Class 11

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.3

Question 1.
Represent the following inequalities in the interval notation:
Solution:

⇒ x ∈ [-1, 4)
[] closed interval, end points are included
() ➝ open interval
end points are excluded

(ii) x ≤ 5 and x ≥ -3[i] x ≤ 5 and x ≥ -3
Solution:

x ∈ [-3, 5)

(iii) x < -1 or x < 3
Solution:

x ∈ (-∞, -1) or x ∈ (-∞, 3)

(iv) – 2x > 0 or 3x – 4 < 11
Solution:
-2x > 0 ⇒ 2x < 0 ⇒ x < 0
x ∈ (-∞, 0)
3x – 4 < 11
⇒ 3x – 4 + 4 < 11 + 4

Question 2.
Solve 23x < 100 when
(i) x is a natural number,
(ii) x is an integer.
Solution:
23x <100

(i.e.,) x > 4.3
(i) x = 1, 2, 3, 4 (x ∈ N)
(ii) x = …. -3, -2, -1, 0, 1, 2, 3, 4 (x ∈ Z)

Question 3.
Solve -2x ≥ 9 when
(i) x is a real number,
(ii) x is an integer,
(iii) x is a natural number.
Solution:
-2x > 9 ⇒ 2x ≤ -9

(ii) x = …. -3, -2, -1, 0, 1, 2, 3, 4
(iii) x = 1, 2, 3, 4

Question 4.

Solution:

(ii)

Solution:

Question 5.
To secure an A grade one must obtain an average of 90 marks or more in 5 subjects each of a maximum of 100 marks. If one scored 84, 87, 95, 91 in the first four subjects, what is the minimum mark one scored in the fifth subject to get an A grade in the course?
Solution:
Given, to secure A grade in 5 subjects required average mark of 90 or more.
The marks scored in the first four subjects are 84, 87, 95, 91
Let the marks scored in the fifth subject be.
Then by the given data, we have

Multiplying both sides by 5, we get
357 + x ≥ 450
x ≥ 450 – 357
x ≥ 93
∴ The person must obtain a minimum of 93 marks to get A grade in the course.

Question 6.
A manufacturer has 600 litres of a 12 percent solution of acid. How many litres of a 30 percent acid solution must be added to it so that the acid content in the resulting mixture will be more than 15 percent but less than 18 percent?
Solution:
12% solution of acid in 600 l ⇒ 600 × \(\frac{12}{100}\) = 72 l of acid
15% of 600 l ⇒ 600 × \(\frac{15}{100}\) = 90 l
18% of 600 l ⇒ 600 × \(\frac{18}{100}\) = 108 l
Let x litres of 18% acid solution be added

(600 + x)15 ≥ 7200 + 30x
9000+ 15x ≥ 7200 + 30x
1800 ≥ 15x
x ≤ 120
Let x litres of 18% acid solution be added

10800 + 18 ≤ 7200 + 30x
3600 ≤ 12x
x > 300
The solution is 120 ≤ x > 300

Question 7.
Find all pairs of consecutive odd natural numbers both of which are larger than 10 and their sum is less than 40.
Solution:
Let the two numbers be x and x + 2
x + x + 2 < 40
⇒ 2x < 38

⇒ x< 19 and x > 10
so x = 11 ⇒ x + 2 = 13
x = 13 ⇒ x + 2 = 15
x = 15 ⇒ x + 2 = 17
When x = 17 ⇒ x + 2 = 19
So the possible pairs are (11, 13), (13, 15), (15, 17), (17, 19)

Question 8.
A model rocket is launched from the ground. The height h of the rocket after t seconds from lift off is given by h(t) = -5t² + 100t; 0 ≤ r ≤ 20. At what time the rocket is 495 feet above the ground?
Solution:
Given h(t) = – 5t² + 100t, 0 ≤ t ≤ 20.
Let the time be ‘t’ sec when the rocket is 495 feet above the ground.
∴ h (t) = 495 for time ‘t’ sec
-5t² + 100t = 495
5t² – 100t + 495 = 0
t² – 20t + 99 = 0
t² – 11t – 9t + 99 = 0
t(t – 11) – 9(t – 11) = 0
(t – 9 ) (t – 11 ) = 0
t – 9 = 0 or t – 11 = 0
t = 9 or t = 11
∴ At t = 11 or 9 seconds, the rocket is 495 feet above the ground.

Question 9.
A Plumber can be paid according to the following schemes: In the first scheme he will be paid Rs. 500 plus Rs.70 per hour, and in the second scheme he will be paid Rs. 120 per hour. If he works x hours, then for what value of x does the first scheme give better wages?
Solution:
Let the number of hours to complete the job = x
Wages for the first scheme = Rs. 500 + Rs. 70 per hour = 500 + 70x
Wages for the second scheme = Rs. 120 per hour = 120x
Let us find the value of x for which the first scheme gives better wages.
500 + 70x > 120x
500 > 120x – 70x
500 > 50x
\(\frac{500}{50}\) > x
x < 10
∴ The value of x so that the first scheme gives better wages is = 1, 2, 3, 4, 5, 6, 7, 8, 9

Question 10.
A and B are working on similar jobs but their annual salaries differ by more than Rs 6000. If B earns Rs. 27000 per month, then what are the possibilities of A’s salary per month?
Solution:
Let A’s salary be x, B’s salary is Rs. 27,000
Given their difference in salary is more than Rs. 6,000
Assume A’s salary is more than B’s salary.
∴ x – 27,000 > 6000
∴ x > 6000 + 27000
x > 33000
Assume B’s salary is more than A’s salary.
∴ 27,000 – x > 6000
∴ 27000 – 6000 > x
x < 21000
The possibilities of A’s salary are greater than Rs. 33,000 or less than Rs. 21,000.

Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.3 Additional Questions

Question 1.

Solution:

Multiplying both sides by 30, we get 15x ≥ 2(4x – 1) ⇒ 15x ≥ 8x – 2⇒ 15x – 8x ≥ -2

Question 2.
Ravi obtained 70 and 75 marks in the first two unit tests. Find the minimum marks he should get in the third test to have an average of at least 60 marks.
Solution:
Let x be the marks obtained by Ravi in the third test.

⇒ 145 + x ≥ 180 ⇒ x >180 – 145
⇒ x ≥ 35
Thus, Ravi must obtain a minimum of 35 marks to get an average of at least 60 marks.
Note. A minimum of 35 marks.
⇒ Marks greater than or equal to 35.

Question 3.
To receive Grade ‘A’ in a course, one must obtain an average of 90 marks or more in five examinations (each of 100 marks). If Sunita’s marks in the first four examinations are 87, 92, 94, and 95, find minimum marks that Sunita must obtain in the fifth examination to get Grade ‘A’ in the course.
Solution:
Let x be the marks obtained by Sunita in the fifth examination. Then,

⇒ 368 + x ≥ 450
⇒ x ≥ 450 – 368
⇒ x ≥ 82
Thus, Sunita must obtain marks greater than or equal to 82,
i.e., a minimum of 82 marks.

Question 4.
Find the pairs of consecutive odd positive integers both of which are smaller than 10 such that their sum is more than 11.
Solution:
Let x be the smaller of the two consecutive odd positive integers, then the other is x + 2.
According to the given conditions.
x < 10, x + 2 < 10 and x + (x + 2) > 11
⇒ x < 10, x < 8

From (1) and (2), we get 9
\(\frac{9}{2}\) < x < 8
Also, x is an odd positive integer. x can take values 5 and 7.
So, the required possible pairs will be (x, x + 2) = (5, 7), (7, 9)

Question 5.
Find all pairs of consecutive even positive integers, both of which are larger than 5 such that their sum is less than 23.
Solution:
Let x be the smaller of the two consecutive even positive integers, then the other is x + 2. According to the given conditions.

