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 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

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 1 Sets Ex 1.1

Question 1.
Write the following in roster form.
(i) {x ∈ N : x² < 121 and x is a prime}.
(ii) the set of all positive roots of the equation (x – 1)(x + 1)(x² – 1) = 0.
(iii) {x ∈ N : 4x + 9 < 52}.
(iv) {x : \(\frac{x-4}{x+2}\) = 3, x ∈ R – {-2}}
Solution:
(i) Let A = { x ∈ N : x² < 121 and x is a prime }
A = {2, 3, 5, 7}
(ii) The set of positive roots of the equations
(x – 1) (x + 1) (x² – 1) = 0
(x – 1 ) (x + 1 ) (x + 1) (x – 1) = 0
(x + 1 )² (x – 1)² = 0
(x + 1)² = 0 or (x – 1)² = 0
x + 1 = 0 or x – 1 = 0
x = -1 or x = 1
A = { 1 }
(iii) Let A = { x ∈ N : 4x + 9 < 52 }
When x = 1, (4) × (1 ) + 9 = 4 + 9 = 13
When x = 2, (4) × (2) + 9 = 8 + 9 = 17
When x = 3, (4) × (3) + 9 = 12 + 9 = 21
When x = 4, (4) × (4) + 9 = 16 + 9 = 25
When x = 5, (4) × (5) + 9 = 20 + 9 = 29
When x = 6, (4) × (6) + 9 = 24 + 9 = 33
When x = 7, (4) × (7) + 9 = 28 + 9 = 37
When x = 8, (4) × (8) + 9 = 32 + 9 = 41
When x = 9, (4) × (9) + 9 = 36 + 9 = 45
When x = 10, (4) × (10) + 9 = 40 + 9 = 49
∴ A = { 1, 2, 3, 4, 5, 6 ,7, 8, 9, 10 }

(i.e.) x – 4 = 3(x + 2)
x – 4 = 3x + 6
– 4 – 6 = 3x – x
2x = -10 ⇒ x = -5
A = {-5}

Question 2.
Write the set {-1, 1} in set builder form.
Solution:
A = {x : x² – 1 = 0, x ∈ R}

Question 3.
State whether the following sets are finite or infinite.

1. {x ∈ N : x is an even prime number}
2. {x ∈ N : x is an odd prime number}
3. {x ∈ Z : x is even and less than 10}
4. {x ∈ R : x is a rational number}
5. {x ∈ N : x is a rational number}

Solution:

1. Finite set
2. Infinite set
3. Infinite
4. Infinite
5. Infinite

Question 4.
By taking suitable sets A, B, C, verify the following results:
(i) A × (B ∩ C) = (A × B) ∩ (A × C).
(if) A × (B ∪ C) = (A × B) ∪ (A × C).
(iii) (A × B) ∩ (B × A) = (A ∩ B) × (B ∩ A).
(iv) C – (B – A) = (C ∩ A) ∪ (C ∩ B).
(v) (B – A) ∩ C = (B ∩ C) – A = B ∩ (C – A).
Solution:
To prove the following results let us take U = {1, 2, 5, 7, 8, 9, 10}
A = {1, 2, 5, 7}
B = {2, 7, 8, 9}
C = {1, 5, 8, 7}
(i) Let A = {1, 2}, B = {3, 4}, C = {4, 5}
B ∩ C = {3, 4} ∩ {4, 5}
B ∩ C = {4}
A × (B ∩ C) = {1, 2} × {4}
A × (B ∩ C) = { (1,4), (2,4) } —– (1)
A × B = {1, 2} × {3, 4}
A × B = { (1,3), (1, 4), (2, 3), (2, 4)}
A × C = {1, 2} × { 4, 5 }
A × C = {(1, 4), (1, 5), (2, 4), (2, 5)}
(A × B) ∩ (A × C) = {(1, 3), (1, 4), (2, 3), (2, 4)} ∩ { (1, 4), (1, 5), (2, 4), (2, 5)}
(A × B) ∩ (A × C) = {(1, 4), (2, 4)} —- (2)
From equations (1) and (2)
A × (B ∩ C) = (A × B) ∩ (A × C)

(ii) To prove A × (B ∪ C) = (A × B) (A × C)
B = {2, 7, 8, 9}, C = {1, 5, 8, 10)
B ∪ C = {1, 2, 5, 7, 8, 9, 10}
A = {1, 2, 5, 7}
A × (B ∪ C) = {(1, 1), (1, 2), (1, 5), (1, 7), (1, 8), (1, 9), (1, 10), (2, 1), (2, 2), (2, 5), (2, 7), (2, 8), (2, 9), (2, 10), (5, 1), (5, 2), (5, 5), (5, 7), (5, 8), (5, 9), (5, 10), (7, 1), (7, 2), (7, 5), (7, 7), (7, 8), (7, 9), (7, 10)) …. (1)
A × B = {(1, 2), (1, 7), (1, 8), (1, 9), (2, 2), (2, 7), (2, 8), (2, 9), (5, 2), (5, 7), (5, 8), (5, 9),
(7, 2), (7, 7), (7, 8), (7, 9)}
A × C = {(1, 1), (1, 5), (1, 8), (1, 10), (2, 1), (2, 5), (2, 8), (2, 10), (5, 1), (5, 5), (5, 8), (5, 10), (7, 1), (7, 5), (7, 8), (7, 10)}
(A × B) ∪ (A × C) = (1, 1), (1, 2), (1, 5), (1, 7), (1, 8), (1,9), (1, 10), (2, 1), (2, 2), (2, 5), (2, 7), (2, 8), (2, 9), (2, 10), (5, 1), (5, 2), (5, 5), (5, 7), (5, 8), (5, 9), (5, 10), (7, 1), (7, 2), (7, 5), (7, 7), (7, 8), (7, 9), (7, 10)} …… (2)
(1) = (2) ⇒ A × (B ∪ C) = (A × B) ∪ (A × C)

(iii) Let A = {1, 2}, B = {2, 3}
A × B = {1, 2} × {2, 3}
A × B = {(1, 2), (1, 3), (2, 2), (2, 3)}
B × A = {2, 3} × {1, 2}
B × A = {(2, 1), (2, 2), (3, 1), (3,2)}
(A × B) ∩ (B × A) = {(1, 2), (1, 3),(2, 2), (2, 3)} ∩ {(2, 1), (2, 2), (3, 1),(3, 2)}
(A × B) ∩ (B × A) = {(2, 2)} ——- (1)
A ∩ B = {1, 2} ∩ {2, 3}
A ∩ B = {2}
B ∩ A = {2, 3} ∩ {1, 2}
B ∩ A = {2}
(A ∩ B) × (B ∩ A) = {2} × {2}
(A ∩ B) × (B ∩ A) = {(2,2)} ———- (2)
From equations (1) and (2)
(A × B) ∩ (B × A) = (A ∩ B) × (B ∩ A)

(iv) To prove C – (B – A) = (C ∩ A) ∪ (C ∩ B)
B – A = {8, 9}
C = {1, 5, 8, 10}
∴ LHS = C – (B – A) = {1, 5, 10} …… (1)
C ∩ A = {1}
U = {1, 2, 5, 7, 8, 9, 10}
B = {2, 7, 8, 9} ∴ B’ = {1, 5, 10}
C ∩ B = {1, 5, 10}
R.H.S. (C ∩ A) ∪ (C ∩ B) = {1} ∪ {1, 5, 10}
= {1, 5, 10} ……. (2)
(1) = (2) ⇒ LHS = RHS

