0 ———- (1)
Sum of the roots 8 + 2 = \(-\frac{a}{1}\) ⇒ a = -10
Product of the roots 8 × 2 = \(\frac{c}{1}\) ⇒ c = 16
Also given 3 , 3 are the roots of
x² + dx + b = 0 ——— (2)
Sum of the roots 3 + 3 = – \(\frac{d}{1}\) ⇒ d = – 6
Product of the roots 3 × 3 = \(\frac{b}{1}\) ⇒ b = 9
Let a and p be roots of the equation x² + ax + b = 0
Sum of the roots α + β = – y
α + β = -(-10)
α + β = 10 ——– (3)
Product of the roots α β = y
α β = 9 ———- (4)
(α – β) = (α + β)² – 4αβ
= 10² – 4 × 9
= 100 – 36 = 64
α – β = 8 ——— (5)
Solving equations (3) and (5)

Substituting in equation (3),
9 + β = 10
⇒ β = 10 – 9 = 1
∴ The required roots are 9, 1

Question 16.
If a and b are the real roots of the equation x² – kx + c = 0 , then the distance between the
points(a, 0) and (b, 0)is
(1) \(\sqrt{\mathbf{k}^{2}-4 \mathbf{c}}\)
(2) \(\sqrt{4 k^{2}-c}\)
(3) \(\sqrt{4 \mathbf{c}-\mathbf{k}^{2}}\)
(4) \(\sqrt{\mathbf{k}-8 \mathbf{c}}\)
Answer:
(1) \(\sqrt{\mathbf{k}^{2}-4 \mathbf{c}}\)

Explanation:
Given that a and b are the roots of the equation
x² – kx + c = 0 ——— (1)
Sum of the roots a + b = \(-\frac{(-k)}{1}\)
a + b = k ———- (2)
Product of the roots ab = \(\frac{c}{1}\)
ab = c ——– (3)
Distance between the points ( a, 0) and (b, 0) is

Question 17.
If then the value of k is
(1) 1
(2) 2
(3) 3
(4) 4
Answer:
(3) 3

Explanation:

kx = 2x – 2 + x + 2
kx = 3x ⇒ k = 3

Question 18.
If , then the value of A + B is
(1) – \(\frac{1}{2}\)
(2) – \(\frac{2}{3}\)
(3) \(\frac{1}{2}\)
(4) \(\frac{2}{3}\)
Answer:
(1) – \(\frac{1}{2}\)

Explanation:

1 – 2x = A (x + 1) + B(3 – x) ——— (1)
Put x = – 1 in equation (1)
1 – 2 (- 1) = A (- 1 + 1) + B (3 + 1)
1 + 2 = 0 + 4B ⇒ B = \(\frac{3}{4}\)
Put x = 3 in equation (1)
1 – 2 × 3 = A(3 + 1) + B(3 – 3)
1 – 6 = 4A + 0
– 5 = 4A

Question 19.
The number of roots of (x + 3)⁴ + (x + 5)⁴ = 16 is
(1) 4
(2) 2
(3) 3
(4) 0
Answer:
(1) 4

Explanation:
The equation is (x + 3)⁴ + (x + 5)⁴ = 16
(x + 3)⁴ + (x + 5)⁴ = 2⁴
This is biquadratic equation. It has 4 roots.

Question 20.
The value of
log₃ 11 . log₁₁ 13 . log ₁₃ 15 . log ₁₅ 27 . log ₂₇ 81 is
(1) 1
(2) 2
(3) 3
(4) 4
Answer:
(4) 4

Explanation:
log₃ 11 . log₁₁ 13 . log ₁₃ 15 . log ₁₅ 27 . log ₂₇ 81
= log₃ 13 . log ₁₃ 15 . log ₁₅ 27 . log ₂₇ 81
= log ₃ 15 . log ₁₅ 27 . log ₂₇ 81
= log ₃ 27 . log ₂₇ 81
= log ₃ 81
= log ₃3⁴
= 4 log ₃3
= 4 × 1
= 4

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 2 Basic Algebra Ex 2.11 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.11

Question 1.
Simplify
(a) (125)^2/3
(b) 16^-3/4
(c) (- 1000)^-2/3
(d) (3^-6)^1/3
(e) \(\frac{27^{-\frac{2}{3}}}{27^{-\frac{1}{3}}}\)
Answer:
(a) (125)^2/3

(b) 16^-3/4

(c) (- 1000)^-2/3

(d) (3^-6)^1/3

(e) \(\frac{27^{-\frac{2}{3}}}{27^{-\frac{1}{3}}}\)

Question 2.
Evaluate
Answer:

Question 3.
If \(\left(x^{\frac{1}{2}}+x^{-\frac{1}{2}}\right)^{2}=\frac{9}{2}\) then find the value of \(\left(x^{\frac{1}{2}}-x^{-\frac{1}{2}}\right)\) for x > 1
Answer:

Question 4.
Simplify and hence find the value of n:
\(\frac{3^{2 n} \cdot 9^{2} \cdot 3^{-n}}{3^{3 n}}\) = 27
Answer:
Given

4 – 2n = 3
2n = 4 – 3
2n = 1
n = \(\frac { 1 }{ 2 }\)

Question 5.
Find the radius of the spherical tank whose volume is \(\frac{32 \pi}{3}\) units.
Answer:
Let r be the radius of the spherical tank.
Given volume of the spherical tank = \(\frac{32 \pi}{3}\)
\(\frac{4}{3}\)πr³ = \(\frac{32 \pi}{3}\)
4r³ = 32
r³ = \(\frac{32}{4}\) = 8
r³ = 2³
r = 2
∴ Radius of the spherical tank r = 2 units.

