Category: 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 4 Combinatorics and Mathematical Induction Ex 4.5

Question 1.
The sum of the digits at the 10th place of all numbers formed with the help of 2, 4, 5, 7 taken all at a time is ……..
(a) 432
(b) 108
(c) 36
(d) 18
Solution:
(b) 108
Hint. Number of digits given = 2, 4, 5, 7
Number of 4 digit numbers formed = 4! =24
So each digit occur \(\frac{24}{4}\) = 6 times
Sum of the digits = 2 + 4 + 5 + 7 = 18
So sum of the digits in each place = 18 × 6 = 108

Question 2.
In an examination there are three multiple choice questions and each question has 5 choices. Number of ways in which a student can fail to get all answer correct is ……..
(a) 125
(b) 124
(c) 64
(d) 63
Solution:
(b) 124
Hint. Each question has 5 options in which 1 is correct
So the number of ways of getting correct answer for all the three questions is 5³ = 125
So the number of ways in which a student can fail to get all answer correct is < 125 (i.e.) 125 – 1 = 124

Question 3.
The number of ways in which the following prize be given to a class of 30 boys first and second in mathematics, first and second in physics, first in chemistry and first in English is ……….
(a) 30⁴ × 29²
(b) 30³ × 29³
(c) 30² × 29⁴
(d) 30 × 29⁵
Solution:
(a) 30⁴ × 29²
Hint.
I and II in maths can be given can be given in 30 × 29 ways.
I and II in physics can be given in 30 × 29 ways.
I and chemistry can be given in 30 ways.
I in English can be given in 30 ways.
So total number of ways = 30 × 29 × 30 × 29 × 30 × 30 = 30⁴ × 29²

Question 4.
The number of 5 digit numbers all digits of which are odd is ………
(a) 25
(b) 5⁵
(c) 5⁶
(d) 625
Solution:
(b) 5⁵
Hint. The odd number are 1, 3, 5, 7, 9
Number of odd numbers = 5
We need a five digit number So the number of five digit number = 5⁵

Question 5.
In 3 fingers, the number of ways four rings can be worn is …… ways.
(a) 4³ – 1
(b) 3⁴
(c) 68
(d) 64
Solution:
(b) 3⁴
Hint. Each letter can be ported in 3 ways
∴ 4 letter is 3⁴ ways

Question 6.

(a) 7 and 11
(b) 6 and 7
(c) 2 and 11
(d) 2 and 6
Solution:
(b) 6 and 7

Question 7.
The product of r consecutive positive integers is divisible by ………
(a) r!
(b) (r – 1)!
(c) (r + 1)!
(d) r!
Solution:
(a) r!
Hint.
1(2) (3) ….. (r) = r! which is ÷ by r!

Question 8.
The number of 5 digit telephone numbers which have none of their digits repeated is
(a) 90000
(b) 10000
(c) 30240
(d) 69760
Solution:
(d) 69760
Hint.
The number of 5 digit telephone numbers which have none of their digits repeated is ¹⁰P₅ = 30240
Thus the required number of telephone number is 10⁵ – 30240 = 69760

Question 9.
If a² – ^aC₂ = a² – ^aC₄ then the value of ‘a’ is ….
(a) 2
(b) 3
(c) 4
(d) 5
Solution:
(b) 3
Hint.
a₂ – a = 2 + 4 = 6
a₂ – a – 6 = 0
(a – 3) (a + 2) = 0 ⇒ a = 3

Question 10.
There are 10 points in a plane and 4 of them are collinear. The number of straight lines joining any two points is ……..
(a) 45
(b) 40
(c) 39
(d) 38
Solution:
(b) 40
Hint.