Also, x is an even positive integer.
x can take the values 6, 8, and 10.
So, the required possible pairs will be (x, x + 2) = (6, 8), (8, 10), (10, 12)

Question 6.
Forensic Scientists use h = 61.4 + 2.3F to predict the height h in centimetres for a female whose thigh bone (femur) measures F cm. If the height of the female lies between 160 to 170 cm find the range of values for the length of the thigh bone?
Solution:
Given h = 61.4 + 2.3 F
Given h = 160
⇒ 160 = 61.4 + 2.3 F
⇒ 2.3 F = 160 – 61.4 = 98.6

Given h = 170 ⇒ 170 = 61.4 + 2.3 F
⇒ 170 – 61.4 = 2.3 F
2.3F = 108.6

So the ranges of values are 42.87 < x < 47.23

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.5

Choose the correct or more suitable answer

Question 1.
The equation of the locus of the point whose distance from y-axis is half the distance from origin is ……..
(a) x² + 3y² = 0
(b) x² – 3y² = 0
(c) 3x² + y² = 0
(d) 3x² – y² = 0
Solution:
(c) 3x² + y² = 0
Hint:
Given that PA = \([\frac{1}{2}/latex]OP
2PA = OP

4PA² = OP²
4(x)² = x² + y² ⇒ 3x² – y² = 0

Question 2.
Which of the following equation is the locus of (at², 2at) ……

Solution:
(d) y² = 4ax
Hint:
Given x = at², y = 2at
y = 2at ⇒ t = [latex]\frac{y}{2 a}\)
x = at2 ⇒ x = a × \(\left(\frac{y}{2 a}\right)^{2}\)
x = a × \(\frac{y^{2}}{4 a^{2}}\) = \(\frac{y^{2}}{4 a^{2}}\)
y² = 4ax

Question 3.
Which of the following point lie on the locus of 3x² + 3y² – 8x – 12y + 17 = 0?
(a) (0, 0)
(b) (-2, 3)
(c) (1, 2)
(d) (0, -1)
Solution:
(c) (1, 2)
Hint:
The equation of the given locus is
3x² + 3y² – 8x – 12y + 17 = 0
(0, 0) does not lie on the locus since the locus contains constant term. Substituting (-2, 3) in the locus
3(- 2)² + 3(3)² – 8 × – 2 – 12 × 3 + 17
= 3 × 4 + 3 × 9 + 16 – 36 + 17
= 12 + 27 + 16 – 36 + 17 ≠ 0
∴ (- 2, 3) does not lie on the locus
Substituting (1, 2) on the locus
3(1)² + 3(2)² – 8 × 1 – 12 × 2 + 17
= 3 + 12 – 8 – 24 + 17
= 32 – 32
= 0
∴ (1, 2 ) lies on the locus

Question 4.

(a) 0
(b) 1
(c) 2
(d) 3
Solution:
(d) 3

Question 5.
Straight line joining the points (2, 3) and (-1, 4) passes through the point (α, β) if
(a) α + 2β = 7
(b) 3α + β = 9
(c) α + 3β = 11
(d) 3α + 3β = 11
Solution:
(c) α + 3β = 11
Hint:
Equation joining (2, 3), (-1, 4)

3y – 12 = – x -1 ⇒ x + 3y – 11 = 0, (α, β) lies on it ⇒ α + 3β – 11 = 0.

Question 6.
The slope of the line which makes an angle 45° with the line 3x – y = – 5 are
(a) 1, -1
(b) \(\frac{1}{2},-2\)
(c) \(1, \frac{1}{2}\)
(d) \(2,-\frac{1}{2}\)
Solution:
(c) \(1, \frac{1}{2}\)
Hint:
Equation of line 3x – y = -5, y = 3x + 5, m₁ = 3

Question 7.
Equation of the straight line that forms an isosceles triangle with coordinate axes in the I-quadrant with perimeter \(4+2 \sqrt{2}\) is
(a)x + y + 2 = 0
(b) x + y – 2 = 0
(c) x + y – \(\sqrt{2}\) = 0
(d) x + y + \(\sqrt{2}\) = 0
Solution:
(b) x + y – 2 = 0
Hint.
Let the sides be x, x

Question 8.
The coordinates of the four vertices of a quadrilateral are (-2, 4), (-1, 2), (1, 2) and (2, 4) taken in order. The equation of the line passing through the vertex (-1, 2) and dividing the quadrilateral into the equal areas is ………
(a) x + 1 = 0
(b) x + y = 1
(c) x + y + 3 = 0
(d) x – y + 3 = 0
Solution:
(b) x + y = 1
Hint:
This equation passes through (-1, 2)
-1 + 2 = 1 ⇒ 1 = 1

Question 9.
The intercepts of the perpendicular bisector of the line segment joining (1, 2) and (3, 4) with coordinate axes are ……….
(a) 5, -5
(b) 5, 5
(c) 5, 3
(d) 5, -4
Solution:
(b) 5, 5

Question 10.
The equation of the line with slope 2 and the length of the perpendicular from the origin equal to \(\sqrt{5}\) is ……
(a) x + 2y = \(\sqrt{5}\)
(b) 2x + y = \(\sqrt{5}\)
(c) 2x + y = 5
(d) x + 2y – 5 = 0
Solution:
(c) 2x + y = 5

The required line is y = 2x + 5 ⇒ 2x – y + 5 = 0

Question 11.
A line perpendicular to the line 5x – y = 0 forms a triangle with the coordinate axes. If the area of the triangle is 5 sq. units, then its equation is …….

Solution:
(a) x + 5y ± 5\(\sqrt{2}\) = 0
Hint:
Equation of a line perpendicular to 5x – y = 0 is

Question 12.
Equation of the straight line perpendicular to the line x – y + 5 = o, through the point of intersection the y-axis and the given line …….
(a) x – y – 5 = 0
(b) x + y – 5 = 0
(c) x + y + 5 = 0
(d) x + y + 10 = 0
Solution:
(b) x + y – 5 = 0
Hint:
x – y + 5 = 0 ⇒ put x = 0, y = 5
The point is (0, 5)
Equation of a line perpendicular to x – y + 5 = 0 is x + y + k = 0
This passes through (0, 5)
k = -5
x + 7 – 5 = 0

Question 13.
If the equation of the base opposite to the vertex (2, 3) of an equilateral triangle is x + y = 2, then the length of a side is ………

Solution:
\(\sqrt{6}\)
Hint:
In an equilateral, ∆ the perpendicular wall bisects the base into two equal parts. Length of the perpendicular drawn from (2, 3) to the line x + 7 – 2 = 0

Question 14.
The line (p + 2q) x + (p – 3q)y = p – q for different values of p and q passes through the point ……

Solution:
(d) \(\left(\frac{2}{5}, \frac{3}{5}\right)\)
Hint:
(p + 2 q)x + (p – 3q)y = p – q
px + 2qx + py – 3qy = p – q
P(x + y) + q (2x – 3y) = p – q
The fourth option x = 2/5, y = 3/5

= p – q = RHS

Question 15.
The point on the line 2x – 3y = 5 is equidistance from (1, 2) and (3, 4) is …
(a) (7, 3)
(b) (4, 1)
(c) (1, -1)
(d) (-2, 3)
Solution:
(b) (4, 1)
Hint:
Let (a, b) be on 2x – 3y = 5 ⇒ 2a – 3b = 5
It is equidistance from (1, 2) and (3, 4)

(a – 1)² + (b – 2)² = (a – 3)² + (6 – 4)²
a² – 2a + 1 + b² – 4b + 4 = a² – 6a + 9 + b² – 8b + 16
4a + 4b = 20
2a+ 2b = 10
2a – 3b = 5
5b = 5
b = 1 ∴ a = 4
∴ The point is (4, 1)

Question 16.
The image of the point (2, 3) in the line y = – x is ………
(a) (-3, -2)
(b) (-3, 2)
(c) (-2, -3)
(d) (3, 2)
Solution:
(a) (-3, -2)

x – 2 = -5, y – 3 = -5
x = -3, y = -2
(-3,-2)

Question 17.
The length of ⊥ from the origin to the line \(\frac{x}{3}-\frac{y}{4}=1\) is ……

Solution:
(c) \(\frac{12}{5}\)
Hint:
4x – 3y = 12 ⇒ 4x – 3y – 12 = 0

Question 18.
The y-intercept of the straight line passing through (1, 3) and perpendicular to 2x – 3y + 1 = 0 is ……..