(v) Let A = {1, 2, 3, 4} , B = {3, 4, 5, 6}, C = { 5, 6, 7, 8 )
B – A = {3, 4, 5, 6} – {1, 2, 3, 4}
B – A = {5, 6}
(B – A) ∩ C = {5, 6} ∩ {5, 6, 7, 8}
(B – A) ∩ C = {5, 6} ——– (1)
(B ∩ C) = {3, 4, 5, 6} ∩ {5, 6, 7, 8}
B ∩ C = {5, 6}
(B ∩ C) – A = {5, 6} – {1,2,3,4}
(B ∩ C) – A = {5, 6} ——- (2)
C – A = {5, 6, 7, 8} – {1, 2, 3, 4}
C – A = {5, 6, 7, 8}
B ∩ (C – A) = {3, 4, 5, 6} ∩ {5, 6, 7, 8}
B ∩ (C – A) = {5, 6} ——– (3)
From equations (1) , (2) and (3)
(B – A) ∩ C = (B ∩ C) – A = B ∩(C – A)

(vi) To prove (B – A) ∪ C ={1, 5, 8, 9, 10}
B – A = {8, 9},
C = {1, 5, 8, 10}
(B – A) ∪ C = {1, 5, 8, 9, 10} ……. (1)
B ∪ C = {1, 2, 5, 7, 8, 9, 10}
A – C = {2, 7}
(B ∪ C) – (A – C) = {1, 5, 8, 9, 10} ……… (2)
(1) = (2)
⇒ (B – A) ∪ C = (B ∪ C) – (A – C)

Question 5.
Justify the trueness of the statement.
“An element of a set can never be a subset of itself.”
Solution:
“An element of a set can never be a subset of itself ”
The statement is correct
Let A = {a, b, c, d} for a ∈ A
‘a’ cannot be a subset of ‘a’

Question 6.
If n(P(A)) = 1024, n(A ∪ B) = 15 and n(P(B)) = 32, then find n(A ∩ B).
Solution:
n(P( A)) = 1024 = 2¹⁰ ⇒ n( A) = 10
n(A ∪ B) = 15
n(P(B)) = 32 = 2⁵ ⇒ n(B) = 5
We know n(A ∪ B) = n{A) + n(B) – n(A ∩ B)
(i.e.) 15 = 10 + 5 – n(A ∩ B)
⇒ n(A ∩ B) = 15 – 15 = 0

Question 7.
If n(A ∩ B) = 3 and n(A ∪ B) = 10, then find n(P(A(A ∆ B)).
Solution:
Given n(A ∩ B) = 3 and n(A ∪ B) = 10
A ∆ B = (A – B) ∪ (B – A)
n(A ∆ B) = n [ (A – B ) ∪ (B – A)]
n(A ∆ B) = n(A – B) + n(B – A) —— (1)
(Since A – B and B – A are disjoint sets)
A ∪ B = (A – B) ∪ (B – A) ∪ (A ∩ B)
n(A ∪ B) = n[(A – B) ∪ (B – A) ∪ (A ∩ B)]
n(A ∪ B) = n (A – B) + n (B – A) + n (A ∩ B)
(Since A – B, B – A and A ∩ B are disjoint sets)
n(A ∪ B) = n(A ∆ B) + n(A ∩ B)
10 = n(A ∆ B) + 3
n(A ∆ B) = 10 – 3 = 7
∴ n(P(A ∆ B)) = 2⁷ = 128

Question 8.
For a set A, A × A contains 16 elements and two of its elements are (1, 3) and (0, 2). Find the elements of A.
Solution:
A × A = 16 elements = 4 × 4
⇒ A has 4 elements
∴ A = {0, 1, 2, 3}

Question 9.
Let A and B be two sets such that n(A) = 3 and n(B) = 2. If (x, 1), (y, 2), (z, 1) are in A × B, find A and B, where x, y, z are distinct elements.
Solution:
Given A and B be two sets such that n (A) = 3 and n(B) = 2.
Also given (x, 1), (y, 2), (z, 1) ∈ A × B
A = { x, y, z }, B = {1, 2}

Question 10.
If A × A has 16 elements, S = {(a, b) ∈ A × A : a < b} ; (-1, 2) and (0, 1) are two elements of S, then find the remaining elements of S.
Solution:
n(A × A) = 16 ⇒ n( A) = 4
S ={(-1, 0), (-1, 1), (0, 2), (1, 2)}

Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.1 Additional Questions

Question 1.
Write the following sets in roster form
(a) {x ∈ N; x³ < 1000}
(b) {The set of positive roots of the equation (x² – 4) (x³ – 27) = 0}
Solution:
(a) A = {1, 2, 3, 4, 5, 6, 7, 8, 9}
(b) B = {2, 3}

Question 2.
By taking suitable sets A, B, C verify the following results
(i) A × (B ∪ C) = (A × B) ∪ (A × C)
(ii) (B – A) ∪ C = (B ∪ C) – (A – C)
Solution:
Prove by yourself

Question 3.
Given n(A) = 7; n(B) = 8 and n(A ∪ B) = 10 find n[P(A ∩ B)].
Solution:
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
(i.e.,) 10 = 7 + 8 – n(A ∩ B)
⇒ n(A ∩ B) = 7 + 8 – 10 = 5
So n[P(A ∩ B)] = 25 = 32

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 6.4

Question 1.
Find the combined equation of the straight lines whose separate equations are x – 2y – 3 = 0 and x + y + 5 = 0.
Solution:
The given separate equations of the lines are
x – 2y – 3 = 0 and x + y + 5 = 0
∴ The combined equation of the straight lines is
(x – 2y – 3 ) (x + y + 5) = 0
x² + xy + 5x – 2xy – 2y² – 10y – 3x – 3y – 15 = 0
x² – xy – 2y² + 2x – 13y – 15 = 0

Question 2.
Show that 4x² + 4xy + y² – 6x – 3y – 4 = 0 represents a pair of parallel lines.
Solution:
Comparing this equation with ax² + 2hxy + by² + 2gx + 2fy + c = 0
we get a = 4, h = \(\frac{4}{2}\) = 2 , b = 1, g = – 3, f = – 3/2, c = – 4
The condition for the lines to be parallel is h² – ab = 0
Now h² – ab = 2² – (4) (1) = 4 – 4 = 0
h² – ab = 0 ⇒ The given equation represents a pair of parallel lines.

Question 3.
Show that 2x² + 3xy – 2y² + 3x + y + 1 = 0 represents a pair of perpendicular lines.
Solution:
The equation of the given pair of straight lines is
2x² + 3xy – 2y² + 3x + y + 1 = 0 ……….. (1)
Compare this equation with the equation
ax² + 2hxy + by² + 2gx + 2f y + c = 0 ……… (2)
a = 2, 2h = 3, b = – 2 , 2g = 3, 2f = 1, c = 1
The condition for pair of straight lines to be perpendicular is a + b = 0.
2 – 2 = 0
Hence the given pair of lines represents a perpendicular straight lines.

Question 4.
Show that the equation 2x² – xy – 3y² – 6x + 19y – 20 = 0 represents a pair of intersecting lines. Show further that the angle between them is tan^-1(5).
Solution:

The given equation represents a pair of straight lines.