Question 11.
The number of ways in which a host lady invite 8 people for a party of 8 out of 12 people of whom two do not want to attend the party together is
(a) 2 × ¹¹C₇ + ¹⁰C₈
(b) ¹¹C₇ + ¹⁰C₈
(c) ¹²C₈ – ¹⁰C₆
(d) ¹⁰C₆ + 2!
Solution:
(c) ¹²C₈ – ¹⁰C₆
Hint.
Number of way of selecting 8 people from 12 in ¹²C₈
∴ out of remaining people 8 can attend in ¹⁰C₈
The number of ways in which two of them do not attend together = ¹²C₈ – ¹⁰C₆

Question 12.
The number of parallelograms that can be formed from a set of four parallel lines intersecting another set of three parallel lines …….
(a) 6
(b) 9
(c) 12
(d) 18
Solution:
(d) 18
Hint.
Number of parallelograms = ⁴C₂ × ³C₂
= 6 × 3 = 18

Question 13.
Everybody in a room shakes hands with everybody else. The total number of shake hands is 66. The number of persons in the room is …….
(a) 11
(b) 12
(c) 10
(d) 6
Solution:
(b) 12
Hint.

Question 14.
Number of sides of a polygon having 44 diagonals is ……….
(a) 4
(b) 4!
(c) 11
(d) 22
Solution:
(c) 11
Hint:

Question 15.
If 10 lines are drawn in a plane such that no two of them are parallel and no three are concurrent, then the total number of points of intersection are ………
(a) 45
(b) 40
(c) 10!
(d) 2¹⁰
Solution:
(a) 45
Hint:

Question 16.
In a plane there are 10 points are there out of which 4 points are collinear, then the number of triangles formed is …….
(a) 110
(b) ¹⁰C₃
(c) 120
(d) 116
Solution:
(d) 116
Hint:

Question 17.
In ²ⁿC₃ : ⁿC₃ = 11 : 1 then n is ………
(a) 5
(b) 6
(c) 11
(d) 1
Solution:
(b) 6
Hint.

Question 18.
^{(n – 1)}C_r + ^{(n – 1)}C_{(r – 1)} is ………
(a) ^{(n + 1)}C_r
(b) ^{(n – 1)}C_r
(c) ⁿC_r
(d) ⁿC_{r – 1}
Solution:
(c) ⁿC_r

Question 19.
The number of ways of choosing 5 cards out of a deck of 52 cards which include at least one king is …….
(a) ⁵²C₅
(b) ⁴⁸C₅
(c) ⁵²C₅ + ⁴⁸C₅
(d) ⁵²C₅ – ⁴⁸C₅
Solution:
(d) ⁵²C₅ – ⁴⁸C₅
Hint.
Selecting 5 from 52 cards = ⁵²C₅
selecting 5 from the (non-king cards 48) = ⁴⁸C₅
∴ Number of ways is ⁵²C₅ – ⁴⁸C₅

Question 20.
The number of rectangles that a chessboard has ……
(a) 81
(b) 99
(c) 1296
(d) 6561
Solution:
(c) 1296
Hint. Number of horizontal times = 9
Number of vertical times = 9
Selecting 2 from 9 horizontal lines = ⁹C₂
Selecting 2 from 9 vertical lines = ⁹C₂

Question 21.
The number of 10 digit number that can be written by using the digits 2 and 3 is ……..
(a) ¹⁰C₂ + ⁹C₂
(b) 2¹⁰
(c) 2¹⁰ – 2
(d) 10!
Solution:
(b) 2¹⁰
Hint.
Selecting the number from (2 and 3)
For till the first digit can be done in 2 ways
For till the second digit can be done in 2 ways ….
For till the tenth digit can be done in 2 ways
So, total number of ways in 10 digit number = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2¹⁰

Question 22.
If P_r stands for ^rP_r then the sum of the series 1 + P₁ + 2P₂ + 3P₃ + … + nP_n is ……..
(a) P_{n + 1}
(b) P_{n + 1} – 1
(c) P_{n – 1} + 1
(d) ^{(n + 1)}P_{(n – 1)}
Solution:
(b) P_{n + 1} – 1
Hint:
1 + 1! + 2! + 3! + … + n!
Now 1 + 1 (1!) = 2 = (1 + 1)!
1 + 1 (1!) + 2(2!) = 1 + 1 + 4 = 6 = 3!
1 + 1(1!) + 2(2!)+ 3(3!) = 1 + 1 + 4 + 18 = 24 = 4!
1 + 1(1!) + 2(2!) + 3(3!) ….+ n(n!) = (n + 1) ! – 1
= ^{n + 1}P_{n + 1} – 1 = P_{n + 1} – 1

Question 23.
The product of first n odd natural numbers equals …….