Solution:
(b) \(\frac{9}{2}\)
Hint:
Equation of a line perpendicular to 2x – 3y + 1 = 0 is 3x + 2y = k. It passes through (1, 3).
3 + 6 = k ⇒ k = 9, 3x + 2y = 9
To find y-intercept x = 0, 2y = 9, y = 9/2

Question 19.
If the two straight lines x + (2k – 7)y + 3 = 0 and 3kx + 9y – 5 = 0 are perpendicular then the value of k is ……

Solution:
(a) k = 3
Hint.

Since the lines are perpendicular m₁m₂ = – 1

Question 20.
If a vertex of a square is at the origin and it’s one side lies along the line 4x + 3y – 20 = 0, then the area of the square is ……..
(a) 20 sq. units
(b) 16 sq. units
(c) 25 sq. units
(d) 4 sq. units
Solution:
(b) 16 sq. units
Hint:
One side of a square = Length of the perpendicular from (0, 0) to the line.

Question 21.
If the lines represented by the equation 6x² + 41xy – 7y² = 0 make angles α and β with the x-axis, then tan α tan β =

Solution:
(a) \(-\frac{6}{7}\)
Hint.
6x² + 41xy – 7y² = 0
⇒ 6x² – xy + 42xy – 7y² = 0
⇒ x (6x – y) + 7y (6x – y) = 0
⇒ (x + 7y) (6x – y) = 0
⇒ x + 7y = 0, 6x – y = 0

Question 22.
The area of the triangle formed by the lines x² – 4y² = 0 and x = a is …….

Solution:
(c) \(\frac{1}{2} a^{2}\)
Hint:
x² – 4y² = 0 , (x – 2y) (x + 2y) = 0 ⇒ x – 2y = 0, x + 2y = 0

Question 23.
If one of the lines given by 6x² – x + 4x² = 0 is 3x + 4y = 0, then c equals to ……
(a) -3
(b) -1
(c) 3
(d) 1
Solution:
(a) -3
Hint.
6x² – xy + 4cy² = 0, 3x + 4y = 0
The other line may be (2x + by)
(3x + 4y) (2x + by) = 6x² – xy + 4cy²
6x² + 3xby + 8xy + 4by² = 6x² – xy + 4cy²
6x² + xy (3b + 8) + 4by² = 6x² – xy + 4cy²
paring, 3b + 8 = -1
3b = -9 ⇒ b = -3
4b = 4c ⇒ 4(-3) = 4c
-12 = 4c ⇒ c = -3

Question 24.


Solution:
(c) \(\frac{5}{9}\)
Hint:

Question 25.
The equation of one the line represented by the equation x² + 2xy cot θ – y² = 0 is ………
(a) x – y cotθ = 0
(b) x + y tan θ = 0
(e) x cos θ + y(sin θ + 1) = 0
(d) x sin θ + y(cos θ + 1) = 0
Solution:
(d) x sin θ + y(cos θ + 1)=0

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 9 Limits and Continuity Ex 9.1

In problems 1-6, complete the table using calculate and use the result to estimate the limit.

Question 1.

Solution:

Question 2.

Solution:

∴ Limit is 0.25

Question 3.

Solution:

∴ Limit is 0.288

Question 4.

Solution:

∴ Limit is -0.25

Question 5.

Solution:

∴ Limit is 1

Question 6.

Solution:

∴ Limit is 0

In exercise problems 7-15, use the graph to find the limits (if it exists). If the limit does not exist, explain why?

Question 7.
\(\lim _{x \rightarrow 3}\)(4 – x)
Solution:

Limit exists and is equal to 1

Question 8.
\(\lim _{x \rightarrow 1}\)(x² + 2)
Solution:

Limit exists and is equal to = 3

Question 9.

Solution:

Question 10.

Solution:

Question 11.

Solution:

The limit does not exist

Question 12.

Solution:
When x → 5, (x – 5) = -(x – 5)
∴ \(\lim _{x \rightarrow 5^{-}} \frac{-(x-5)}{(x-5)}\) = -1
When x → 5⁺, (x – 5) = (x – 5)

Question 13.
\(\lim _{x \rightarrow 1}\) sin(πx)
Solution:

Question 14.
\(\lim _{x \rightarrow 0}\) (sec x)
Solution:

Question 15.
\(\lim _{x \rightarrow \frac{\pi}{2}}\) tan x
Solution:

The limit does not exist

Sketch the graph of f, then identify the values of x₀ for which \(\lim _{x \rightarrow x_{0}}\) f(x) exists.

Question 16.

Solution:

Question 17.

Solution:

Limit exists except at x₀ = π

Question 18.
Sketch the graph of a function f that satisfies the given values:

Solution:

Question 19.
Write a brief description of the meaning of the notation \(\lim _{x \rightarrow 8}\) f(x) = 25
Solution:

Question 20.
If f(2) = 4, can you conclude anything about the limit of f(x) as x approaches 2?
Solution:
Given f(2) = 4
Here at x = 2 the value of the function is given.
Therefore, we cannot conclude anything about the limit of f(x) as x approaches 2.

Question 21.
If the limit of f(x) as z approaches 2 is 4, can you conclude anything about f(2)?
Explain reasoning.
Solution:
Given \(\lim _{x \rightarrow 2}\) f(x) = 4
Since the limit of the function need not be equal to the value of the function, we cannot conclude anything about f(2).

Question 22.
Evaluate: \(\lim _{x \rightarrow 3} \frac{x^{2}-9}{x-3}\) if it exists by finding f(3^–) and f(3⁺).
Solution:

Question 23.
Verify the existence of
Solution:

The limit does not exist

Samacheer Kalvi 11th Maths Solutions Chapter 9 Limits and Continuity Ex 9.1 Additional Questions

Question 1.
Suppose . What are the possible values of a and b?
Solution:

Question 2.

Solution:
We have,

Question 3.

Solution:
We have,

Question 4.

Solution:

Question 5.
Let a₁, a₂ …………… a_n be fixed real numbers such that f(x) = (x – a₁) , (x – a₂), ………. (x – a_n) what \(\lim _{x \rightarrow a}\) f(x) For a ≠ a₁, a₂, ………… a_n compute \(\lim _{x \rightarrow a}\) f(x).
Solution:
We have,

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.1

Question 1.

Solution:


Similarly (a + b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴
∴ (a – b)⁴ + (a + b)⁴ = 2 [a⁴ + 6a²b² + b⁴]
Substituting the value of a and b we get

= 2[16x⁸ + 6(4x⁴)(9(1 – x²)) + 81(1 – x²)²]
= 2[16x⁸ + 216x⁴(1 – x²) + 81(1 – x²)²]
= 2[16x⁸ + 216x⁴ – 216x⁶ + 81 + 81x⁴ – 162x²]
= 2[16x⁸ – 216x⁶ + 297x⁴ – 162x² + 81]
= 32x⁸ – 432x⁶ + 594x⁴ – 324x² + 162

Question 2.
Compute
(i) 102⁴
(ii) 99⁴
(iii) 9⁷
Solution:

(i) 102⁴ = (100 + 2)⁴ = (10² + 2)⁴

= 1(10⁸) + 4(10⁶)(2) + 6(10⁴)(4) + 4(10²)(8) + 16
= 100000000 + 8000000 + 240000 + 3200 + 16
= 108243216

(ii) 994 = (100 – 1)⁴ = (10² – 1)⁴

= 1(10⁸) + 4(10⁶)(-1) + 6 (10⁴)(1) + 4( 10⁴)(-1) + (-1)⁴
= 100000000 – 4000000 + 60000 – 400 + 1
= 100060001 – 4000400 = 96059601

(iii) 9⁷ = (10 – 1)⁷

= 1(10000000) + 7(1000000)(-1) + 21(100000)(1) + 35(10000)(-1) + 35(1000)(1) + 21(100)(-1) + 7(10)(1) + 1(-1)
= 10000000 – 7000000 + 2100000 – 350000 + 35000 – 2100 + 70 – 1
= 12135070 – 7352101 = 4782969

Question 3.
Using binomial theorem, indicate which of the following two number is larger: (1.01)^1000000, 10000.
Solution:
(1.01)^1000000 = (1 + 0.01)^1000000

which is > 10000
So (1.01)^1000000 > 10000 (i.e.) (1.01)^1000000 is larger

Question 4.