Question 5.
Prove that the equation to the straight lines through the origin, each of which makes an angle α with the straight line y = x is x² – 2xy sec 2α + y² = 0
Solution:
Slope of y = x is m = tan θ = 1
⇒ θ = 45°
The new lines slopes will be
m = tan(45 + α) and m = tan (45 – α)
∴ The equations of the lines passing through the origin is given by
y = tan(45 + α)x and y = tan(45 – α)x
(i.e) y = tan(45 + α)x = 0 and y = tan(45 – α)x = 0
The combined equation is [y – tan (45 + α)x] [y – tan (45 – α)x] = 0
y² + tan(45 + α)tan(45 – α)x² – xy[tan(45 – α) + tan(45 + α)] = 0


Let the equation of lines passes through the origin
So the equations are y = m₁x = 0 and y = m₂x = 0
So the combined equations is (y – m₁x) (y – m₂x) = 0
(i.e)y² – xy(m₁ + m₂) + m₁m₂x = 0
(i.e) y² – xy(2sec α) + x²(1) = 0
(i.e) y² – 2xy sec 2α + x² = 0

Question 6.
Find the equation of the pair of straight lines passing through the point (1, 3) and perpendicular to the lines 2x – 3y + 1 = 0 and 5x + y – 3 = 0
Solution:
The equation of the given lines are
2x – 3y + 1 = 0 ……….. (1)
5x + y – 3 = 0 ……….. (2)
Equation of any line perpendicular to 2x – 3y + 1 = 0 is
– 3x – 2y + k = 0
3x + 2y – k = 0
This line passes through the point (1, 3)
∴ 3(1) + 2(3) – k = 0
3 + 6 – k = 0 ⇒ k= 9
Substituting the value of k in the above equation we have
3x + 2y – 9 = 0 ………. (3)
Equation of any line perpendicular to 5x + y – 3 = 0 is
x – 5y + k₁ = 0
This line passes througi the point (1 , 3)
∴ 1 – 5 (3) + k₁ = 0
1 – 15 + k₁ ⇒ k₁ = 14
Substituting the value of k₁ in the above equation we have
x – 5y + 14 = 0 ……….. (4)
The combined equation of (3) and (4) is
( 3x + 2y – 9) (x – 5y + 14 ) = 0
3x² – 15xy + 42x + 2xy – 10y² + 28y – 9x + 45 y – 126 = 0
3x² – 13xy – 10y² + 33x + 73y – 126 = 0

Question 7.
Find the separate equation of the following pair of straight lines
(i) 3x² + 2xy – y² = 0
(ii) 6 (x – 1)² + 5(x – 1)(y – 2) – 4(y – 2)² = 0
(iii) 2x² – xy – 3y² – 6x + 19y – 20 = 0
Solution:
(i) 3x² + 2xy – y² = 0
The given equation is
3x² + 2xy – y² = 0 ……. (1)
3x² + 3xy – xy – y² = 0
3x (x + y) – y (x + y) = 0
(3x – y) (x + y) = 0
3x – y = 0 and x + y = 0
∴ The separate equations are
3x – y = 0 and x + y = 0

(ii) 6 (x – 1)² + 5 (x – 1)(y – 2) – 4(y – 2)² = 0
⇒ 6(x² – 2x +1) + 5(xy – 2x – y + 2) – 4( y² – 4y + 4) = 0
(i.e) 6x² – 12x + 6 + 5xy – 10x – 5y + 10 – 4y² + 16y – 16 = 0
(i.e) 6x² + 5xy – 4y² – 22x + 11y = 0
Factorising 6x² + 5xy – 4y² we get
6x² – 3xy + 8xy – 4y² = 3x (2x – y) + 4y (2x – y)
= (3x + 4y)(2x – y)
So, 6x² + 5xy – 4y² – 22x + 11y = (3x + 4y + l )(2x – y + m)
Equating coefficient of x ⇒ 3m + 21 = -22 …….. (1)
Equating coefficient of y ⇒ 4m – l = 11 ……. (2)
Solving (1) and (2) we get l = -11, m = 0
So the separate equations are 3x + 4y – 11 = 0 and 2x – y = 0

(iii) 2x² – xy – 3y² – 6x + 19y – 20 = 0
Factorising 2x² – xy – 3y² we get
2x² – xy – 3y² = 2x² + 2xy – 3xy – 3y²
= 2x(x + y) – 3y(x + y) = (2x – 3y) (x + y)
∴ 2x² – xy – 3y² – 6x + 19y – 20 = (2x – 3y + l)(x + y + m)
Equating coefficient of x 2m + l = -6 ……. (1)
Equating coefficient of y -3m + l = 19 …….. (2)
Constant term -20 = lm
Solving (1) and (2) we get l = 4 and m = – 5 where lm = – 20.
So the separate equations are 2x – 3y + 4 = 0 and x + y – 5 = 0

Question 8.
The slope of one of the straight lines ax² + 2hxy + by² = 0 is twice that of the other, show that 8h² = 9ab.
Solution:
ax² + 2hxy + by² = 0
We are given that one slope is twice that of the other.
So let the slopes be m and 2m.
Now sum of the slopes = m + 2m

Question 9.
The slope of one of the straight lines ax² + 2hxy + by² = 0 is three times the other, show that 3h² = 4ab.
Solution:
Let the slopes be m and 3m.

Question 10.
A ∆OPQ is formed by the pair of straight lines x² – 4xy + y² = 0 and the line PQ. The equation of PQ is x + y – 2 = 0. Find the equation of the median of the triangle ∆OPQ drawn from the origin O.
Solution:
Equation of pair of straight lines is x² – 4xy + y² = 0 ….. (1)
Equation of the given line is x + y – 2 = 0 ⇒ y = 2 – x ……… (2)
On solving (1) and (2) we get x² – 4x (2 – x) + (2 – x)² = 0
(i.e) x² – 8x + 4x² + 4 + x² – 4x = 0
(i.e) 6x² – 12x + 4 = 0
(÷ by 2) 3x² – 6x + 2 = 0

The midpoint of PQ is

Question 11.
Find p and q, ¡f the following equation represents a pair of perpendicular lines 6x² + 5xy – py² + 7x + qy – 50
Solution:
6x² + 5xy – py² + 7x + qy – 50
The given equation represents a pair of perpendicular lines
⇒ coefficient of x² + coefficient of y² = 0
(i.e) 6 – p = 0 ⇒ p = 6
Now comparing the given equation with the general form
ax² + 2hxy + by² + 2gx + 2fy + c = 0
we get a = 6, b = -6 and c = -5, f = q/2, g = 7/2 and h = 5/2
The condition for the general form to represent a pair of straight lines is abc + 2fgh – af² – bg² – ch² = 0

Question 12.
Find the value of k, if the following equation represents a pair of straight lines. Further, find whether these lines are parallel or intersecting, 12x² + 7xy – 12y² – x + 7y + k = 0.
Solution:
Comparing the given equation with the general form ax² + 2hxy + by² + 2gx + 2fy + c = 0
we get a = 12, b = -12, c = k, f = 7/2, g = – 1/2, h = 7/2
Here a + b = 0 ⇒ the given equation represents a pair of perpendicular lines
To find k: The condition for the given equation to represent a pair of straight lines is abc + 2fgh – af² – bg² – ch² = 0

Question 13.
For what value of k does the equation 12x² + 2kxy + 2y², + 11x – 5y + 2 = 0 represent two straight lines.
Solution:
12x² + 2 kxy + 2y² + 11x – 5y + 2 = 0
Comparing this equation with the general form we get

4k² + 55k + 175 = 0
4k² + 20k + 35k + 175 = 0
4k(k + 5) + 35(k + 5) = 0
(4k + 35) (k + 5) = 0
k = -5 or -35/4