Question 6.
Simplify by rationalizing the denominator \(\frac{7+\sqrt{6}}{3-\sqrt{2}}\)
Answer:
\(\frac{7+\sqrt{6}}{3-\sqrt{2}}\)
Multiply the numerator and denominator by 3 + √2

Question 7.
Simplify:

Answer:

Question 8.
If x = √2 + √3 find \(\frac{x^{2}+1}{x^{2}-2}\)
Answer:

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 2 Basic Algebra Ex 2.10 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.10

Determine the region in the plane determined by the inequalities

Question 1.
x ≤ 3y, x ≥ y
Answer:
Consider the line x = 3y
When y = 0 ⇒ x = 0
When y = 1 ⇒ x = 3
When y = – 1 ⇒ x = – 3

Consider the line x = y
When x = 0 ⇒ y = o
When x = 1 ⇒ y = 1
When x = – 1 ⇒ y = – 1

To find the region of x ≤ 3y: The line x = 3y divides the cartesian plane into two half planes. Consider the point (1,1) this point satisfies the inequality x < 3y. This point (1,1) lies above the line x = 3y.

Hence all points satisfying the inequality lie above the line x = 3y.

Therefore x < 3y represents the upper half plane of the Cartesian plane bounded by the line x = 3y. Since x ≤ 3y, this region also contains all the points on the straight line x = 3y.

To find the region x ≥ y: The line x = y divides the cartesian plane into two half planes. Consider the point (2,1). This point (2,1) satisfies the inequality x > y and this point (2,1) lies below the line x = y. ∴ All points satisfying the inequality x > y will lie below the line x = y. Therefore x > y represents the lower half plane of the cartesian plane bounded by the line x = y. Since x > y this region also contains all the points on the straight line x = y.

The required region is the region common to the regions x ≤ 3y and x ≥ y

Question 2.
y ≥ 2x, – 2x + 3y ≤ 6
Answer:
Consider the straight line y = 2x
When x = 0 ⇒ y = 2 × 0 = 0
When x = 0 ⇒ y = 2 × 1 = 2
When x = – 1 ⇒ y = 2 × – 1 = – 2

Consider the line – 2x + 3y = 6
When x = 0 ⇒ – 2 × 0 + 3y = 6 ⇒ 3y = 6 ⇒ y = 2
When y = 0 ⇒ – 2x + 3 × 0 = 6 ⇒ -2x = 6 ⇒ x = – 3

To find the region of y ≥ 2x: The straight line y = 2x divides the cartesian plane into two half planes. Consider the point (1, 3) satisfying the inequality y > 2x. This point (1, 3) lies above the straight line y = 2x. ∴ All points satisfying the inequality y > 2x will lie above the line y = 2x.

To find the region of -2x + 3y ≤ 6:
The straight line – 2x + 3y = 6 divides the cartesian plane into two half planes one half plane contains the origin and the other half plane does not contain the origin.

Substitute the origin (0, 0) in the inequality – 2x + 3y < 6
we get – 2 × 0 + 3 × 0 < 6 ⇒ 0 < 6
∴ (0,0) satisfies the inequality – 2x + 3y < 6
∴ The inequality – 2x + 3y < 6 represents the half plane containing the origin. Since – 2x + 3y ≤ 6, this region contains all the points on the straight line – 2x + 3y = 6.

The required region is the region common to y ≥ 2x and – 2x + 3y ≤ 6

Question 3.
3x + 5y ≥ 45, x ≥ 0, y ≥ 0
Answer:
Consider the line 3x + 5y = 45
when x = 0, 3 × 0 + 5y = 45 ⇒ y = \(\frac{45}{5}\) = 9
when y = 0, 3x + 5 × 0 = 45 ⇒ x = \(\frac{45}{3}\) = 15

To find the region of x ≥ 0, y ≥ 0:
x ≥ 0 and y ≥ 0 denote the first quadrant of the Cartesian plane.
Since x ≥ 0 and y ≥ 0 this region contains all the points on the lines x = 0 and y = 0.

To find the region of 3x + 5y ≥ 45: The straight-line 3x + 5y = 45 divides the cartesian plane into two half-planes, one-half plane containing the origin and the other half-plane does not contain the origin.

Substitute the origin (0,0) in the inequality 3x + 5y > 45 we get 3 × 0 + 5 × 0 > 45 ⇒ 0 > 45 which is impossible. Therefore, (0, 0) does not satisfy the inequality 3x + 5y > 45 represents the half plane that does not contain the origin bounded by the straight line 3x + 5y = 45. Since 3x + 5y ≥ 45, x ≥ 0, y ≥ 0, this region contains all the points on the straight lines 3x + 5y = 45.