Solution:

Hint:

Question 24.
If ⁿC₄, ⁿC₅, ⁿC₆ are in AP the value of n can be ………..
(a) 14
(b) 11
(c) 9
(d) 5
Solution:
(a) 14
Hint:

30 + n² – 9n + 20 – 12n + 48 = 0
n² – 21 n + 98 = 0
(n – 7) (n – 14) = 0
n = 7 (or) 14

Question 25.
1 + 3 + 5 + 7 + + 17 is equal to ………
(a) 101
(b) 81
(c) 71
(d) 61
Solution:
(b) 81
Hint:

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 8 Vector Algebra – I Ex 8.1

Question 1.
Represent graphically the displacement of
(i) 45 cm 30 ° north of east
(ii) 80 km, 60° south of west
Solution:
(i) 45 cm 30 0 north of east

(ii) 80 km 60° south of west

Question 2.
Prove that the relation R defined on the set V of all vectors by \(\vec{a}\) R \(\vec{b}\) if \(\vec{a}=\vec{b}\) is an equivalence relation on V.
Solution:
\(\vec{a}\) R \(\vec{b}\) is given as \(\vec{a}=\vec{b}\).
(i) \(\vec{a}\) = \(\vec{a}\) ⇒ \(\vec{a}\) R \(\vec{a}\)
(i.e.,) the relation is reflexive.

(ii) \(\vec{a}=\vec{b}\) ⇒ \(\vec{b}\) = \(\vec{a}\)
(i.e.,) \(\vec{a}\) R \(\vec{b}\) – \(\vec{b}\) R \(\vec{a}\)
So, the relation is symmetric.

(iii) \(\vec{a}=\vec{b} ; \vec{b}=\vec{c} \Rightarrow \vec{a}=\vec{c}\)
(i.e„) \(\vec{a}\) R \(\vec{b}\) ; \(\vec{b}\) R \(\vec{c}\) ⇒ \(\vec{a}\) R \(\vec{c}\)
So the given relation is transitive
So, it is an equivalence relation.

Question 3.
Let \(\vec{a}\) and \(\vec{a}\) be the position vectors of the points A and B. Prove that the position vectors of the points which trisects the line segment AB are
Solution:

Question 4.
If D and E are the midpoints of the sides AB and AC of a triangle ABC, prove that

Solution:

Question 5.
Prove that the line segment joining the midpoints of two sides of a triangle is parallel to the third side whose length is half of the length of the third side.
Solution:

Question 6.
Prove that the line segments joining the midpoints of the adjacent sides of a quadrilateral form a parallelogram.
Solution:

In a quadrilateral when opposite sides are equal and parallel it is a parallelogram So, PQRS is a parallelogram, from (1) and (2).

Question 7.
If \(\vec{a}\) and \(\vec{b}\) represent a side and a diagonal of a parallelogram, find the other sides and the other diagonal.
Solution:
OABC is a parallelogram where

Question 8.
If \(\overrightarrow{\mathrm{PO}}+\overrightarrow{\mathrm{OQ}}=\overrightarrow{\mathrm{QO}}+\overrightarrow{\mathrm{OR}}\), prove that the points P, Q, R are collinear.
Solution:

But Q is a common point.
⇒ P, Q, R are collinear.

Question 9.
If D is the midpoint of the side BC of a triangle ABC, prove that \(\overrightarrow{\mathbf{A B}}+\overrightarrow{\mathbf{A C}}=2 \overrightarrow{\mathbf{A D}}\)
Solution:
D is the midpoint of ∆ ABC.