Solution:

To find a coefficient of x¹⁵ we have to equate x power to 15
i.e. 20 – 5r = 15
20 – 15 = 5r ⇒ 5r = 5 ⇒ r = 5/5 = 1
So the coefficient of x¹⁵ is ¹⁰C₁ = 10

Question 5.

Solution:

To find coefficient of x⁶
12 – 5r = 6
12 – 6 = 5r ⇒ 5r = 6 ⇒ r = 6/5 which is not an integer.
∴ There is no term involving x⁶.
To find coefficient of x²
12 – 5r = 2
5r = 12 – 2 = 10 ⇒ r = 2

Question 6.

Solution:

when multiplying these terms, we get x⁴ terms

∴ The co-eff of x⁴ is 26325

Question 7.

Solution:

Question 8.
Find the last two digits of the number 3⁶⁰⁰.
Solution:
3⁶⁰⁰ = 3^{2 × 300} = (9)300 = (10 – 1)³⁰⁰

All the terms except last term are ÷ by 100. So the last two digits will be 01.

Question 9.
If n is a positive integer, show that, 9^{n + 1} – 8n – 9 is always divisible by 64.
Solution:

∴ 9^{n + 1} – 8n – 9 = 64 [an integer]
⇒ 9^{n + 1} – 8n – 9 is divisible by 64

Question 10.
If n is an odd positive integer, prove that the coefficients of the middle terms in the expansion of (x + y)ⁿ are equal.
Solution:
Given n is odd. So let n = 2n + 1, where n is an integer.
The expansion (x + y)ⁿ has n + 1 terms.
= 2n + 1 + 1 = 2(n + 1) terms which is an even number.

⇒ The coefficient of the middle terms in (x + y)ⁿ are equal.

Question 11.
If n is a positive integer and r is a non – negative integer, prove that the coefficients of x^r and x^{n – r} in the expansion of (1 + x)ⁿ are equal.
Solution:

Question 12.
If a and b are distinct Integers, prove that a – b is a factor of aⁿ – bⁿ, whenever n is a positive integer. [Hint: write aⁿ = (a – b + b)ⁿ and expand]
Solution:

= (a – b)[aⁿ integer]
⇒ aⁿ – bⁿ is divisible by (a – b)

Question 13.
In the binomial expansion of (a + b)aⁿ, the coefficients of the 4^th and 13^th terms are equal to each other, find n.
Solution:
In the expansion of (a + b)ⁿ,
The general term is T_r+1 = nC_r . a^n-r . b^r ……….. (1)
To find the coefficient of 4th term, Put r = 3 in equation (1)
∴ T₃₊₁ = nC₃ a^n-3 . b³
To find the coefficient of 13th term , Put r = 12 in equation (1)
∴ T₁₂₊₁ = nC₁₂ a^n-12 . b¹²
Given nC₃ = nC₁₂
nC_x = nC_y ⇒ x = y or x + y = n
∴ 3 + 12 = n ⇒ n = 15

Question 14.
If the binomial coefficients of three consecutive terms in the expansion of (a + x)ⁿ are in the ratio 1 : 7 : 42, then find n.
Solution:
In (a + x)ⁿ general term is t_{r + 1} = ⁿC_r
So, the coefficient of t_{r + 1} is ⁿC_r
We are given that the coefficients of three consecutive terms are in the ratio 1 : 7 : 42.

Question 15.
In the binomial coefficients of (1 + x)ⁿ, the coefficients of the 5^th, 6^th and 7 terms are in AP. Find all values of n.
Solution:

⇒ (n – 1)(n – 14) = 0
∴ n = 7 , 14

Question 16.

Solution:

Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.1 Additional Questions Solved

Question 1.

Solution:
Let T_{r + 1} be the term in which x³² and x^-17 occurs,

(i) Since x³² occurs in this term
∴ Exponent of x = 32
⇒ 60 – 7r = 32 ⇒ 7r = 28
r = 28 ÷ 7 = 4
∴ Coefficient of the term containing x³² = ¹⁵C₄ (-1)⁴ = 1365

(ii) Since x^-17 occurs in this term .
∴ Exponent of x = -17
⇒ 60 – 7r = -17 ⇒ 7r = 77, ∴ r = 11

Question 2.
Find a positive value of m for which the coefficient of x² in the expansion of (1 + x)^m is 6.
Solution:

Also, coefficient of x² in the expansion of (1 + x)^m is 6

⇒ m (m – 1) = 4.3 ⇒ m = 4

Question 3.
In the binomial expansion of (1 + a)^{m + n}, prove that the coefficients of a^m and aⁿ are equal.
Solution:

Question 14.
The coefficient of (r – 1)^th, r^th, and (r + 1)^th terms in the expansion of (x + 1)ⁿ are in the ratio 1 : 3 : 5. Find both n and r.
Solution:
We know that co-effcients of (r – 1)^th, r^th and (r + 1)^th terms in the expansion of (x + 1)ⁿ are ⁿC_{r – 2}: ⁿC_{r – 1} and ⁿC_r respectively

⇒ 3n – 8r + 3 = 0 and n – 4r + 5 = 0
Solving these for n, r we get,
n = 7 and r = 3

Question 5.

Solution:

Using the binomial theorem,

Question 6.
Show that the coefficient of the middle term in the expansion of (1 + x)²ⁿ is equal to the sum of the coefficients of the two middle terms in the expansion of (1 + x)^{2n – 1}.
Solution:
In the expansion of (1 + x)²ⁿ,
Number of terms = 2n + 1, which is odd

Question 7.
If three consecutive coefficients in the expansion of (1 + x)ⁿ are in the ratio 6 : 33 : 110, find n.
Solution:
Let the consecutive coefficients ⁿC_r, ⁿC_{r + 1} and ⁿC_{r + 2} be the coefficients of T_{r + 1}, T_{r + 2} and T_{r + 3} then ⁿC_r : ⁿC_{r + 1} : ⁿC_{r + 2} = 6 : 33 : 110

Question 8.
If the sum of the coefficients in the expansion of (x + y)ⁿ is 4096. Then find the greatest coefficient in the expansion.
Solution:
Given that, Sum of the coefficients in the expansion of (x + y)ⁿ = 4096
∴ ⁿC₀ + ⁿC₁ + ⁿC₂+…+ ⁿC_n = 4096
[∴ Sum of binomial coefficients in the expansion of (x + a)ⁿ is 2ⁿ]
⇒ 2ⁿ = 4096 = 2¹²

Question 9.

Solution:
The general term in the expansion of

Question 10.

Solution:

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 9 Limits and Continuity Ex 9.3

Question 1.
(a) Find the left right limits of f(x) = \(\frac{x^{2}-4}{\left(x^{2}+4 x+4\right)(x+3)}\) at x = -2
(b) f(x) = tan x at x = \(\frac{\pi}{2}\)
Solution:
(a) f(x) = \(\frac{x^{2}-4}{\left(x^{2}+4 x+4\right)(x+3)}\) at x → -2

Evaluate the following limits
Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

f(x) → ∞ as x → ∞

Question 6.

Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.
An important problem in fishery science is to estimate the number of fish presently spawning in streams and use this information to predict the number of mature fish or “recruits” that will return to the rivers during the reproductive period. If S is the number of spawners and R the number of recruits, “Beverton-Holt spawner recruit function” is R(S) = \(\frac{S}{(\alpha S+\beta)}\) where α and β are positive constants. Show that this function predicts approximately constant recruitment when the number of spawners is sufficiently large.
Solution:

Question 10.
A tank contains 5000 litres of pure water. Brine (very salty water) that contains 30 grams of salt per litre of water is pumped into the tank at a rate of 25 litres per minute. The concentration of saltwater after t minutes (in grams per litre) is C(t) = \(\frac{30 t}{200+t}\). What happens to the concentration as t → ∞?
Solution:

Samacheer Kalvi 11th Maths Solutions Chapter 9 Limits and Continuity Ex 9.3 Additional Problems

Question 1.
Evaluate
Solution:
The given expression is of the form ∞ – ∞. So we first write it in the rational form \(\frac{f(x)}{g(x)}\). So that it reduces to either \(\frac{0}{0}\) form or \(\frac{\infty}{\infty}\) form.