∴ The required region is the common region bounded by x ≥ 0, y ≥ 0, and 3x + 5y ≥ 45.

Question 4.
2x + 3y ≤ 35, y ≥ 2, x ≥ 5.
Answer:
Consider the straight line 2x + 3y = 35
When x = 0, 2 × 0 + 3y = 35 ⇒ \(\frac{35}{3}\)
When y = 0, 2x + 3 × 0 = 35 ⇒ \(\frac{35}{2}\)

To find the region of 2x + 3y ≤ 35: The straight-line 2x + 3y = 35 divides the cartesian plane into two half-planes, one-half plane contains the origin and the other half-plane does not contain the origin.

Substitute the origin (0, 0) in the inequality 2x + 3y < 35, we get 2 × 0 + 3 × 0 < 35 ⇒ 0 < 35. (0, 0) satisfies the inequality.

Therefore, the inequality 2x + 3y < 35 represents the half plane that contains the origin (0, 0) bounded by the straight line 2x + 3y = 35. Since 2x + 3y ≤ 35, this region contains all the points on the straight line 2x + 3y = 35.

To find the region y ≥ 2: The straight line y = 2 divides the cartesian plane into two half-planes, one-half plane contains the origin and the other half-plane does not contain the origin. Substitute the point (0, 0) in the inequality y ≥ 2 we get 0 >2 which is impossible. Hence (0, 0) does not satisfy the inequality y > 2.

∴ The inequality y > 2 represents the half-plane that does not contain the origin bounded by the straight line y = 2. Since y ≥ 2, this region contains all the points on the straight line y = 2.

To find the region x ≥ 5: The straight line x – 5 divides the cartesian plane into two half-planes, one half-plane containing the origin and the other half-plane does not contain the origin.

Substitute the origin (0,0) in the inequality x > 5 we get 0 > 5 which is impossible. Hence (0, 0) does not satisfy the inequality x > 5.

∴ The inequality x > 5 represents the half-plane that does not contain the origin bounded the straight line x = 5.
Since x ≥ 5, this region contains all the points on the straight line x = 5

∴ The required region is the common region bounded by 2x + 3y ≤ 35, y ≥ 2, x ≥ 5.

Question 5.
2x + 3y ≤ 6, x + 4y ≤ 4, x ≥ 0, y ≥ 0
Answer:
Consider the straight line 2x + 3y = 6
When x = 0, 2 × 0 + 3y = 6 ⇒ y = \(\frac{6}{3}\) = 2
When y = 0, 2x + 3 × 0 = 6 ⇒ x = \(\frac{6}{2}\) = 3

Consider the straight line x + 4y = 4
When x = 0 ⇒ 0 + 4y = 4 ⇒ y = \(\frac{4}{4}\) = 1
When y = 0 ⇒ x + 4 × 0 = 4 ⇒ x = 4

To find the region of 2x + 3y ≤ 6: The straight line 2x + 3y = 6 divides the cartesian plane into two half planes, one half plane contains the origin and the other half plane does not contain the origin. Substitute the origin (0, 0) in the inequality 2x + 3y < 6 we get 2 × 0 + 3 × 0 < 6 ⇒ 0 < 6

∴ The origin (0,0) satisfies the inequality 2x + 3y < 6. Hence the inequality 2x + 3y < 6 represents the half plane that contains the origin (0, 0). Since 2x + 3y ≤ 6, this region contains all the points on the straight line 2x + 3y = 6.

To find the region of x + 4y ≤ 4 : The straight line x + 4y = 4 divides the cartesian plane into two half planes, one half plane contains the origin and the other half plane does not contain the origin. Substitute the origin (0, 0) in the inequality x + 4y < 4, we get 0 + 4 × 0 < 4 ⇒ 0 < 4.

Therefore the origin (0, 0) satisfies the inequality x + 4y < 4. Therefore the inequality x + 4y < 4 represents the half-plane that contains the origin bounded by the line x + 4y = 4. Since x + 4y ≤ 4, this region contains all the points on the straight line.

To find the region x ≥ 0, y ≥ 0: x ≥ 0 and y ≥ 0 denote the first quadrant of the cartesian plane.
Since x ≥ 0 and y ≥ 0 this region contains all the points on the lines x = 0 and y = 0.

The required region is the common region bounded by 2x + 3y ≤ 6, x + 4y ≤ 4, x ≥ 0 and y ≥ 0.

Question 6.
x – 2y ≥ 0 , 2x – y ≤ – 2 , x ≥ 0, y ≥ 0
Answer:
Consider the straight line x – 2y = 0
When x = 0
⇒ -2y = 0
⇒ y = 0

When x = 2
⇒ 2 – 2y = 0
⇒ 2y = 2
⇒ y = 1

When x = -2
⇒ – 2 – 2y = 0
⇒ -2y = 2
⇒ y = – 1

Consider the line 2x – y = – 2
When x = 0 ⇒ – y = – 2 ⇒ y = 2
When y = 0 ⇒ 2x – 0 = – 2 ⇒ x = – 1

To find the region of x – 2y ≥ 0: The straight line x – 2y = 0 divides the cartesian plane into two half-planes. Consider the point (3,1) satisfying the inequality x > 2y. The point (3, 1) lies below the straight line x = 2y in the cartesian plane.