Question 14.
Show that the equation 9x² – 24xy + 16y² – 12x + 16y – 12 = 0 represents a pair of parallel lines. Find the distance between them.
Solution:
Comparing the given equation with ax² + 2kxy + by² = 0 we get a = 9, h = -12, b = 16.
Now h² = (-12)² = 144, ab = (9) (16) = 144
h² = ab ⇒ The given equation represents a pair of parallel lines.
To find their separate equations:
9x² – 24xy + 16y² = (3x – 4y)²
So, 9x² – 24xy +16y² – 12x + 16y – 12 = (3x – 4y + l )(3x – 4y + m)
Here coefficient of x ⇒ 3m + 3l = -12 ⇒ m + l = -4
coefficient of y ⇒ -4m – 4l = 16 ⇒ m + l = -4
Constant term l m = -12
Now l + m = -4 and lm = -12 ⇒ l = -6 and m = 2
So the separate equations are 3x – 4y – 6 = 0 and 3x – 4y + 2 = 0

Question 15.
Show that the equation 4x² + 4xy + y² – 6x – 3y – 4 = 0 represents a pair of parallel lines. Find the distance between them.
Solution:
4x² + 4xy + y² – 6x – 3y – 4 = 0
a = 4,
b = 1,
h = 4/2 = 2
h² – ab = 2² – (4) (1) = 4 – 4 = 0
⇒ The given equation represents a pair of parallel lines.
To find the separate equations 4x² + 4xy + y² = (2x + y)²
So, 4x² + 4xy + y² – 6x – 3y – 4 = (2x + y + l )(2x + y + m)
Coefficient of x ⇒ 2m + 2l = -6 ⇒ l + m = – 3 ……. (1)
Coefficient of y ⇒ l + m = – 3 ……… (2)
Constant term ⇒ l m = – 4 ……… (3)
Now l + m = -3 and lm = -4 ⇒ l = -4, m = 1
So the separate equations are 2x + y + 1 = 0 and 2x + y – 4 = 0

Question 16.
Prove that one of the straight lines given by ax² + 2hxy + by² = 0 will bisect the angle between the co-ordinate axes if (a + b)² = 4h².
Solution:
Let the slopes be l and m
∵ One line bisects the angle between the coordinate axes ⇒ θ = 45°
So tan θ = 1
The slopes are l and m

Question 17.
If the pair of straight lines x² – 2kxy – y² = 0 bisect the angle between the pair of straight lines x² – 2lxy – y² = 0, show that the latter pair also bisects the angle between the former.
Solution:
Given that x² – 2kxy – y² = 0 …….. (1)
Bisect the angle between the lines x² – 11xy – y² = 0 …… (2)

x² – 2kxy – y² = 0

Question 18.
Prove that the straight lines joining the origin to the points of intersection of 3x² + 5xy – 3y² + 2x + 3y = 0 and 3x – 2y – 1 = 0 are at right angles.
Solution:
The equation of the pair of straight lines is
3x² + 5xy – 3y² + 2x + 3y = 0 ……… (1)
The given line is 3x – 2y – 1 = 0
3x – 2y = 1 ……….. (2)
The equation of the straight lines joining the origin to the points of intersection of the pair of lines (1) and the line (2) is obtained by homogeneous using equation (1) by using equation (2)
(1) ⇒ (3x² + 5xy – 3y²) + (2x + 3y)(1) = 0
(3x² + 5xy – 3y²) + (2x + 3y) (3x – y) = 0
3x² + 5xy – 3y² + 6x² – 4xy + 9xy – 6y² = 0
9x² + 10xy – 9y² = 0 ………. (3)
Coefficient of x² + coefficient of y² = 9 – 9 = 0
∴ The pair of straight line (3) represents a perpendicular straight lines.

Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.4 Additional Questions Solved

Question 1.
Find the angle between the pair of straight lines given by (a² – 3b²)x² + 8abxy + (b² – 3a²)y² = 0.
Solution:

Question 2.
Show that 9x² + 24xy + 16y² + 21x + 28y + 6 = 0 represents a pair of parallel straight lines and find the distance between them.
Solution:
9x² + 24xy + 16y² + 21x + 28y + 6 = 0
Here a = 9.6,
b = 16,
g = \(\frac{21}{2}\),
f = 14,
c = 6,
h = 12
h² – ab = (12)² – 9(16) = 144 – 144 = 0
∴ The lines are parallel.
9x² + 24xy + 16y² = (3x + 4y)(3x + 4y)
Let 9x² + 24xy + 16y² + 21x + 28y + 6 = (3x + 4y + l)(3x + 4y + m)
Equating the coefficients of x and constant term
3l + 3m = 21
lm = 6
Solving we get, l = 1 or 6
m = 6 or 1
∴ The separate equations are 3x + 4y + 1 = 0 and 3x + 4y + 6 = 0

Question 3.
If the equation 12x² – 10xy + 2y² + 14x – 5y + c = 0 represents a pair of straight lines, find the value of c. Find the separate equations of the straight lines and also the angle
between them.
Solution:
12x² – 10xy – 2y² + 14x – 5y + c = 0
ax² + 2hxy + by² +2gx + 2fy – c = 0
Here a = 12,
b = 2,
g = 7,
f = 5/2,
c = c,
h = -5
af² + bg² + ch² – 2fgh – abc = 0 is the condition

The equation is 12x² – 10y + 2y² + 14x – 5y + 2 = 0
12x² – 10xy + 2y = (3x – y)(4x – 2y)
Let 12x² – 10y + 2y² + 14x – 5y + 2(3x – y + l)(4x – 2y + m)
So that 4l + 3m = 14 , -2l – m = -5

Question 4.
For what value of k does 12x² + 7xy + ky² + 13x – y + 3 = 0 represents a pair of straight lines? Also write the separate equations.
Solution:
12x² + 7xy + ky² + 13x – y + 3 = 0
a = 12,
h = \(\frac{7}{2}\),
f = \(-\frac{1}{2}\) ,
c = 3
af² + bg² + ch² – abc – 2fgh = 0

⇒ 12 + 169k + 147 – 144k + 91 = 0
25k = – 250 ⇒ k = -10
The equation is 12x² + 7xy – 10y² + 13x – y + 3 = 0
To find separate equations: 12x² + 7xy – 10y² = (3x – 2y)(4x + 5y)
Let 12x² + 7xy – 10y² + 13x – y + 3 = 0(3x – 2y + l)(4x + 5y + m)
Equating the coefficient of x ⇒ 4l + 3m = 13 …… (1)
Equating the coefficient of y ⇒ 5l – 2m = -1 …… (2)
(1) × 2 ⇒ 8l + 6m = 26
(2) × 3 ⇒ 15l – 6m = -3
23l = 23 ⇒ l = 1
4 + 3 m = 13
3 m = 9 ⇒ m = 3
The separate equations are 3x – 2y + 1 = 0 and 4x + 5y + 3 = 0

Question 5.
Show that 3x² + 10xy + 8y² + 14x + 22y + 15 = 0 represents a pair of straight lines and the angle between them is tan^-1\(\left(\frac{2}{11}\right)\)
Solution:
3x² + 10xy + 8y² + 14x + 22y + 15 = 0
a = 3,
A = 5,
b = 8,
g = 7,
f = 11,
c = 15
The condition is af² + bg² + ch² – abc – 2fgh = 0
3(11)² + 8(7)² + 15 (5)² – (3)(8)(15) – 2(11)(7)(5) = 363 + 392 + 375 – 360 – 770 = 0
The angle between the pair of straight line is given by

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 4.1

Question 1.