Question 2.
Evaluate:
Solution:
Here the expression assumes the form ∞ – ∞ as x → ∞. So, we first reduce it to the rational form \(\frac{f(x)}{g(x)}\)

Question 3.
Evaluate
Solution:
We have

Question 4.
Evaluate:
Solution:
We have,

Question 5.
Evaluate:
Solution:

Question 6.
Evaluate:
Solution:

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 9 Limits and Continuity Ex 9.2

Evaluate the following limits:
Question 1.

Solution:

Question 2.
m and n are integers.
Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.

Solution:

Question 10.

Solution:

= \(\frac{1}{4}-\frac{1}{2}\)
= \(\frac{-1}{4}\)

Question 11.

Solution:

Question 12.

Solution:

Question 13.

Solution:

∴ Limit does not exist

Question 14.

Solution:

Question 15.

Solution:

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.3

Solve the following problems by using factor theorem

Question 1.
Show that < = (x – a)² (x + 2a)
Solution:

⇒ (x + 2d) is a factor of A.
Now degree of Δ is 3 (x × x × x = x³) and we have 3 factors for A
∴ There can be a constant as a factor for A.
(i.e.,) Δ = k(x – a)² (x + 2d)
equating coefficient of x³ on either sides we get k = 1

∴ Δ = (x – a)² (x + 2a)

Question 2.
Show that
Solution:

Similarly b and c are factors of Δ.
The product of the leading diagonal elements is (b + c) (c + a) (a + b)
The degree is 3. And we got 3 factors for Δ ∴ m = 3 – 3 = 0
∴ there can be a constant k as a factor for Δ.
>

Question 3.

Solution:

⇒ x = 0, 0 are roots.
Now the degree of the leading diagonal elements is 3.
∴ the equation is of degree 3, so the roots are 0, 0, – (a + b + c)

Question 4.
Show that = (a + b + c) (a – b) (b – c) (c – a)
Solution:

⇒ (a – b) is a factor of Δ.
Similarly, (b – c) and (c – a) are factors of Δ.
The degree of the product of elements along the leading diagonal is 1 + 1 + 2 = 4 and we got 3 factors for Δ.
m = 4 – 3 = 1
∴ There can be one more factor symmetric with a, b, c which is of the form k (a + b + c).

Question 5.
Solve
Solution:

Question 6.
Show that = (x – y) (y – z) (z – x)
Solution:

⇒ (x – y) is a factor of Δ.
Similarly (y – z) and (z – x) are factors of Δ.
Now degree of Δ = 0 + 1 + 2 = 3 and we have 3 factors of Δ.
and so there can be a constant k as a factor of Δ.

Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.3 Additional Problems

Question 1.
Prove that = (a – b) (b – c) (c – a) (a + b + c).
Solution:

∴ (a – b) is a factor of Δ.
Similarly, we observe that Δ is symmetric in a, b, c by putting b = c, c = a, we get Δ = 0.
Hence (b – c) and (c – a) are also factors of Δ.
∴ The product (a – b) (b – c) (c – a) is a factor of Δ. The degree of this product is 3.
The product of leading diagonal elements is 1. bc³. The degree of this product is 4.
∴ By cyclic and symmetric properties, the remaining symmetric factor of the first degree must be k (a + b + c), where k is any non-zero constant.

Question 2.
Using factor method show that = (a – b) (b – c) (c – a)
Solution:

⇒ (a – b) is a factor of Δ.
similarly, (b – c) and (c – a) are factors of Δ.
The product of leading diagonal elements is bc². The degree of the product is 1 + 2 = 3.
∴ there will be three factors for Δ.
We got 3 factors for Δ as (a – b), (b – c) and (c – a). Its degree = 3.
∴ m = 3 – 3 = 0
∴ there can be a constant k as a factor of Δ.

Question 3.

Solution:

⇒ (a – b) is a factor of A.
Similarly, (b – c) and (c – a) are factors of Δ.
The degree of Δ = 5 and degree of product of factors = 3.

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3

Question 1.
If ⁿC₁₂ = ⁿC₉ find ²¹C_n.
Solution:
ⁿC_x = ⁿC_y ⇒ x = y or x + y = n
Here ⁿC₁₂ = ⁿC₉ ⇒ 12 ≠ 9 so 12 + 9 = n (i.e) n = 21

Question 2.
If ¹⁵C_{2r – 1} = ¹⁵C_{2r + 4}, find r.
Solution:

Question 3.
If ⁿP_r = 720 and ⁿC_r = 120, find n, r.
Solution:

Question 4.
Prove that ¹⁵C₃ + 2 × ¹⁵C₄ + ¹⁵C₅ = ¹⁷C₅
Solution:

Question 5.

Solution:

Question 6.
If ^{(n + 1)}C₈ : ^{(n – 3)}P₄ = 57 : 16, find the value of n.
Solution:

Question 7.

Solution:

Question 8.
Prove that if 1 ≤ r ≤ n then n × ^{(n – 1)}C_{r – 1} = (n – r + 1)C_{r – 1}.
Solution:


(1) = (2) ⇒ LHS = RHS

Question 9.
(i) A Kabaddi coach has 14 players ready to play. How many different teams of 7 players could the coach put on the court?
Solution:
No. of players in the team = 14
We need 7 players
So selecting 7 from 14 players can be done is ¹⁴C₇ = 3432 ways

(ii) There are 15 persons in a party and if, each 2 of them shakes hands with each other, how many handshakes happen in the party?
Solution:
Total No. of persons = 15
Every two persons shake hands

(iii) How many chords can be drawn through 20 points on a circle?
Solution:
A chord is a line join of 2 points
No. of points given = 20
Selecting 2 from 20 can be done in ²⁰C₂ ways

(iv) In a parking lot one hundred, one-year-old cars are parked. Out of the five are to be chosen at random to check its pollution devices. How many different sets of five cars are possible?
Solution:
Number of cars =100
Select 5 from 100 cars can be done in ¹⁰⁰C₅ ways

(v) How many ways can a team of 3 boys, 2 girls and 1 transgender be selected from 5 boys, 4 girls and 2 transgenders?
Solution:
We have 5 boys, 4 girls, and 2 transgenders. We need 3 boys, 2 girls and 1 transgender The selection can be done as follows Selecting 3 boys from 5 boys can be done in ⁵C₃ ways

Selecting 2 girls from 4 girls can be done in ⁴C₂ ways

Selecting 1 transgender from 2 can be done in ²C₁ = 2 ways
∴ Selecting 3 boys, 2 girls and 1 transgender can be done in 10 × 6 × 2 = 120 ways

Question 10.
Find the total number of subsets of a set with
(i) 4 elements
(ii) 5 elements
(iii) n elements

Solution:
If a set has n elements then the number of its subsets = 2ⁿ

(i) Here n = 4
So number of subsets = 2⁴ = 16

(ii) n = 5
So number of subsets = 2⁵ = 32

(iii) n = n
So number of subsets = 2

Question 11.
A trust has 25 members.