∴ All the points satisfying the inequality x > 2y will lie in the half-plane below the straight line x = 2y. Since x ≥ 2y this region contains all the points on the straight line x = 2y.

To find the region of 2x – y ≤ – 2:
-(2x – y) ≥ 2 ⇒ – 2x + y ≥ 2
Consider the straight line – 2x + y = 2. This line divides the cartesian plane in to two half planes, one half plane contains the origin and the other half plane does not contain the origin. Substitute the origin (0, 0) in the inequality – 2x + y > 2 we get- 2 × 0 + 0 > 2 ⇒ 0 > 2 which is impossible. Therefore (0, 0) does not satisfy the inequality -2x + y > 2. Hence the inequality -2x + y > 2 represents the half plane that does not contain the origin bounded by the straight line. Since -2x + y > 2, this region contains all the points on the straight line -2x + y = 2

To find the region of x ≥ 0, y ≥ 0: x ≥ 0 and y ≥ 0 denote the first quadrant of the cartesian plane. Since x ≥ 0 and y ≥ 0 this region contains all the points on the lines x = 0 and y = 0.

∴ The required region is the region common to x – 2y ≥ 0 , 2x – y ≤ – 2 , x ≥ 0, y ≥ 0.

Question 7.
2x + y ≥ 8, x + 2y ≥ 8, x + y ≤ 6
Answer:
Consider the straight line 2x + y = 8
When x = 0 ⇒ 2 × 0 + y = 8 ⇒ y = 8
When y = 0 ⇒ 2x + 0 = 8 ⇒ x = \(\frac{8}{2}\) = 4

Consider the straight line x + 2y = 8
When x = 0 ⇒ 0 + 2y = 8 ⇒ y = \(\frac{8}{2}\) = 4
When y = 0 ⇒ x + 2 × 0 = 8 ⇒ x = 8

Consider the straight line x + y = 6
When x = 0 ⇒ 0 + y = 6 ⇒ y = 6
When y = 0 ⇒ x + 0 = 6 ⇒ x = 6

To find the region of 2x + y ≥ 8: The straight-line 2x + y = 8 divides the cartesian plane into two half-planes one half plane contains the origin and the other half plane does not contain the origin. Substitute the origin (0, 0) in the inequality 2x + y > 8 we get 2 × 0 + 0 > 8 ⇒ 0 > 8 which is impossible.
∴ The origin (0, 0) does not satisfy the inequality 2x + y > 8. Hence the inequality 2x + y > 8 represents the half plane that does not contain the origin bounded by the straight line 2x + y = 8. Since 2x + y ≥ 8, this region contains all the points on the straight line 2x + y = 8.

To find the region of x + 2y ≥ 8: The straight line x + 2y = 8 divides the cartesian plane into two half planes, one half plane contains the origin and the other half plane doesnot contain the origin. Substitute the origin (0, 0) in the inequality x + 2y > 8 we get 0 + 2 × 0 > 8 ⇒ 0 > 8 which is impossible. ∴ The origin (0, 0) does not satisfy the inequality x + 2y > 8. Therefore, the inequality x + 2y > 8 represents the half plane that does not contain the origin bounded by the straight line x + 2y = 8. Since x + 2y ≥ 8 this region contains all the points on the straight line x + 2y = 8.

To find the region of x + y ≤ 6: The straight line x + y = 6 divides the cartesian plane into two half-planes, one half-plane contains the origin and the other half-plane does not contain the origin. Substitute the origin (0, 0) in the inequality x + y < 6 we get 0 + 0 < 6 ⇒ 0 < 6 which is true.
∴ The origin (0, 0) satisfies the inequality x + y < 6. Hence, the inequality x + y < 6 represents the half plane that contains the origin bounded by the straight line x + y = 6. Since x + y ≤ 6, this region contains all the points on the straight line x + y = 6.

Thus the required region is the region common to 2x + y ≥ 8, x + 2y ≥ 8, x + y ≤ 6

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 2 Basic Algebra Ex 2.9 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.9

Question 1.
Resolve the following rational expressions into partial fractions : \(\frac{1}{x^{2}-a^{2}}\)
Answer:

1 = A (x – a) + B (x + a) ——– (1)
Put x = a in equation (1)
1 = A (0) + B (a + a)
1 = B(2a)
⇒ B = \(\frac{1}{2 a}\)
Put x = – a in equation (1)
1 = A(- a – a) + B(- a + a)
1 = – 2a A + 0
⇒ A = \(-\frac{1}{2 a}\)
∴ The required partial fraction is

Question 2.
\(\frac{3 x+1}{(x-2)(x+1)}\)
Answer:

3x + 1 = A(x + 1) + B (x – 2) ——— (1)
Put x = 2 in equation (1)
3(2) + 1 = A (2 + 1) + B (2 – 2)
6 + 1 = 3A + 0
⇒ A = \(\frac{7}{3}\)
Put x = – 1 in equation (1)
3(-1) + 1 = A (-1 + 1 ) + B (- 1 – 2)
– 3 + 1 = A × 0 – 3B
– 2 = 0 – 3B
⇒ B = \(\frac{2}{3}\)
∴ The required partial fractions are

Question 3.
\(\frac{x}{\left(x^{2}+1\right)(x-1)(x+2)}\)
Answer:

x = Ax (x + 1) (x + 2) + B(x – 1)(x + 2) + C(x² + 1)(x + 2) + D(x² + 1)(x – 1) ——— (1)
Put x = 1 in equation (1)
1 = A(1)(1 – 1) (1 + 2) + B ( ( 1 – 1 ) (1 + 2) + C(1² + 1 ) ( 1 + 2) + D(1² + 1) (1 – 1)
1 = A × 0 + B × 0 + C(2)(3) + D × 0
1 = 6C
⇒ C = \(\frac{1}{6}\)

Put x = – 2 in equation (1)
-2 = A(- 2)(- 2 – 1)(- 2 + 2) + B(- 2 – 1)(- 2 + 2) + C ((- 2)² + 1) (- 2 + 2) + D((-2)² + 1 ) (- 2 – 1)
– 2 = A × 0 + B × 0 + C × 0 + D(4 + 1)(-3)
-2 = D(5)(-3)
⇒ – 2 = – 15 D
⇒ D = \(\frac{2}{15}\)

Put x = 0 in equation (1)
0 = A(0) (0 – 1) (0 + 2) + B(0 – 1)(0 + 2)+ C(0² + 1) (0 + 2) + D(0² + 1) (0 – 1)
0 = 0 + B (- 2 ) + C (2) + D (- 1)
0 = – 2B + 2C – D

In equation (1), equate the coefficient of x³ on both sides
0 = A + C + D

Question 4.
\(\frac{x}{(x-1)^{3}}\)
Answer:

X = A(x – 1)² + B(x – 1) + C ——— (1)
Put x = 1 in equation (1)
⇒ 1 = A(1 – 1)² + B(1 – 1) + C
1 = 0 + 0 + C
⇒ C = 1

In equation (1), equating the coefficient of x² on both sides
0 = A ⇒ A = 0
Put x = 0 in equation (1) ⇒ 0 = A(0 – 1)² + B(0 – 1) + C
⇒ 0 = A – B + C
0 = 0 – B + 1
⇒ B = 1

Question 5.
\(\frac{1}{x^{4}-1}\)
Answer:

1 = Ax (x + 1)(x – 1) + B (x + 1) (x – 1) + C (x<sup2 + 1)(x – 1) + D(x + 1)(x² + 1) —— (1)
Put x = 1 in equation (1)
1 = A(1 ) (1 + 1) (1 – 1) + B(1 + 1) (1 – 1) + C(1<sup2 + 1) (1 – 1) + D(1 + 1) (1<sup2 + 1)
1 = A × 0 + B × 0 + C × 0 + D(2)(2)
⇒ 1 = – 4D
⇒ D = \(\frac{1}{4}\)

Put x = -1 in equation (1)
1 = A(- 1)(- 1 + 1)(- 1 – 1) + B(- 1 + 1)(- 1 – 1) + C ((- 1)² + 1)(- 1 + 1) + D(- 1 + 1)((-1)² + 1)
1 = A × 0 + B × 0 + C (2) (-2) + D × 0
⇒ 1 = -4C
⇒ C = \(-\frac{1}{4}\)

Put x = 0 in equation (1)
I = A(0) (0 + 1 )(0 – 1) + B(0 + 1 )(0 – 1) + C(0² + 1 )(0 – 1) + D(0 + 1)(0² + 1)
1 = A × O + B(-1) + C(- 1) + D(1)
⇒ 1 = – B – C + D
1 = – B + \(\frac{1}{4}\) + \(\frac{1}{4}\)
⇒ B = \(\frac{1}{2}\) – 1 = – \(\frac{1}{2}\)
⇒ B = – \(\frac{1}{4}\)
In equation (1), equate the coefficient of x3 on both sides
0 = A + C + D
⇒ 0 = A – \(\frac{1}{4}+\frac{1}{4}\)
⇒ A = 0

Question 6.
\(\frac{(x-1)^{2}}{x^{3}+x}\)
Answer:

(x – 1)² = A(x² + 1) + Bx² + Cx ——- (1)
Put x = 0 in equation (I)
(0 – 1)² = A(0² + 1) + B × 0 + C × 0
1 = A + 0 + 0
⇒ A = 1
Equating the coefficient of x² on both sides
1 = A + B
1 = 1 + B
⇒ B = 0
Put x = 1 in equation (1)
(1 – 1)² = A(1² + 1) + B × 1² + C × 1
0 = 2A + B + C
0 = 2 × 1 + 0 + C
⇒ C = – 2
∴ The required partial fraction is