Question 10.
If G is the centroid of a triangle ABC, prove that \(\overrightarrow{\mathrm{GA}}+\overrightarrow{\mathrm{GB}}+\overrightarrow{\mathrm{GC}}=0\)
Solution:
For any triangle ABC,
\(\overrightarrow{\mathrm{GA}}+\overrightarrow{\mathrm{GB}}+\overrightarrow{\mathrm{GC}}=0\)
Now G is the centroid of ∆ABC, which divides the medians (AD, BE and CF) in the ratio 2 : 1.

Question 11.
Let A, B and C be the vertices of a triangle. Let D, E and F be the midpoints of the sides BC, CA, and AB respectively. Show that \(\overrightarrow{\mathrm{AD}}+\overrightarrow{\mathrm{BE}}+\overrightarrow{\mathrm{CF}}=\overrightarrow{0}\)
Solution:
In ∆ABC, D, E, F are the midpoints of BC, CA and AB respectively.

Question 12.
If ABCD is a quadrilateral and E and F are the midpoints of AC and BD respectively, then prove that \(\overrightarrow{\mathbf{A B}}+\overrightarrow{\mathbf{A D}}+\overrightarrow{\mathbf{C B}}+\overrightarrow{\mathbf{C D}}=4 \overrightarrow{\mathrm{EF}}\)
Solution:
ABCD is a quadrilateral in which E and F are the midpoints of AC and BD respectively.

Samacheer Kalvi 11th Maths Solutions Chapter 8 Vector Algebra – I Ex 8.1 Additional Problems

Question 1.
Shown that the points with position vectors are collinear.
Solution:
To prove the points P, Q, R are collinear we have to prove that
\(\overrightarrow{\mathrm{PQ}}\) = t \(\overrightarrow{\mathrm{PR}}\) where t is a scalar.
Let the given points be P, Q, R.

So, the points P, Q, R are collinear (i.e,) the given points are collinear.

Question 2.
If ABC and A’B’C’ are two triangles and G, G’ be their corresponding centroids, prove that \(\overrightarrow{\mathrm{AA}^{\prime}}+\overrightarrow{\mathrm{BB}^{\prime}}+\overrightarrow{\mathrm{CC}^{\prime}}=3 \overrightarrow{\mathrm{GG}}\)
Solution:
Let O be the origin.
We know when G is the centroid of ∆ ABC,

Question 3.
Prove using vectors the mid-points of two opposite sides of a quadrilateral and the mid-points of the diagonals are the vertices of a parallelogram.
Solution:
ABCD is a quadrilateral with position vectors
OA = \(\vec{a}\), OB = \(\vec{b}\), OC = \(\vec{c}\) and OD = \(\vec{d}\)
P is the midpoint of BC and R is the midpoint of AD.
Q is the midpoint of AC and S is the midpoint of BD.
To prove PQRS is a parallelogram. We have to prove that \(\overrightarrow{\mathrm{PQ}}=\overrightarrow{\mathrm{SR}}\)

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)

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)

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.
IfA 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) 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 sides
(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 same order. Then which one of the following statement 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)

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:

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.
A= {0, -1, 1, 2}
|x² + y²| ≤ 2
The values of x and y can be 0, -1 or 1
So range = {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 relation
(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.
(0, 1), (1, 2) it is not an equivalence relation
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.
n(M ∪ C) = n(M) + n(C) – n(M ∩ C)
= 700 + 500 – 70 = 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.
n[(A × B) ∪ (A × C)] = n[ A × (B ∪ 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.

When A ⊂ B, (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.
(4, 4} ∉ R ⇒ R is not reflexive
(1, 4), (4, 1) ∈ R ⇒ R is symmetric
(1, 4), (4, 1) ∈ R but (4, 4) ∉ R
So R is not transitive

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: R ➝ R defined by
f(x) = 1 – |x|
For example,
f(1) = 1 – 1 = 0
f(8) = 1 – 8 = -1
f(-9) = 1 – 9 = -8
f(-0.2) = 1 – 0.2 = 0.8
so range = (-∞, 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 7.3

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 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 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.