(i) How many ways 3 officers can be selected?
Solution:
Selecting 3 from 25 can be done in ²⁵C₃ ways

(ii) In how many ways can a President, Vice President, and secretary be selected?
Solution:
The number of ways of selecting a president from 25 members = 25C₁ = 25
After the selection of the president, the remaining number of members in the trust is 24
The number of ways of selecting a vice president
from the remaining 24 members of the trust is = 24C₁ 24
After the selection of the president and vice president, the number of remaining members in the trust = 23
The number of ways of selecting a secretary from the remaining 23 members of the trust is = 23 C₁ = 23
∴ Total number of ways of selection = 25 × 24 × 23 = 13800

Question 12.
How many ways a committee of six persons from 10 persons can be chosen along with a chairperson and a secretary?
Solution:
Selecting a chairperson from the 10 persons can be done in 10 ways
After the selection of chairperson, only 9 persons are left out so selecting a secretary (from the remaining persons) can be done in 9 ways.
The remaining persons = 8
Totally we need to select 6 persons
We have selected 2 persons.
So we have to select 4 persons
Selecting 4 from 8 can be done in ⁸C₄ ways

Question 13.
How many different selections of 5 books can be made from 12 different books if,
Solution:
No. of books given = 12
No. of books to be selected = 5

(i) Two particular books are always selected?
Solution:
So we need to select 3 more books from (12 – 2) 10 books which can be done in ¹⁰C₃ ways

(ii) Two particular books are never selected?
Solution:
Two particular books never to be selected.
So only 10 books are there and we have to select 5 books which can be done in ¹⁰C₅ ways

Question 14.
There are 5 teachers and 20 students. Out of them, a committee of 2 teachers and 3 students is to be formed. Find the number of ways in which this can be done. Further, find in how many of these committees
(i) a particular teacher is included?
(ii) a particular student is excluded?
Solution:
No. of teachers = 5
No of students = 20
We need to select 2 teachers and 3 students

(i) A particular teacher should be included. So from the remaining 4 teachers, one teacher is to be selected which can be done in ⁴C₁ = 4 ways

So selecting 2 teachers and 3 students can be done in 4 × 1140 = 4560 ways

(ii) the particular student should be excluded.
So we have to select 3 students from 19 students which can be done in ¹⁹C₃ ways

∴ 2 teachers and 3 students can be selected in 969 × 10 = 9690 ways

Question 15.
In an examination, a student has to answer 5 questions, out of 9 questions in which 2 are compulsory. In how many ways students can answer the questions?
Solution:
No. of questions given = 9
No. of questions to be answered = 5
But 2 questions are compulsory
So the student has to answer the remaining 3 questions (5 – 2 = 3) from the remaining 7 (9 – 2 = 7) questions which can be done in ⁷C₃ ways

Question 16.
Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly three aces in each combination.
Solution:
No. of cards = 52
In that number of aces = 4
No. of cards needed = 5
In that 5 cards number of aces needed = 3
So the 3 aces can be selected from 4 aces in ⁴C₃ = ⁴C₁ = 4 ways
So the remaining = 5 – 3 = 2
These 2 cards can be selected in ⁴⁸C₂ ways

Question 17.
Find the number of ways of forming a committee of 5 members out of 7 Indians and 5 Americans, so that always Indians will be the majority’ in the committee.
Solution:
We need a majority of Indian’s which is obtained as follows.

The possible ways are (5I) or (4I and 1A) or (3I and 2A)

Question 18.
A committee of 7 peoples has to be formed from 8 men and 4 women. In how many ways can this be done when the committee consists of
(i) exactly 3 women?
(ii) at least 3 women?
(iii) at most 3 women?
Solution:

We need a committee of 7 people with 3 women and 4 men.
This can be done in (⁴C₃) (⁸C₄) ways

The number of ways = (70) (4) = 280

(ii) Atleast 3 women

So the possible ways are (3W and 4M) or (4W and 3M)

The number of ways (4) (70) + (1) (56) = 280 + 56 = 336

(iii) Atmost 3 women

Question 19.
7 relatives of a man comprises 4 ladies and 3 gentlemen, his wife also has 7 relatives; 3 of them are ladies and 4 gentlemen. In how many ways can they invite a dinner party of 3 ladies and 3 gentlemen so that there are 3 of man’s relative and 3 of the wife’s relatives?
Solution:

We need 3 ladies and 3 gentlemen for the party which consist of 3 Husbands relative and 3 wife’s relative.
This can be done as follows

Question 20.
A box contains two white balls, three black balls, and four red balls. In how many ways can three balls be drawn from the box, if at least one black ball is to be included in the draw?
Solution:
The box contains 2 white, 3 black, and 4 red balls
We have to draw 3 balls in which there should be at least 1 black ball
The possible draws are as follows
Black balls = 3
Red and White = 2 + 4 = 6

Question 21.
Find the number of strings of 4 letters that can be formed with the letters of the word EXAMINATION.
Solution:
EXAMINATION
(i.e.) A, I, N are repeated twice. So the number of distinct letters = 8
From the 8 letters, we have to select and arrange 4 letters to form a 4 letter word which can
be done in ⁸P₄ = 8 × 7 × 6 × 5 = 1680
From the letters A, A, I, I, N, N when any 2 letters are taken as AA, II or AA, NN or II, NN

From AA, II, NN we select one of them and from the remaining we select and arrange 3 which can be done in ways

Total number of 4 letter word = 1680 + 18 + 756 = 2454

Question 22.
How many triangles can be formed by joining 15 points on the plane, in which no line joining any three points?
Solution:
No. of non-collinear points = 15
To draw a Triangle we need 3 points
∴ Selecting 3 from 15 points can be done in ¹⁵C₃ ways.

Question 23.
How many triangles can be formed by 15 points, in which 7 of them lie on one line and the remaining 8 on another parallel line?
Solution:

7 points lie on one line and the other 8 points parallel on another paraller line.
A triangle is obtained by taking one point from one line and second points from the other parallel line which can be done as follows.

∴ Number of triangles = (7) (28) + (21) (8) = 196 + 168 = 364

Question 24.
There are 11 points in a plane. No three of these lie in the same straight line except 4 points, which are collinear. Find,
(i) The number of straight lines that can be obtained from the pairs of these points?
Solution:
4 points are collinear

Total number of points 11.
To get a line we need 2 points

But in that 4 points are collinear

From (1) Joining the 4 points we get 1 line
∴ Number of lines = ¹¹C₂ – ⁴C₂ + 1 = 55 – 6 + 1 = 50

(ii) The number of triangles that can be formed for which the points as their vertices?
A triangle is obtained by joining 3 points.
So selecting 3 from 11 points can be

But of the 11 points, 4 points are collinear. So we have to subtract ⁴C₃ = ⁴C₁ = 4
∴ Number of triangles = 165 – 4 = 161

Question 25.
A polygon has 90 diagonals. Find the number of its sides?
Solution:

∴ n = 15

Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.3 Additional Questions Solved

Question 1.
A group consists of 4 girls and 7 boys. In bow many ways can a team of 5 members be selected, if the team has
(i) no girls
(ii) atleast one boy and one girl.
(iii) at least three girls
Solution:
We have 4 girls and 7 boys and a team of 5 members is to be selected.

(i) If no girl is selected, then all the 5 members are to be selected out of 7 boys

(ii) When at least one boy and one girl are to be selected, then


Hence the required number of ways are (i) 21 ways (ii) 441 ways (iii) 91 ways

Question 2.
A committee of 6 is to be chosen from 10 men and 7 women so as to contain atleast 3 men and 2 women. In how many different ways can this be done it two particular women refuse to serve on the same committee?
Solution:
We have 10 men and 7 women out of which a committee of 6 is to be formed which contain atleast 3 men and 2 women

∴ Total number of committee = 8610 – 810 = 7800
Hence, the value of the filler is 7800

Question 3.
Using the digits 1, 2, 3,4, 5, 6, 7 a number of 4 different digits is formed. Find

Solution:
(a) Total of 4 digit number formed with 1, 2, 3, 4, 5, 6, 7

(b) When a number is divisible by 2 = 4 × 5 × 6 × 3 = 360
(c) Total numbers which are divisible by 25 = 40
(d) Total numbers which are divisible by 4 (last two digits is divisble by 4) = 200
Hence, the required matching is (a) ⟷ (z), (b) ⟷ (iii), (c) ⟷ (iv), (d) ⟷ (ii)

Question 4.
If ²²P_{r + 1} : ²⁰P_{r + 2} = 11 : 52, find r.
Solution:

Question 5.

Solution:

Question 6.
A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of:
(i) exactly 3 girls?
(ii) at least 3 girls?
(iii) almost 3 girls?
Solution:

(ii) We have to select at least 3 girls. So the committee consists of 3 girls and 4 boys or 4 girls and 3 boys.
∴ Number of ways of selection = ⁴C₃ × ⁹C₄ + ⁴C₄ × ⁹C₃

(iii) We have to select at most 3 girls. So the committee consists of no girl and 7 boys or 1 girl and 6 boys or 2 girls and 5 boys or 3 girls and 4 boys.