Question 7.
\(\frac{x^{2}+x+1}{x^{2}-5 x+6}\)
Answer:
\(\frac{x^{2}+x+1}{x^{2}-5 x+6}\)
Here the degree of the numerator is equal to the degree of the denominator. Let us divide the numerator by the

6x – 5 = A(x – 3) + B(x – 2) ——- (3)
Put x = 3 in equation (3)
6(3) – 5 = A(3 – 3) + B(3 – 2)
18 – 5 = 0 + B
⇒ B = 13
Put x = 2 in equation (3)
6(2) – 5 = A(2 – 3) + B(2 -2)
12 – 5 = – A + 0
7 = -A
⇒ A = – 7
Substituting the values of A and B in equation (2)

Question 8.
\(\frac{x^{3}+2 x+1}{x^{2}+5 x+6}\)
Answer:
\(\frac{x^{3}+2 x+1}{x^{2}+5 x+6}\)
Since the numerator is of degree greater than that of the denominator divide the numerator by the denominator.

21x + 31 = A(x + 3) + B(x + 2) ——- (3)
Put x = – 3 . in equation (3)
21(- 3) + 31 = A(- 3 + 3) + B(- 3 + 2)
– 63 + 31 = 0 – B
– 32 = – B
⇒ B = 32
Put x = – 2 , in equation (3)
21(- 2) + 31 = A(- 2 + 3) + B(- 2 + 2)
– 42 + 31 = A + 0
⇒ A = – 11
Substituting the values of A and B in equation (2)

Question 9.
\(\frac{x+12}{(x+1)^{2}(x-2)}\)
Answer:

x + 12 = A(x + 1 ) (x – 2) + B(x – 2) + C(x + 1)² ——– (1)
Put x = 2 in equation (l)
2 + 12 = A(2 + 1 ) (2 – 2) + B(2 – 2) + C(2 + 1)²
14 = A(3) (0) + B × 0 + C (3 )²
4 = 0 + 0 + 9C
⇒ C = \(\frac{14}{9}\)

Put x = – 1 in equation (1)
-1 + 12 = A(- 1 + 1)(- 1 – 2) + B(- 1 – 2) + C(- 1 + 1)²
11 = A × 0 + B (- 3 ) + C × 0
11 = -3 B
⇒ B = \(-\frac{11}{3}\)

Put x = 0 in equation (1)
0 + 12 = A (0 + 1)(0 – 2) + B(0 – 2) + C(0 + 1)²
12 = – 2A – 2B + C

Question 10.
\(\frac{6 x^{2}-x+1}{x^{3}+x^{2}+x+1}\)
Answer:

6x² – x + 1 = Ax (x + 1) + B (x + 1) + C(x² + 1) ——— (1)
Put x = – 1 in equation (1)
6 x (- 1)² – (- 1) + 1 = A (- 1 ) (- 1 + 1 ) + B(- 1 + 1) + C ( (- 1)² + 1 )
6 + 1 + 1 = A × 0 + B × 0 + C (2)
8 = 2C
⇒ C = 4

Put x = 0 in equation (1)
6 × 0² – 0 + 1 = A(0)(0 + 1 ) + B(0 + 1) + C(0² + 1)
1 = 0 + B + C
1 = B + 4
B = 1 – 4 = – 3
Equating the coefficient of x² in equation (1) we have
6 = A + C
6 = A + 4
⇒ A = 6 – 4
⇒ A = 2
∴ The required partial fraction is

Question 11.
\(\frac{2 x^{2}+5 x-11}{x^{2}+2 x-3}\)
Answer:
\(\frac{2 x^{2}+5 x-11}{x^{2}+2 x-3}\)
Since the degree of the numerator is equal to the degree of the denominator divide the numerator by the denominator

Put x = 1 in equation (3)
1 – 5 = A(1 + 3) + B(1 – 1)
– 4 = 4A + 0
⇒ A = – 1

Put x = – 3 in equation (3)
– 3 – 5 = A (- 3 + 3) + B(- 3 – 1)
– 8 = 0 – 4B
⇒ B = 2

Substituting the values of A and B in equation (2) we have

∴ The required partial fraction is

Question 12.
\(\frac{7+x}{(1+x)\left(1+x^{2}\right)}\)
Answer:

7 + x = A( 1 + x²) + Bx (1 + x) + C(1 + x) ——- (1)
Put x = -1 , in equation (1)
7 – 1 = A(1 + (-1)²) + B (- 1) (1 – 1) + C(1 – 1)
6 = A(1 + 1) + 0 + 0
A = \(\frac{6}{2}\) = 3
⇒ A = 3
Put x = 0 , in equation (1)
7 + 0 = A(1 + 0²) + B × 0 (1 + 0) + C(1 + 0)
7 = A + 0 + C
7 = 3 + C
⇒ C = 4
Equating the coefficient of x² in equation (I) we have
0 = A + B
0 = 3 + B
⇒ B = – 3
∴ The required partial fraction is

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 2 Basic Algebra Ex 2.8 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.8