(i) A person went to a restaurant for dinner. In the menu card, the person saw 10 Indian and 7 Chinese food items. In how many ways the person can select either an Indian or Chinese food?
Solution:
Number of Indian food items = 10
Number of Chinese food items = 7
Number of ways of selecting 10 Indian food items = 10 ways
Number of ways of selecting 7 Chinese food items = 7 ways
∴ By the fundamental principle of addition, the number of ways of selecting 10 Indian food items or 7 Chinese food items is = (10 + 7) ways = 17 ways

(ii) There are 3 types of a toy cars and 2 types of toy trains available in a shop. Find the number of ways a baby can buy a toy car and a toy train?
Solution:
Given, Number of toy cars = 3
Number of toy trains = 2
∴ A baby buying a toy car from 3 can be done in 3 ways
∴ A baby buying a toy train from 2 can be done in 2 ways
∴ Buying a toy car and a toy train together can be done in 3 × 2 = 6 ways

(iii) How many two-digit numbers can be formed using 1, 2, 3, 4, 5 without repetition of digits?
Solution:
The given digits are 1, 2, 3, 4, 5 The one’s place can be filled up in 5 ways using 1, 2, 3, 4, 5 and the ten’s place can be filled up in 4 ways.
The number of two-digit numbers using the digits 1, 2, 3, 4, 5 is 4 × 5 = 20

(iv) Three persons enter into a conference hall in which there are 10 seats. In how many ways they can take their places?
Solution:
Given, Number of persons = 3 and Number of seats = 10
The first person can take his place (from 10 seats) in 10 ways
The second person can take his place (from the remaining 9 seats) in 9 ways
The third person can take his place (from the remaining 8 seats) in 8 ways
∴ The three persons together can take their places in 10 × 9 × 8 = 720 ways

(v) In how many ways 5 persons can be seated in a row?
Solution:
Number of ways of 1^st person can be seated in a row = 5
Number of ways of 2^nd person can be seated in a row = 4
Number of ways of 3^rd person can be seated in a row = 3
Number of ways of 4^th person can be seated in a row = 2
Number of ways of 5^th person can be seated in a row = 1
∴ By fundamental principle of multiplication, number of ways of 5 persons can be seated in a row
= 5 × 4 × 3 × 2 × 1
= 5!
= 120

Question 2.
(i) A mobile phone has a passcode of 6 distinct digits. What is the maximum number of attempts one makes to retrieve the passcode?
Solution:
Number of digits = 10
∴ Number of attempts made = 10 × 9 × 8 × 7 × 6 × 5 = 151200 ways

(ii) Given four flags of different colours, how many different signals can be generated if each signal requires the use of three flags, One below the other?
Solution:
Number of flags given = 4
Number of flag needed (to show a signal) = 3
The first flag can be chosen in 4 ways (from the 4 flags)
The second flag can be chosen (from the remaining 3 flags) in 3 ways
The third flag can be chosen (from the remaining 2 flags) in 2 ways
So the first, second and the third flags together can be chosen in (to generate a signal) 4 × 3 × 2 = 24 ways
(i.e) 24 signals can be generated

Question 3.
Four children are running a race.
(i) In how many ways can the first two places be filled?
Solution:
First place can be given to any one of the 4 children and second place can be given to any one of the remaining 3 children.
Number of ways of filling the first place = 4
Number of ways of filling the second place = 3
Therefore, by the fundamental principle of multiplication total number of ways of filling the first two places is = 4 × 3 =12 ways

(ii) In how many different ways could they finish the race?
Solution:
In how many different ways could they finish the race?
The race can be finished in = 4 × 3 × 2 × 1 ways = 24 ways

Question 4.
Count the number of three-digit numbers which can be formed from the digits 2, 4, 6, 8? if.

(i) repetitions of digits is allowed
Solution:
Number of digit given = 4 (2,4, 6, 8)
So the unit place can be filled in 4 ways, 10’s place can be filled in 4 ways and 100’s place can be filled in 4 ways
∴ The unit place, 10’s place and 100’s place together can be filled (i.e) So the Number of 3 digit numbers = 4 × 4 × 4 = 64 ways

(ii) repetitions of digits is not allowed.
Solution:
Repetitions of digits is not allowed
Hundred’s Ten’s Unit
The number of ways of filling the unit place using the 4 digits 2,4,6,8 in 4 ways. A number of ways of filling the tens place using the remaining 3 digits 3 ways. The number of ways of filling the hundred’s place using the remaining 2 digits is 2 ways.
Therefore, by the fundamental principle of multiplication, the total number of 3 digit numbers without repetitions of digits is = 4 × 3 × 2 = 24 ways

Question 5.
How many three-digit numbers are there with 3 in the unit place?
(i) with repetition
Solution:
with repetition
The unit place is filled (by 3) in 1 way
The 10’s place can be filled in 10 ways
The 100’s place can be filled in 9 ways (excluding 0)
So the number of 3 digit numbers with 3 unit – place = 9 × 10 × 1 = 90

(ii) without repetition
Solution:
The unit place can be filled in only one way using the digit 3. The hundred’s place can be filled in 8 ways using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 excluding 0 and 3. The ten’s place can be filled in 8 ways using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 excluding the digit 3 and the digit placed in the hundred’s place.
Therefore, by the fundamental principle of multiplication, the total number of 3 digit numbers = 1 × 8 × 8 = 64

Question 6.
How many numbers are there between 100 and 500 with the digits 0, 1, 2, 3, 4, 5 if
(i) repetition of digits allowed
Solution:
repetition of digits allowed
The given digits are 0, 1, 2, 3, 4, 5
We have to find numbers between 100 and 500. So the 100’s place can be filled (by the numbers 1, 2, 3, 4) in 4 ways.
The 10’s place can be filled in (using 0, 1, 2, 3, 4, 5) 6 ways
and the unit-place can be filled in (using 0,1, 2, 3, 4, 5) 6 ways
But the number 100 should be excluded
So the number of numbers between 100 and 500 = 4 × 6 × 6 = 144

(ii) the repetition of digits is not allowed
Solution:

The hundred’s place can be filled in 4 ways using the digits 0, 1, 2, 3, 4, 5 excluding 0 and 5. Ten’s place can be filled in 5 ways using the digits 0, 1, 2, 3, 4, 5 excluding the digit placed in the hundred’s place. The unit place can be filled in 4 ways using the digits 0, 1, 2, 3, 4, 5 excluding the digits placed in hundred’s place’ and ten’s place.

Therefore, by the fundamental principle of multiplication, the number of 3 digit numbers between 100 and 500 with repletion of digits using the digits 0, 1, 2, 3, 4, 5 is = 4 × 5 × 4 = 80

Question 7.
How many three-digit odd numbers can be formed by using the digits 0, 1, 2, 3, 4, 5 if

(i) The repetition of digits is not allowed
Solution:
The repetition of digits is not allowed
The given digits are 0, 1, 2, 3, 4, 5. Here the odd number are 1, 3, 5.
So the unit place can be filled in 3 ways (using the 3 odd number)
After filling the unit place since 0 is a given digit be fill the 100’s place which can be
filled in

Then the 10’s place can be filled in (6 – 2) 4 ways.
So the number of 3 digit odd numbers = 3 × 4 × 4 = 48

(ii) The repetition of digits is allowed
Solution:
Since we need 3 – digit odd numbers the unit place can be filled in 3 ways using the digits 1,3 or 5. Hundred’s place can be filled in 5 ways using the digits 0, 1, 2, 3, 4, 5 excluding 0. Ten’s place can be filled in 6 ways using the digits 0 , 1 , 2 , 3 , 4 , 5.