Question 7.
Determine n if
(i) ²ⁿC₃ : ⁿC₂ = 12 : 1
(ii) ²ⁿC₃ : ⁿC₃ = 11 : 1
Solution:

Question 8.
Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour.
Solution:
There are 6 red balls, 5 white balls and 5 blue balls.
We have to select 3 balls of each colour.

Question 9.

Solution:

Question 10.
If ⁿC₄, ⁿC₅ and ⁿC₄ are in A.P. then find n.
[Hint: ²ⁿC₅ = ⁿC₆ + ⁿC₄]
Solution:

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.1

Question 1.
Find the locus of P, if for all values of a, the co-ordinates of a moving point P is
(i) (9 cos α, 9 sin α)
(ii) (9 cos α, 6 sin α)
Solution:
(i) Let P(h, k) be the moving point.
We are given h = 9 cos α and k = 9 sin α and

∴ locus of the point is x² + y² = 81

(ii) Let P(h , k) be a moving point.
We are given h = 9 cos α and k = 6 sin α

Question 2.
Find the locus of a point P that moves at a constant distance of
(i) Two units from the x-axis
(ii) Three units from the y-axis.
Solution:
(i) Let the point (x, y) be the moving point.
The equation of a line at a distance of 2 units from the x-axis is k = 2
So the locus is y = 2 (i.e.) y – 2 = 0

(ii) Equation of a line at a distance of 3 units from y-axis is h = 3
So the locus is x = 3 (i.e.) x – 3 = 0

Question 3.
If θ is a parameter, find the equation of the locus of a moving point, whose coordinates are x = a cos³ θ, y = a sin³ θ
Solution:

Question 4.
Find the value of k and b, if the points P (-3, 1) and Q (2, b) lie on the locus of x² – 5x + ky = 0.
Solution:
Given P (-3, 1) lies on the locus of
x² – 5x + ky = 0
∴ (- 3)² – 5 (-3) + k(1) = 0
9 + 15 + k = 0
⇒ k = -24
Also given Q(2 , b) lies on the locus of
x² – 5x + ky = 0
x² – 5x – 24y = 0
∴ (2)² – 5(2) – 24(b) = 0
4 – 10 – 24b = 0 ⇒ – 6 – 24b = 0
⇒ 24b = -6 ⇒ b = \(-\frac{6}{24}\) = \(-\frac{1}{4}\)
Thus k = -24, b = \(-\frac{1}{4}\)

Question 5.
A straight rod of length 8 units slides with its ends A and B always on the x and y-axis respectively. Find the locus of the midpoint of the line segment AB.
Solution:
Let P (h, k) be the moving point A (a, 0) and B (0, b) P is the midpoint of AB.

Question 6.
Find the equation of the locus of a point such that the sum of the squares of the distance from the points (3, 5), (1, -1) is equal to 20.
Solution:
Let the given points be A (3, 5) and (1, -1).
Let P (h, k ) be the point such that
PA² + PB² = 20 ………….. (1)
PA² = (3 – h)² + (5 – k)²
PB² = (1 – h)² + (- 1 – k)²
(1) ⇒ (3 – h)² + (5 – k)² + (1 – h)² + (1 + k)² = 20
9 – 6h + h² + 25 – 10k + k² + 1 – 2h + h² + 1 + 2k + k² = 20
2h² + 2k² – 8h – 8k + 36 = 20
2h² + 2k² – 8h – 8k + 16 = 0
h² + k² – 4h – 4k + 8 = 0
The locus of P ( h , k ) is obtained by replacing h by x and k by y
∴ The required locus is x² + y² – 4x – 4y + 8 = 0

Question 7.
Find the equation of the locus of the point P such that the line segment AB, joining the points A (1, -6) and B (4, -2), subtends a right angle at P.
Solution:
Let P (h, k) be the moving point

Given A (1, – 6) and B (4, – 2),
Since ∆APB = 90°, PA² + PB² = AB²
(i.e.) (h – 1)² + (k + 6)² + (h – 4)² + (k + 2)² = (4 – 1)² + (-2 + 6)²
(i.e) h² + 1 – 2h + k² + 36 + 12k + h² + 16 – 8h + k² + 4 + 4k = 3² + 4² = 25
2h² + 2k² -10h + 16k + 57 – 25 = 0
2h² + 2k² – 10h + 16k + 32 = 0
(÷ by 2)h² + k² – 5h + 8k + 16 = 0
So the locus of P is x² + y² – 5x + 8y + 16 = 0

Question 8.
If O is origin and R is a variable point on y² = 4x, then find the equation of the locus of the mid-point of the line segment OR.
Solution:
Let P(h, k) be the moving point
We are given O (0, 0). Let R = (a, b)

Substituting a, b values is y² = 4x
we get (2k)² = 4 (2h)
(i.e) 4k² = 8h
(÷ by 4) k² = 2h
So the locus of P is y² = 2x

Question 9.

Solution:

Question 10.
If P (2, -7) is a given point and Q is a point on 2x² + 9y² = 18, then find the equations of the locus of the mid-point of PQ.
Solution:
P = (2, -7); Let (h, k) be the moving point Q = (a, b)

⇒ a = 2h – 2,
b = 2k + l
Q is a point on 2x² + 9y² = 18 (i.e) (a, b) is on 2x² + 9y² = 18
⇒ 2(2h – 2)² + 9 (2k + 7)² = 18
(i.e) 2 [4h² + 4 – 8h] + 9 [4k² + 49 + 28k] – 18 = 0
(i.e) 8h² + 8 – 16h + 36k² + 441 + 252k – 18 = 0
8h² + 36k² – 16h + 252k + 431 = 0
The locus is 8x² + 36y² – 16x + 252y + 431 = 0

Question 11.
If R is any point on the x-axis and Q is any point on the y-axis and Pis a variable point on RQ with RP = b, PQ = a. then find the equation of locus of P.
Solution:
P = (x, 0), Q = (0, y), R (h, k) be a point on RQ such that PR : RQ = b : a

From the right-angled triangle OQR, OR² + OQ² = QR²

Question 12.
If the points P (6, 2) and Q (-2, 1) and R are the vertices of a ∆PQR and R is the point on the locus y = x² – 3x + 4, then find the equation of the locus of the centroid of ∆PQR.
Solution:
P (6, 2), Q (-2, 1). Let R = (a, b) be a point on y = x² – 3x + 4.

But (a, b) is a point on y = x² – 3x + 4
b = a² – 3a + 4
(i.e) 3k – 3 = (3h – 4)² – 3(3h – 4) + 4
(i.e) 3k – 3 = 9h² + 16 – 24h – 9h + 12 + 4
⇒ 9h² – 24h – 9h + 32 – 3k + 3 = 0
(i.e) 9h² – 33h – 3k + 35 = 0,
Locus of (h, k) is 9x² – 33x – 3y + 35 = 0

Question 13.
If Q is a point on the locus of x² + y² + 4x – 3y + 7 = 0 then find the equation of locus of P which divides segment OQ externally in the ratio 3 : 4, where O is origin.
Solution:
Let (h, k) be the moving point O = (0, 0);
Let PQ = (a, b) on x² + y² + 4x – 3y + 7 = 0

Question 14.
Find the points on the locus of points that are 3 units from the x-axis and 5 units from the point (5, 1).
Solution:
A line parallel to the x-axis is of the form y = k.
Here k = 3 ⇒ y = 3
A point on this line is taken as P (a, 3).
The distance of P (a, 3) from (5, 1) is given as 5 units
⇒ (a – 5)² + (3 – 1)² = 5²
a² + 25 – 10a + 9 + 1 – 6 = 25
a² – 10a + 25 + 4 – 25 = 0
a² – 10a + 4 = 0

Question 15.
The sum of the distance of a moving point from the points (4, 0) and (-4, 0) is always 10 units. Find the equation of the locus of the moving point.
Solution:
Let P (h, k) be a moving point
Here A = (4, 0) and B = (-4, 0)
Given PA + PB = 10

Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.1 Additional Questions

Question 1.
If the sum of the distance of a moving point in a plane from the axis is 1, then find the locus of the point.
Solution:
Let coordinates of a moving point P be (x, y).
Given that the sum of the distances from the axis to the point is always 1.