Question 1.
Find all values of x for which \(\frac{x^{3}(x-1)}{x-2}\) > 0
Answer:
The given inequality is f(x) = \(\frac{x^{3}(x-1)}{x-2}\) > 0
[The critical numbers of f(x) are those values of x for which f(x) = 0, and those values of x for which f(x) is not defined.
When x = 2 , f(x) = ∞ ⇒ f(x) is not defined.]
The critical numbers are x = 0, 1, 2
Divide the number line into 4 intervals
(- ∞, 0), (0, 1), (1, 2) and (2, ∞)

(1) (- ∞, 0)
When x < 0 say x = – 1
The factor x³ = (- 1)³ = – 1 < 0
The factor x – 1 = – 1 – 1 = – 2 < 0
The factor x – 2 = – 1 – 2 = – 3 < 0
∴ \(\frac{x^{3}(x-1)}{x-2}\) < 0
Thus \(\frac{x^{3}(x-1)}{x-2}\) > 0 is not true in the interval (- ∞, 0)
Therefore, it has no solution in the interval (- ∞, 0)

(2) (0, 1)
When 0 < x < 1 say x = 0.5
The factor x³ = (0.5 )³ > 0
The factor x – 1 = 0.5 – 1 = – 0.5 < 0
The factor x – 2 = 0.5 – 2 = – 1.5 < 0
Thus x³ > 0, x – 1 < 0 and x – 2 < 0
∴ \(\frac{x^{3}(x-1)}{x-2}\) > 0
Thus \(\frac{x^{3}(x-1)}{x-2}\) > 0 is true in the interval (0, 1)
Therefore it has solution in (0,1)

(3) (1, 2)
When 1 < x < 2 say x = 1.5
The factor x³ = 0
The factor x – 1 = 1.5 – 1 = 0.5 > 0
The factor x – 2 = 1.5 – 2 = – 0.5 < 0
Thus x³ > 0, x – 1 > 0 and x – 2 < 0
∴ \(\frac{x^{3}(x-1)}{x-2}\) < 0
Thus \(\frac{x^{3}(x-1)}{x-2}\) > 0 is not true in the interval (1, 2).
Therefore it has no solution in (1, 2).

(4) (2, ∞)
When x > 2 say x = 3
The factor x³ = 3³ > 0
The factor x – 1 = 3 – 1 = 2 > 0
The factor x – 2 = 3 – 2 = 1 > 0
Thus x³ > 0, x – 1 > 0 and x – 2 > 0
∴ \(\frac{x^{3}(x-1)}{x-2}\) > 0
Thus \(\frac{x^{3}(x-1)}{x-2}\) > 0 is true in the interval (2, ∞).
Therefore it has a solution in (2, ∞)

∴ \(\frac{x^{3}(x-1)}{x-2}\) > 0 has solution in the intervals (0, 1) and (2, ∞)
∴ The solution set is given by (0, 1) ∪ (2, ∞)

Question 2.
Find all values of x that satisfies the inequality \(\frac{2 x-3}{(x-2)(x-4)}\) < 0.
Answer:
The given inequality is

[The critical numbers of f(x) are those values of x for which f(x) = 0, and those values of x for which f(x) is not defined. When x = 2, f(x) = ∞ ⇒ f(x) is not defined.]
The critical numbers are x = \(\frac{3}{2}\), x = 2 , x = 4
Divide the number into 4 intervals

(1) \(\left(-\infty, \frac{3}{2}\right)\)
When x < \(\frac{3}{2}\) say x = 0
The factor x – \(\frac{3}{2}\) = 0 – \(\frac{3}{2}\) < 0
The factor x – 2 = 0 – 2 < 0
The factor x – 4 = 0 – 4 < 0
Thus x – \(\frac{3}{2}\) < 0, x – 2 < 0 and x – 4 < 0

(2) \(\left(\frac{3}{2}, 2\right)\)

(3) (2, 4)
When 2 < x < 4 say x = 3 The factor x – \(\frac{3}{2}\) = 3 – \(\frac{3}{2}\) = \(\frac{3}{2}\) > 0
The factor x – 2 = 3 – 2 = 1 > 0
The factor x – 4 = 3 – 4 = – 1 < 0 Thus x – \(\frac{3}{2}\) > 0, x – 2 > 0 and x – 4 < 0

Thus \(\frac{2 x-3}{(x-2)(x-4)}\) < 0 is true in the interval (2, 4) ∴ It has solution in (2, 4). (4) (4, ∞) When x > 4 say x = 5
The factor x – \(\frac{3}{2}\) = 5 – \(\frac{3}{2}\) = \(\frac{7}{2}\) > 0
The factor x – 2 = 5 – 2 = 3 >0
The factor x – 4 = 5 – 4 = 1 > 0
Thus x – \(\frac{3}{2}\) > 0, x – 2 > 0 and x – 4 > 0

Thus \(\frac{2 x-3}{(x-2)(x-4)}\) < 0 is not true in the interval (4, ∞)
∴ It has a solution in (4, ∞)

Question 3.
Solve: \(\frac{x^{2}-4}{x^{2}+4 x-15}\) ≤ 0
Answer:

[The critical numbers of f(x) are those values of x for which f(x) = 0, and those values of x for which f(x) is not defined. When x = – 3, 5. f(x) = ∞ ⇒ f(x) is not defined.]