Therefore, by the fundamental principle of multiplication, the number of 3 – digit odd numbers formed by using the digits 0, 1 , 2 , 3 , 4, 5 with repetition of digits is = 3 × 5 × 6 = 90

Question 8.
Count the numbers between 999 and 10000 subjects to the condition that there are

(i) no restriction
Solution:
no restriction
We have to find 4 digit numbers
The 1000’s place can be filled in 9 ways (excluding zero) and the 100’s, 10’s and unit places respectively can be filled in 10, 10, 10 ways (including zero)
So the number of numbers between 999 and 10000 = 9 × 10 × 10 × 10 = 9000

(ii) no digit is repeated
Solution:
Since 0 is given as a digit we have to start filling 1000’s place.
Now 1000’s place can be filled in 9 ways (excluding 0)
Then the 100’s place can be filled in 9 ways (excluding one digit and including 0)
10’s place can be filled in (9 – 1) 8 ways and unit place can be filled in (8 – 1) 7 ways So the number of 4 digit numbers are 9 × 9 × 8 × 7 = 4536 ways

(iii) at least one of the digits is repeated
Solution:
Required number of numbers = 9000 – 4536 = 4464 numbers

Question 9.
How many three-digit numbers, which are divisible by 5, can be formed using the digits 0, 1, 2, 3, 4, 5 if

(i) The repetition of digits are not allowed?
Solution:
The repetition of digits are not allowed.
The given digits are 0, 1, 2, 3, 4, 5. A number will be divisible by 5 if the digit in the unit place is 0 or 5
So the unit place can be filled by 0 or 5

(a) When the unit place is 0 it is filled in 1 way
And so 10’s place can be filled in 5 ways (by using 1, 2, 3, 4, 5) and 100’s place can be filled in (5 – 1) 4 ways
So the number of 3 digit numbers with unit place 0 = 1 × 5 × 4 = 20

(b) When the unit place is 5 it is filled in 1 way
Since 0 is given as a digit to fill 100’s place 0 should be excluded
So 100’s place can be filled in (excluding 0 and 5) 4 ways and 10’s place can be filled in (excluding 5 and one digit and including 0) 4 ways So the number of 3 digit numbers with unit place 5 = 1 × 4 × 4 = 16
∴ Number of 3 digit numbers ÷ by 5 = 20 + 16 = 36

(ii) The repetition of digits are allowed.
Solution:
Since the 3 – digit number is divisible by 5, the unit place can be filled in 2 ways using the digits 0 and 5. Since the repetition of digits is allowed the ten’s place is filled in 6 ways using the digits 0 , 1, 2, 3, 4,5 and the hundred’s place is filled in 5 ways using the digits 0, 1, 2, 3, 4, 5 excluding 0 .

Therefore, by the fundamental principle of multiplication, the number of 3 digit numbers formed by using the digits 0,1, 2, 3 , 4, 5 with repetition of digits is = 2 × 6 × 5 = 60

Question 10.
To travel from place A to place B, there are two different bus routes B₁, B₂ two different train routes T₁, T_2, and one air route A₁. From place B to place C, there is one bus route say B₁‘, two different train routes say T₁‘, T₂‘ and one air route A₁‘. Find the number of routes of commuting from place A to place C via place B without using a similar mode of transportation.
Solution:

From the above diagram the number of routes from A to C
= (2 × 2 + 2 × 1) + [(2 × 1) + (2 × 1)] + [(1 × 1) + (1 × 2)]
= 4 + 2 + 2 + 2 + 1 + 2 = 13

Question 11.
How many numbers are there between 1 and 1000 (both inclusive) which are divisible neither by 2 nor by 5?
Solution:
Given digits are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Numbers which are neither divisible by 2 nor 5 should have unit place 1, 3, 7, 9.
One digit numbers:
1, 3, 7, 9 are the one-digit numbers which are neither divisible by 2 nor by 5
Therefore, the required number of one-digit numbers = 4

Two-digit numbers:
The unit place can be filled in 4 ways using the digits 1, 3, 7, 9. Ten’s place can be filled in 9 ways using all the digits excluding 0. Therefore, the required number of 2 – digit numbers = 9 × 4 = 36

Question 12.
How many strings can be formed using the letters of the word LOTUS if the word

(i) either start with L or end with S?
Solution:
either start with L or end with S?
To find the number of words starting with L
Number of letters in LOTUS = 5 when the first letter is L it can be filled in 1 way only. So the remaining 4 letters can be arranged in 4! =24 ways = n(A). When the last letter is S it can be filled in the 1 way and the remaining 4 letters can be arranged is 4! = 24 ways = n(B)

(1) (1) 3! = 6 = n(A ∩ B)
Now n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
= 24 + 24 – 6 = 42
Now, neither words starts with L nor ends with S = 42

(ii) neither starts with L nor ends with S?
Solution:
Number of letters of the word LOTUS = 5.
They can be arranged in 5 ! = 120 ways
Number of words starting with L and ending with S = 42
So the number of words neither starts with L nor ends with S = 120 – 42 = 78

Question 13.
(i) Count the total number of ways of answering 6 objective type questions, each question having 4 choices.
Solution:
Number of choices for each question = 4
Total number of questions = 6
Each question can be answered in 4 ways.
∴ The total number of ways of answering 6 questions is = 4 × 4 × 4 × 4 × 4 × 4 = 4⁶

(ii) In how many ways 10 Pigeons can be placed in 3 different Pigeonholes?
Number of Pigeons = 10
Number of Pigeonholes = 3
Each Pigeon can occupy any of these 3 holes
∴ Total number of ways of placing 10 Pigeons
= 3 × 3 × 3 × …………….. 10 times
= 3¹⁰

(iii) Find the number of ways of distributing 12 distinct prizes to 10 students?
Each price can be distributed to any one of the 10 students.
Therefore, by the rule of product, the number of ways of distributing 12 distinct prizes to 10 students are
= 10 × 10 × 10 × …………. 12 times
= 10¹²

Question 14.
Find the value of

(i) 6!
Solution:
6! = 6 × 5 × 4 × 3 × 2 × 1 = 720

(ii) 4! + 5!
Solution:
4! + 5! = (4 × 3 × 2 × 1) + (5 × 4 × 3 × 2 × 1)
= 24 + 120 = 144

(iii) 3! – 2!
Solution:
3! – 2! = (3 × 2 × 1) – (2 × 1)
= 6 – 2 = 4

(iv) 3! × 4!
Solution:
3! × 4! = (3 × 2 × 1) × (4 × 3 × 2 × 1) = 6 × 24 = 144
12!