∴ |x| + |y| = 1 ⇒ x + y = 1
⇒ -x – y = 1 ⇒ x – y = 1
Hence, these equations give us the locus of the point P which is a square.

Question 2.
A point moves so that square of its distance from the point (3, -2) is numerically equal to its distance from the line 5x -12y = 3. The equation of its locus is ……..
Solution:
The given equation of line is 5x – 12y = 3 and the given point is (3, -2).
Let (a, b) be any moving point.


⇒ 13a² + 13b² – 78a + 52b + 169 = 5a – 12b – 3
⇒ 13a² + 13b² – 83a + 64b + 172 = 0
So, the locus of the point is 13x² + 13y² – 83x + 64y + 172 = 0

Question 3.
Find the Locus of the midpoints of the portion of the line x cos θ + y sin θ = p intercepted between the axis.
Solution:
Given the equation of the line is x cos θ + y sin θ = p … (i)
Let C (h, k) be the midpoint of the given line AB where it meets the two-axis at A (a, 0) and B (0, b).
Since (a, 0) lies on eq (i) then “a cos θ + θ = p”


B (0, b) also lies on the eq (i) then 0 + b sin θ = p

Since C (h, k) is the midpoint of AB

Putting the values of a and b is eq (ii) and (iii) we get P

Squaring and adding eq (iv) and (v) we get

Question 4.

Solution:

Here, α is a variable. To find the locus of P (h, k), we have to eliminate α.
From (i), we obtain

Question 5.

Solution:

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.5

Choose the correct or the most suitable answer from the given four alternatives.

Question 1.
If a_ij = \(\frac{1}{2}\) (3i – 2j) and A = [a_ij]_2×2 is

Solution:

Question 2.
What must be the matrix X, if 2X + \(\left[\begin{array}{ll}{1} & {2} \\ {3} & {4}\end{array}\right]=\left[\begin{array}{ll}{3} & {8} \\ {7} & {2}\end{array}\right]\) ?

Solution:

Question 3.
Which one of the following is not true about the matrix \(\left[\begin{array}{lll}{1} & {0} & {0} \\ {0} & {0} & {0} \\ {0} & {0} & {5}\end{array}\right]\)?
(a) a scalar matrix
(b) a diagonal matrix
(c) an upper triangular matrix
(d) A lower triangular matrix
Solution:
(b) a diagonal matrix

Question 4.
If A and B are two matrices such that A + B and AB are both defined, then …………
(a) A and B are two matrices not necessarily of same order.
(b) A and B are square matrices of same order.
(c) Number of columns of a is equal to the number of rows of B.
(d) A = B.
Solution:
(b) A and B are square matrices of same order.

Question 5.
If A = \(\left[\begin{array}{rr}{\lambda} & {1} \\ {-1} & {-\lambda}\end{array}\right]\), then for what value of λ, A² = 0?
(a) 0
(b) ±1
(c) -1
(d) 1
Solution:

Question 6.
If and (A + B)² = A² + B², then the values of a and b are ……………….
(a) a = 4, b = 1
(b) a = 1, b = 4
(c) a = 0, b = 4
(d) a = 2, b = 4
Solution:

Question 7.
If is a matrix satisfying the equation AA^T = 9I, where I is 3 × 3 identity matrix, then the ordered pair (a, b) is equal to ………….
(a) (2, -1)
(b) (-2, 1)
(c) (2, 1)
(d) (-2, -1)
Solution:

Question 8.
If A is a square matrix, then which of the following is not symmetric?
(a) A + A^T
(b) AA^T
(c) A^TA
(d) A – A^T
Solution:
(b) AA^T

Question 9.
If A and B are symmetric matrices of order n, where (A ≠ B), then …………….
(a) A + B is skew-symmetric
(b) A + B is symmetric
(c) A + B is a diagonal matrix
(d) A + B is a zero matrix
Solution:
(b) A + B is symmetric

Question 10.
If and if xy = 1, then det (AA^T) is equal to …………..
(a) (a – 1)²
(b) (a² + 1)²
(c) a² – 1
(d) (a² – 1)²
Solution:

Question 11.
The value of x, for which the matrix is singular is ………….
(a) 9
(b) 8
(c) 7
(d) 6
Solution:
(b) Hint: Given A is a singular matrix ⇒ |A| = 0

⇒ e^x-2.e^2x+3 – e^2+x.e^7+x = 0
⇒ e^3x+1 – e^9+2x = 0
⇒ e^3x+1 = e^9+2x
⇒ 3x + 1 = 9 + 2x
⇒ 3x – 2x = 9 – 1
⇒ x = 8

Question 12.
If the points (x, -2), (5, 2), (8, 8) are collinear, then x is equal to …………
(a) -3
(b) \(\frac{1}{3}\)
(c) 1
(d) 3
Solution:
(d) Hint: Given that the points are collinear
So, area of the triangle formed by the points = 0

Question 13.

Solution:

Question 14.
If the square of the matrix is the unit matrix of order 2, then α, β and γ should satisfy the relation.
(a) 1 + α² + βγ = 0
(b) 1 – α² – βγ = 0
(c) 1 – α² + βγ = 0
(d) 1 + α² – βγ = 0
Solution:

Question 15.

(a) Δ
(b) kΔ
(c) 3kΔ
(d) k³Δ
Solution:

Question 16.
A root of the equation is …………….
(a) 6
(b) 3
(c) 0
(d) -6
Solution:

Question 17.
The value of the determinant of is ……………
(a) -2abc
(b) abc
(c) 0
(d) a² + b² + c²
Solution:

Question 18.
If x₁, x₂, x₃ as well as y₁, y₂, y₃ are in geometric progression with the same common ratio, then the points (x₁, y₁), (x₂, y₂), (x₃, y₃) are
(a) vertices of an equilateral triangle
(b) vertices of a right-angled triangle
(c) vertices of a right-angled isosceles triangle
(d) collinear
Solution:
(d)

Question 19.
If \(\lfloor.\rfloor\) denotes the greatest integer less than or equal to the real number under consideration and -1 ≤ x < 0, 0 ≤ y < 1, 1 ≤ z ≤ 2, then the value of the determinant is …………..
(a) \(\lfloor z\rfloor\)
(b) \(\lfloor y\rfloor\)
(c) \(\lfloor x\rfloor\)
(d) \(\lfloor x\rfloor+ 1\)
Solution:
(a) Hint: From the given values
>

Question 20.
If a ≠ b, b, c satisfy then abc = ……………..
(a) a + b + c
(b) 0
(c) b³
(d) ab + bc
Solution:
(c) Hint: Expanding along R₁,
a(b² – ac) – 2b (3b – 4c) + 2c (3a – 4b) = 0
(b² – ac) (a – b) = 0
b² = ac (or) a = b
⇒ abc = b(b²) = b³

Question 21.
If then B is given by ………………..
(a) B = 4A
(b) B = -4A
(c) B = -A
(d) B = 6A
Solution:

Question 22.
If A is skew-symmetric of order n and C ¡s a column matrix of order n × 1, then C^T AC is ……………..
(a) an identity matrix of order n
(b) an identity matrix of order 1
(e) a zero matrix of order I
(d) an Identity matrix of order 2
Solution:
(c) a zero matrix of order I
Hint: Given A is of order n × n
C is of order n × 1
so, CT is of order 1 × n

Let it be equal to (x) say
Taking transpose on either side
(C^T, AC)^T (x)^T .
(i.e.) C^T(A^T)(C) = x
C^T(-A)(C) = x
⇒ C^TAC = -x
⇒ x = -x
⇒ 2x = 0
⇒ x = 0

Question 23.
The matrix A satisfying the equation is ……………

Solution:

Question 24.
If A + I = , then (A + I) (A – I) is equal to …………….

Solution:

Question 25.
Let A and B be two symmetric matrices of the same order. Then which one of the following statements is not true?
(a) A + B ¡s a symmetric matrix
(b) AB ¡s a symmetric matrix
(c) AB = (BA)^T
(d) A^TB = AB^T
Solution:
(b) AB ¡s a symmetric matrix