The critical numbers are x = – 2 , 2, – 3, 5
Divide the number line into five intervals
(- ∞, – 3), (- 3, – 2), (- 2, 2), (2, 5) ,(5, ∞)

(a) (- ∞, – 3)
When x <- 3 say x = – 4
The factor x + 2 = – 4 + 2 = – 2 < 0
The factor x – 2 = – 4 – 2 = 6 < 0
The factor x + 3 = – 4 + 3 = – 1 < 0
The factor x – 5 = – 4 – 5 = – 9 < 0
Thus x + 2 < 0, x + 3 < 0, x – 2 < 0, x – 5 < 0
∴ \(\frac{x^{2}-4}{x^{2}+4 x-15}\) > 0
Thus \(\frac{x^{2}-4}{x^{2}+4 x-15}\) ≤ 0 is not true in the interval (- ∞, – 3).
∴ It has no solution in (- ∞, – 3)

(b) (- 3, – 2)
When – 3 < x ≤ – 2 say x = – 2.5
The factor x + 2 = – 2.5 + 2 = – 0.5 < 0
The factor x – 2 = – 2.5 – 2 = – 4.5 < 0 The factor x + 3 = – 2.5 + 3 = 0.5 > 0
The factor x – 5 = – 2.5 – 5 = – 7.5 < 0
Thus x + 2 < 0, x + 3 > 0
and
x – 2 < 0
x – 5 < 0
∴ \(\frac{x^{2}-4}{x^{2}+4 x-15}\) < 0
Thus \(\frac{x^{2}-4}{x^{2}+4 x-15}\) ≤ 0 is not true in the interval (- 3, – 2).
∴ It has no solution in (- 3, – 2)

(c) (-2, 2)
When – 2 ≤ x ≤ 2 say x = 0
The factor x + 2 = 0 + 2 = 2 > 0
The factor x – 2 = 0 – 2 = – 2 < 0
The factor x + 3 = 0 + 3 = 3 > 0
The factor x – 5 = 0 – 5 = – 5 < 0

Thus x + 2 > 0 ,
x + 3 > 0
and
x – 2 < 0
x – 5 < 0
∴ \(\frac{x^{2}-4}{x^{2}+4 x-15}\) > 0
Thus \(\frac{x^{2}-4}{x^{2}+4 x-15}\) ≤ 0 is not true in the interval (- 2, – 2).
∴ It has no solution in (- 2, – 2)

(d) (2, 5)
When 2 ≤ x < 5 say x = 3 The factor x + 2 = 3 = 3 + 2 = 5 > 0
The factor x – 2 = 3 – 2 = 1 > 0
The factor x + 3 = 3 + 3 = 6 > 0
The factor x – 5 = 3 – 5 = – 2 < 0 Thus x + 2 > 0,
x + 3 > 0
and
x – 2 > 0
x – 5 < 0
∴ \(\frac{x^{2}-4}{x^{2}+4 x-15}\) < 0
Thus \(\frac{x^{2}-4}{x^{2}+4 x-15}\) ≤ 0 is not true in the interval (2, 5).
∴ It has no solution in (2, 5)

(e) (5, ∞)
When 5 < x < ∞ say x = 6 The factor x + 2 = 6 + 2 = 8 > 0
The factor x – 2 = 6 – 2 = 4 > 0
The factor x + 3 = 6 + 3 = 9 > 0
The factor x – 5 = 6 – 5 = 1 > 0
Thus
x + 2 > 0,
x + 3 > 0
and
x – 2 > 0,
x – 5 > 0
∴ \(\frac{x^{2}-4}{x^{2}+4 x-15}\) > 0
Thus \(\frac{x^{2}-4}{x^{2}+4 x-15}\) ≤ 0 is not true in the interval (5, ∞).
∴ It has no solution in (5, ∞)
The given inequality f(x) = \(\frac{x^{2}-4}{x^{2}+4 x-15}\) ≤ 0 has solution in the intervals (-3, – 2]
∴ The solution set is (-3, 2] ∪ [2, 5)

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 2 Basic Algebra Ex 2.7 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.7

Question 1.
Factorize x⁴ + 1
Answer:
The given equation is x⁴ + 1
x⁴ + 1 = (x²)² + 1²
= (x² + 1)² – 2 (x²) (1)
[ a² + b² = (a + b)² – 2ab]
= (x² + 1)² – (√2x)²
= (x² + 1 + √2x) (x² + 1 – √2x)
x⁴ + 1 = (x² + 2x + 1) (x² – √2x + 1)

Question 2.
If x⁴ + x + 1 is a factor of the polynomial 3x³ + 8x² + 8x + a, then find the value of a.
Answer:
Given that x² + x + 1 is a factor of the polynomial 3x³ + 8x² + 8x + a.
∴ 3x³ + 8x² + 8x + a is divisible by x² + x + 1

Since 3x³ + 8x² + 8x + a is divisible by x² + x + 1, the remainder must be zero.
a – 5 = 0
⇒ a = 5