(v) \(\frac{12 !}{9 ! \times 3 !}\)
Solution:

(vi) \(\frac{(n+3) !}{(n+1) !}\)
Solution:

Question 15.
Evaluate \(\frac{n !}{r !(n-r) !}\) when

(i) n = 6,
r = 2
Solution:

(ii) n = 10,
r = 3
Solution:

(iii) For any n with r = 2
Solution:

Question 16.
Find the value of n if

(i) (n + 1)! = 20(n – 1)!
Solution:

(ii) \(\frac{1}{8 !}+\frac{1}{9 !}=\frac{n}{10 !}\)
Solution:

Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.1 Additional Questions Solved

Question 1.
If the letter of the word ‘RACHIT’ are arranged in all possible ways as listed in the dictionary, then what is the rank of the word ‘RACHIT’?
Solution:
The alphabetical order of RACHIT is A, C, H, I, R, and T
Number of words beginning with A = 5!
Number of words beginning with C = 5!
Number of words beginning with H = 5!
Number of words beginning with 1 = 5!
and Number of words beginning with R (i.e) RACHIT = 1
∴ The rank of the word ’RACHIT’ in the dictionary = 5! + 5! + 5! + 5! + 1 = 4 × 5! + 1
= 4 × 5 . 4 . 3 . 2 . 1 + 1 = 4 × 120 + 1 = 480 + 1 = 481

Question 2.
Find the number of positive integers greater than 6000 and less than 7000 which are divisible by 5, provided that no digit is to be repeated.
Solution:
Any number divisible by 5, its unit place must have 0 or 5. We have to find 4-digit numbers greater than 6000 and less than 7000.
So, the unit place can be filled with 2 ways (0 or 5) since, repetition is not allowed.
∴ Tens place can be filled with 7 ways and hundreds place can be filled with 8 ways.
But the required number is greater than 6000 and less than 7000. So, thousand places can be filled with 1 digit (i.e) 6.

So, the total number of integers =1 × 8 × 7 × 2 = 112
Hence, the required number of integers = 112

Question 3.
Find the number of integers greater than 7000 that can be formed with the digits 3, 5, 7, 8 and 9 where no digits are repeated.
Solution:
Given that all the 5 digit numbers are greater than 7000.
So, the ways of forming 5-digit numbers = 5 × 4 × 3 × 2 × 1 = 120
Now, all the four-digit numbers greater than 7000 can be formed as follows.
Thousand places can be filled with 3 ways
Hundred places can be filled with 4 ways
Tenths place can be filled with 3 ways
Units place can be filled with 2 ways
So, the total number of 4-digits numbers = 3 × 4 × 3 × 2 = 72
∴ Total number of integers = 120 + 72 = 192
Hence, the required number of integers = 192

Question 4.
How many words (with or without dictionary meaning) can be made from the letters of the word MONDAY, assuming that no letter is repeated, if

Solution:

(c) All letters are used but the first is a vowel = 2 × 5! = 2 × 120 = 240
Hence, the required matching is
(a) ↔ (iii), (b) ↔ (i), (c) ↔ (ii)

Question 5.
Five boys and 5 girls form a line. Find the number of ways of making the seating arrangement under the following condition.

Solution:
(a) Total number of arrangement when boys and girls alternate : = (5!)² + (5!)²
(b) No two girls sit together = 5! 6!
(c) All the girls sit together = 2! 5! 5!
(d) All the girls sit never together = 10! – 5! 6!
Hence, the required matching is (a) ↔ (iii), (b) ↔ (i), (c) ↔ (iv), (d) ↔ (ii)

Question 6.
How many automobile license plates can be made if each plate contains two different letters followed by three different digits?
Solution:
We have 26 English alphabet and 10 digits (0 to 9)
Since it is given that each plate contains 2 different letters followed by 3 different digits.
∴ Number of the arrangement of 26 letters taken 2 at a time

A three-digit number can be formed out of 10 digits = ¹⁰P₃

Total number of license plates = 650 × 720 = 468000
Hence, the required number of plates = 468000.

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 1 Sets Ex 1.5

Choose the correct or the most suitable answer.

Question 1.
If A = {(x, y) : y = e^x ; x ∈ R } and B = {(x, y) : y = e^-x, x ∈ R } then n(A ∩ B)
(a) Infinity
(b) 0
(c) 1
(d) 2
Solution:
(c) 1
Hint.
A∩B = (0, 1)
n(A∩B) = 1

Question 2.
IfA {(x, y) : y = sin x, x ∈ R) and 8= (x, y) : y = cos x, x ∈ R) then A∩B contains ……..
(a) no element
(b) infinitely many elements
(c) only one element
(d) cannot be determined.
Solution:
(b) infinitely many elements

Question 3.
The relation R defined on a set A = {0, -1, 1, 2} by xRy if |x² +y²| ≤ 2, then which one of the following is true?
(a) R = {(0, 0), (0, -1), (0, 1), (-1, 0), (-1, 1), (1, 2), (1, 0)}
(b) R = {(0, 0), (0, -1), (0, 1), (-1, 0), (1, 0)
(c) Domain of R is {0, -1, 1, 2}
Solution:
(a) Range of R is {0, -1, 1}
Hint.
Given A = {0, -1, 1, 2 }
the relation R is given by x R y = |x² + y²| ≤ 2
∴ x, y must be 0 or 1
∴ Range of R is {0, – 1 , 1 }

Question 4.
If f(x) = |x – 2| + |x + 2|, x ∈ R, then

Solution:

Hint.
f(x) = |x – 2| + |x + 2|

Question 5.
Let R be the set of all real numbers. Consider the following subsets of the plane R x R: S = {(x, y): y = x + 1 and 0 < x < 2} and T = {(x, y) : x – y is an integer} Then which of the following is true?
(a) T is an equivalence relation but S is not an equivalence relation.
(b) Neither S nor T is an equivalence relation
(c) Both S and T are equivalence relations
(d) S is an equivalence relation but T is not an equivalence relation.
Solution:
(a) T is an equivalence relation but S is not an equivalence relation.
Hint.

Given R is the set of all real numbers
S = { (x, y): y = x + 1 and 0 < x < 2}
T = { (x, y): x – y is an integer} are subsets of R × R
s = { (x, y): y = x + 1 and 0 < x < 2} for x ∈ R,
x = x + 1 is not possible. ∴ (x , x) ∉ S
Hence S does not satisfy the reflexive property
∴ S is not an equivalence relation
T = {(x, y): x – y is an integer}

Reflexive:
For x ∈ R, we have x – x = 0 is an integer.
∴ (x,x)∈T forall X∈ R
Hence T satisfies reflexive property

Symmetric:
Let (x, y) ∈ T, then x – y
⇒ – (x – y) is an integer
⇒ y – x is an integer
⇒ (y, x) ∈ T
∴ T satisfies the symmetric property

Transitive:
Let (x, y), (y, z) ∈ T then x – y and y – z are integers.
⇒ x – y + y – z is an integer
⇒ x – z is an integer
⇒ (x, z) ∈ T
∴ T satisfies the transitive property
we have proved T is reflexive, symmetric, and transitive.
Thus T is an equivalence relation.

Question 6.
Let A and B be subsets of the universal set N, the set of natural numbers. Then
A’ ∪ [(A ∩ B) ∪ B’] is ………
(a) A
(b) A’
(c) B
(d) N
Solution:
(d) N
Hint.

Question 7.
The number of students who take both the subjects Mathematics and Chemistry is 70. This represents 10% of the enrollment in Mathematics and 14% of the enrollment in Chemistry. How many students take at least one of these two subjects?
(a) 1120
(b) 1130
(c) 1100
(d) insufficient data
Solution:
(b) 1130
Hint.
Let M denotes Mathematics students
C denotes Chemistry students
Given n(M ∩ C) = 70
10 % of the enrolement in Mathematics
Out of 100 enrolement 10 students take mathematics
∴ Number of Mathematics students n (M) = \(\frac{100}{10}\) × 70
n (M ) = 700
Number of Chemistry students n(C) = \(\frac{100}{14}\) × 70
n (C) = 500
∴ n(M ∪ C) = n(M) + n(C) – n(M ∩ C)
= 700 + 500 – 70
= 1200 – 70
= 1130
The number of students take atleast one of the subject mathematics or Chemistry = 1130

Question 8.
If n[(A × B) ∩ (A × C)] = 8 and n(B ∩ C) = 2 , then n(A) is
(a) 6
(b) 4
(c) 8
(d) 16
Solution:
(b) 4

Question 9.
If n(A) = 2 and n(B ∪ C) = 3, then n[(A × B) ∪ (A × C)] is …….
(a) 2³
(b) 3²
(c) 6
(d) 5
Solution:
(c) 6
Hint.
Given n (A) = 2 and n(B ∪ C) = 3
n[(A × B) ∪ (A × C)] = n[A × ( B ∪ C ) ]
A × (B ∪ C) = (A × B) ∪ (A × C)
= n(A) . n(B ∪ C)
= 2 × 3 = 6

Question 10.
If two sets A and B have 17 elements in common, then the number of elements common to the set A × B and B × A is
(a) 2¹⁷
(b) 17²
(c) 34
(d) insufficient data
Solution:
(b) 17²
Hint.
n (A ∩ B) = 17
So n [(A × B) ∩ (B × A)]
= n(A ∩ B) × n(B ∩ A) = 17 × 17 = 17²

Question 11.
For non-empty sets A and B, if A ⊂ B then (A × B) ∩ (B × A) is equal to ……….
(a) A ∩ B
(b) A × A
(c) B × B
(d) None of these
Solution:
(b) A × A
Hint.
Given A ⊂ B, take A = { 1, 2 } and B = { 1, 2 , 3 }
A × B = {1, 2} × {1, 2, 3}
A × B = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3)}
B × A = {1, 2, 3} × {1, 2}
B × A = {(1, 1), (1, 2), (2, 1), (2, 2), (3, 1), (3, 2)}
(A × B) ∩ (B × A) = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3)} ∩ {(1, 1), (1, 2), (2, 1),
(2, 2), (3, 1), (3, 2)}
(A × B) ∩ (B × A) = {(1, 1), (1, 2), (2, 1), (2, 2)}
A × A = {1, 2} × {1, 2}
A × A = { (1, 1), (1, 2), (2, 1), (2, 2) }
(A × B) × (B × A) = A × A

Question 12.
The number of relations on a set containing 3 elements is
(a) 9
(b) 81
(c) 512
(d) 1024
Solution:
(c) 512
Hint.
Number of relations = 2ⁿ² = 2³² = 2⁹ = 512

Question 13.
Let R be the universal relation on a set X with more than one element. Then R is
(a) Not reflexive
(b) Not symmetric
(c) Transitive
(d) None of the above
Solution:
(c) Transitive

Question 14.
Let X = {1, 2, 3, 4} and R = {(1, 1), (1, 2), (1, 3), (2, 2), (3, 3), (2, 1), (3, 1), (1,4), (4, 1)}. Then R is ……..
(a) Reflexive
(b) Symmetric
(c) Transitive
(d) Equivalence
Solution:
(b) Symmetric
Hint.
x = {1, 2, 3, 4}
R = { (1, 1) , (1, 2), (1, 3), (2, 2), (3, 3), (2, 1), (3, 1), (1, 4), (4, 1)}

Question 15.

Solution:

Question 16.

Solution:
(c) [0, 1)

Question 17.
The rule f(x) = x² is a bijection if the domain and the co-domain are given by …..
(a) R,R
(b) R, (0, ∞)
(c) (0, ∞), R
(d) [0, ∞), [0, ∞)
Solution:
(d) [0, ∞), [0, ∞)

Question 18.
The number of constant functions from a set containing m elements to a set containing n elements is
(a) mn
(b) m
(c) n
(d) m + n
Solution:
(c) n

Question 19.
The function f: [0, 2π] ➝ [-1, 1] defined by f(x) = sin x is
(a) One to one
(b) Onto
(c) Bijection
(d) Cannot be defined
Solution:
(b) Onto

So it is not one-to-one
So it is an onto function

Question 20.
If the function f : [-3, 3] ➝ S defined by f(x) = x² is onto, then S is ………
(a)[-9, 9]
(b) R
(c) [-3, 3]
(d) [0, 9]
Solution:
(d) [0, 9]

Question 21.
Let X = {1, 2, 3, 4}, Y = {a, b, c, d) and f = {(1, a), (4, b), (2, c), (3, d) (2, d)}. Then f is ………
(a) An one-to-one function
(b) An onto function
(c) A function which is not one-to-one
(d) Not a function
Solution:
(d) Not a function
Hint.

Since the element 2 has two images, it is not a function

Question 22.

Solution:

Question 23.
Let f : R ➝ R be defined by f(x) = 1 – |x|. Then the range of f is ………
(a) R
(b) (1, ∞)
(c) (-1, ∞)
(d) (-∞, 1]
Solution:
(d) (-∞, 1]
Hint.
f(x) = 1 – |x|
When x = 0, f(0) = 1 – 0 = 1
When x = – 2 , f(-2) = 1 – |- 2| = 1 – 2 = -1
When x = – 5 , f(-5) = 1 – |- 5| = 1 – 5 = -4
∴ Range of f is (- ∞, 1]

Question 24.
The function f : R ➝ R is defined by f(x) = sin x + cos x is ……
(a) An odd function
(b) Neither an odd function nor an even function
(c) An even function
(d) Both odd function and even function
Solution:
(b) Neither an odd function nor an even function

Question 25.

(a) An odd function
(b) Neither an odd function nor an even function
(c) An even function
(d) Both odd function arid even function
Solution:
(c) An even function

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 7.4

Question 1.
Find the area of the triangle whose vertices are (0, 0), (1, 2) and (4, 3)
Solution:
Area of triangle with vertices

∴ Area of A with vertices (0, 0), (1, 2) and (4, 3) is

(as the area cannot be negative).

Question 2.
If (k, 2), (2, 4) and (3, 2) are vertices of the triangle of area 4 square units then determine the value of k.
Solution:

Question 3.
Identify the singular and non-singular matrices:

Solution:
(i) For a given square matrix A,
1. If |A| = 0 then it is a singular matrix.
2. If |A| ≠ 0 then it is a non singular matrix.

⇒ A is a singular matrix.

Which is a skew symmetric matrix
∴ |A| = 0 ⇒ A is a singular matrix.

Question 4.
Determine the value of a and b so that the following matrices are singular:

Solution:

expanding along R₁
b(4 + 4) + 7 (-6 – 1) = 0 (given)
8b + 7 (-7) = 0
(i.e.,) 8b – 49 = 0 ⇒ 8b = 49 ⇒ b = 49/8

Question 5.
If cos 2θ = 0, determine \(\left[\begin{array}{ccc}{\theta} & {\cos \theta} & {\sin \theta} \\ {\cos \theta} & {\sin \theta} & {0} \\ {\sin \theta} & {0} & {\cos \theta}\end{array}\right]^{2}\)
Solution:

Question 6.
Find the value of the product;
Sol:

Samacheer Kalvi 11th Maths Solutions Chapter 7 Matrices and Determinants Ex 7.4 Additional Problems

Question 1.
Identify the singular and non-singular matrix.

Solution:
If the determinant value of a square matrix is zero it is called a singular matrix. Otherwise it is non-singular.

Question 2.
Show that
Solution: