RS Aggarwal Solution Class 12 Chapter 23 Algebra of Vectors Exercise 23.2

 Exercise 23.2

Question 1

If P, Q and R are three collinear points such that $\overrightarrow{P Q}=\vec{a}$ and $\overrightarrow{Q R}=\vec{b}$. Find the vector $\overrightarrow{P R}$.

Sol :

As P, Q and R are three collinear points.

Hence, $\overrightarrow{\mathrm{PR}}=\overrightarrow{\mathrm{PQ}}+\overrightarrow{\mathrm{QR}}$ as shown in above fig

And given $\overrightarrow{\mathrm{PQ}}=\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{QR}}=\overrightarrow{\mathrm{b}}$

Therefore, $\overrightarrow{\mathrm{PR}}=\overrightarrow{\mathrm{PQ}}+\overrightarrow{\mathrm{QR}}=\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}$

Question 2

Give condition that three vectors $\vec{a}, \vec{b}$, and $\vec{c}$ form the three sides of a triangle. What are the other possibilities?

Sol :

Given that, $\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{c}}$ are three sides of a triangle.

Hence from the above figure we get,

$\mathrm{AB}=\overrightarrow{\mathrm{a}}, \mathrm{BC}=\overrightarrow{\mathrm{b}}$ and $\mathrm{AC}=\overrightarrow{\mathrm{c}}$

So, $\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}$=AB+BC+CA=AC+CA

[Since AB+BC=AC]

=AC-AC=0 [Since CA=-AC]

Triangle law says that, if vectors are represented in magnitude and direction by the two sides of a triangle is same order, then their sum is represented by the third side took in reverse order. Thus,

$\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}=-\overrightarrow{\mathrm{c}}$ or $\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{c}}=-\overrightarrow{\mathrm{b}}$ or $\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=-\overrightarrow{\mathrm{a}}$

Question 3

If $\vec{a}$ and $\vec{b}$ are two non-collinear vectors having the same initial point. What are the vectors represented by $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ ?

Sol :

Here, it is given that $\vec{a}$ and $\vec{b}$ are two non-collinear vectors having the same initial point.

Let $\vec{a}=\overrightarrow{A B}$ and $\vec{b}=\overrightarrow{A D}$, So we can draw a parallelogram ABCD as above.

By the properties of parallelogram

$\overrightarrow{B C}=\vec{b}$ and $\overrightarrow{D C}=\vec{a}$

In ΔABC,

Using triangle law,

$\overrightarrow{A B}+\overrightarrow{B C}=\overrightarrow{A C}$

$\vec{a}+\vec{b}=\overrightarrow{A C}$...(i)

In ΔABD,

Using triangle law,

$\overrightarrow{A D}+\overrightarrow{D B}=\overrightarrow{A B}$

$\vec{b}+\overrightarrow{D B}=\vec{a}$

$\overrightarrow{D B}=\vec{a}-\vec{b}$...(ii)

From equation (i) and (ii), we get that

$\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ are diagonals of a parallelogram whose adjacent sides are $\vec{a}$ and $\vec{b}$

Question 4

If $\vec{a}$ is a vector and m is a scalar such that $m \vec{a}=\overrightarrow{0}$, then what are the alternatives for m and $\vec{a}$ ?

Sol :

Given $\vec{a}$ is a vector and m is a scalar such that $m \vec{a}=\overrightarrow{0}$

Let $\vec{a}=a_{1} \hat{i}+b_{1} \hat{j}+c_{1} \hat{k}$ then according to the given question

$m \vec{a}=\overrightarrow{0}$

$\Rightarrow \mathrm{m}\left(\mathrm{a}_{1} \hat{\imath}+\mathrm{b}_{1} \hat{\jmath}+\mathrm{c}_{1} \hat{\mathrm{k}}\right)=0 \hat{\imath}+0 \hat{\jmath}+0 \hat{\mathrm{k}}$

$\Rightarrow\left(\mathrm{ma}_{1} \hat{\imath}+\mathrm{mb}_{1} \hat{\jmath}+\mathrm{mc}_{1} \hat{\mathrm{k}}\right)=0 \hat{\imath}+0 \hat{\jmath}+0 \hat{\mathrm{k}}$

Compare the coefficients of $\hat{1}, \hat{\jmath}, \hat{k}$ we get

$\mathrm{ma}_{1}=0 \Rightarrow \mathrm{m}=0$ or $\mathrm{a}_{1}=0$

Similarly, $\mathrm{mb}_{1}=0 \Rightarrow \mathrm{m}=0$ or $\mathrm{b}_{1}=0$

And $m c_{1}=0 \Rightarrow m=0$ or $c_{1}=0$

From the above three conditions ,

m=0 or $a_{1}=b_{1}=c_{1}=0$

$\Rightarrow \mathrm{m}=0$ or $\overrightarrow{\mathrm{a}}=\mathrm{a}_{1} \hat{\imath}+\mathrm{b}_{1} \hat{\jmath}+c_{1} \hat{\mathrm{k}}$

$=0 \hat{\imath}+0 \hat{\jmath}+0 \hat{\mathrm{k}}=0$

Hence the alternatives for m and $\vec{a}$ are m=0 or $\vec{a}=0$

Question 5

If $\vec{a} \vec{b}$ are two vectors, then write the truth value of the following statements:

(i) $\vec{a}=-\vec{b} \Rightarrow|\vec{a}|=|\vec{b}|$

(ii) $|\vec{a}|=|\vec{b}| \Rightarrow \vec{a}=\pm \vec{b}$

(iii) $|\vec{a}|=|\vec{b}| \Rightarrow \vec{a}=\vec{b}$

Sol :

(i)

Let $\vec{a}=a_{1} \hat{i}+b_{1} \hat{j}+c_{1} \hat{k}$

$\vec{b}=a_{2} \hat{i}+b_{2} \hat{j}+c_{2} \hat{k}$

Given that, a=-b

$a_{1} \hat{i}+b_{1} \hat{j}+c_{1} \hat{k}=-a_{2} \hat{i}-b_{2} \hat{j}-c_{2} \hat{k}$

Comparing the coefficients of i, j, k in LHS and RHS,

$a_{1}=-a_{2}$...(1)

$b_{1}=-b_{2}$...(2)
$c_{1}=-c_{2}$...(3)
$|\vec{a}|=\sqrt{{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}}$

Using (1),(2) and (3),

$|\vec{a}|=\sqrt{\left(-a_{2}\right)^{2}+\left(-b_{2}\right)^{2}+\left(-c_{2}\right)^{2}}$

$|\vec{a}|=\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}$

$\therefore|\vec{a}|=|\vec{b}|$

(ii)

Given a and b are two vectors such that $|\vec{a}|=|\vec{b}|$

It means magnitude of vector $\vec{a}$ is equal to the magnitude of vector $\vec{b}$, but we cannot conclude anything about the direction of the vector.

So,it is false that

$|\vec{a}|=|\vec{b}| \Rightarrow \vec{a}=\pm \vec{b}$

(iii)

Given for any vector $\vec{a}$ and $\vec{b}$ 

$|\vec{a}|=|\vec{b}|$

It means magnitude of the vector $\vec{a}$ and $\vec{b}$ are equal but we cannot say any thing about the direction of the vector $\vec{a}$ and $\vec{b}$. And we know that $\vec{a}=\vec{b}$ means magnitude and same direction. So, it is false.

Question 6

ABCD is a quadrilateral. Find the sum of the vectors $\overrightarrow{B A}, \overrightarrow{B C}, \overrightarrow{C D}$ and $\overrightarrow{D A}$.

Sol :

Here it given that ABCD is a quadrilateral.

In ΔADC, using triangle law, we get

$\overrightarrow{C D}+\overrightarrow{D A}=\overrightarrow{C A}$...(i)

In ΔABC, using triangle law, we get

$\overrightarrow{B C}+\overrightarrow{C A}=\overrightarrow{B A}$...(ii)

Put value of $\overrightarrow{C A}$ in equation (ii)

$\overrightarrow{B C}+\overrightarrow{C D}+\overrightarrow{D A}=\overrightarrow{B A}$

Adding $\overrightarrow{B A}$ on both the sides,

$\overrightarrow{B A}+\overrightarrow{B C}+\overrightarrow{C D}+\overrightarrow{D A}=\overrightarrow{B A}+\overrightarrow{B A}$

$\therefore \overrightarrow{B A}+\overrightarrow{B C}+\overrightarrow{C D}+\overrightarrow{D A}=2 \overrightarrow{B A}$

Question 7

ABCDE is a pentagon, prove that

(i) $\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{C D}+\overrightarrow{D E}+\overrightarrow{E A}=0$

Sol :

Given: ABCDE is a pentagon as shown below

Consider ΔABC and apply triangle law of vector, we get

$\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{AC}}$....(i)

Similarly, consider ΔACD and apply triangle law of vector, we get

$\overrightarrow{\mathrm{AC}}+\overrightarrow{\mathrm{CD}}=\overrightarrow{\mathrm{AD}}$...(ii)

And, consider ΔADE and apply triangle law of vector, we get

$\overrightarrow{\mathrm{AD}}+\overrightarrow{\mathrm{DE}}=\overrightarrow{\mathrm{AE}}$...(iii)

Adding (i), (ii) and (iii), we get

$\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{AC}}+\overrightarrow{\mathrm{CD}}+\overrightarrow{\mathrm{AD}}+\overrightarrow{\mathrm{DE}}=\overrightarrow{\mathrm{AE}}+\overrightarrow{\mathrm{AD}}+\overrightarrow{\mathrm{AC}}$

$\Rightarrow \overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CD}}+\overrightarrow{\mathrm{AD}}+\overrightarrow{\mathrm{DE}}=\overrightarrow{\mathrm{AE}}+\overrightarrow{\mathrm{AD}}$

$\Rightarrow \overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CD}}+\overrightarrow{\mathrm{DE}}=\overrightarrow{\mathrm{AE}}$

$\Rightarrow \overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CD}}+\overrightarrow{\mathrm{DE}}=-\overrightarrow{\mathrm{EA}}~[\text{ as } \overrightarrow{\mathrm{AE}}=-\overrightarrow{\mathrm{EA}}]$

$\Rightarrow \overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CD}}+\overrightarrow{\mathrm{DE}}+\overrightarrow{\mathrm{EA}}=0$

Hence Proved

(ii) $\overrightarrow{A B}+\overrightarrow{A E}+\overrightarrow{B C}+\overrightarrow{D C}+\overrightarrow{E D}+\overrightarrow{A C}=3 \overrightarrow{A C}$

Sol :

Given: ABCDE is a pentagon as shown below

Consider ΔABC and apply triangle law of vector, we get

$\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{AC}}$....(i)

RS Aggarwal Solution Class 12 Chapter 23 Algebra of Vectors Exercise 23.1

Exercise 23.1

Question 1

(i) Represent graphically a displacement of 40 km, 30° east of north.

Sol :

(i) a displacement of 40 km, 30° east of north

Step 1: Draw north, south, east and west as shown below:

Step 2: Plot a line  30° east of north as shown below:

Step 3: Define scale and mark 40km on line $\overrightarrow{\mathrm{OP}}$

Let the scale be 10km = 1cm

∴$\overrightarrow{\mathrm{OP}}$represents the displacement of 40 km, 30° East of North

(ii) Represent graphically a displacement of 50 km, south-east

Sol :

(ii) a displacement of 50 km south - east

Step 1: Draw north, south, east and west as shown below:

Step 2: As the displacement should be south - east, the angle between the displacement and east (or south) will be 45°. Now, plot a line $\overrightarrow{\mathrm{OP}}$ 45° east of south as shown below:

Step 3: Define scale and mark point R such that OR = 50km on line $\overrightarrow{\mathrm{OP}}$. Let the scale be 10km = 1cm

∴$\overrightarrow{\mathrm{OR}}$ represents the displacement of 50 km south – east

(iii) Represent graphically a displacement of 70 km, 40° north of west.

Sol :

(iii) A displacement of 70 km, 40° north of west.

Step 1: Draw north, south, east and west as shown below:

Step 2: Plot a line $\overrightarrow{\mathrm{OP}}$ 40° north of west as shown below:

Step 3: Define scale and mark point R such that OR = 70km on line $\overrightarrow{\mathrm{OP}}$.Let the scale be 10km = 1cm

∴$\overrightarrow{\mathrm{OP}}$  represents the displacement of 70 km, 40o north of west

Question 2

Classify the following measures as scalars and vectors.

(i) 15 kg

Sol :

15 kg - is a scalar quantity as this involves only mass. A scalar quantity is a one - dimensional measurement of a quantity, like temperature, or mass.

(ii) 20 kg weight

Sol :

20 kg weight - is a vector quantity as it involves both magnitude and direction. Weight is a force which is a vector and has a magnitude and direction.

(iii) 45°

Sol :

45° is a scalar quantity as it involves the only magnitude. A scalar quantity is a one - dimensional measurement of a quantity, like temperature, or mass.

(iv) 10 metres south-east

Sol :

10 meters south - east is a vector quantity as it involves both magnitude and direction.

(v) 50 m/s²

Sol :

50 m/sec² is a scalar quantity as it involves a magnitude of acceleration. A scalar quantity is a one - dimensional measurement of a quantity.

Question 3

Classify the following as scalar and vector quantities.

(i) Time period

Sol :

Time period - is a scalar quantity as it involves only magnitude. A scalar quantity is a one - dimensional measurement of a quantity. Eg: 10 seconds has only magnitude, i.e., 10 and no direction.

(ii) Distance

Sol :

Distance - is a scalar quantity as it involves only magnitude. A scalar quantity is a one dimensional measurement of a quantity. Eg: 5meters has only magnitude 5 and no direction.

(iii) Displacement

Sol :

Displacement - is vector quantity as it involves both magnitude and direction. Vector quantity has both magnitude and direction.

(iv) Force

Sol :

Force - is a vector quantity as it involves both magnitude and direction. Vector quantity has both magnitude and direction. Eg., 5N downward has magnitude of 5 and direction is downward.

(v) Work

Sol :

Work done - is a scalar quantity as it involves only magnitude and no particular direction. A scalar quantity is a one dimensional measurement of a quantity.

(vi) Velocity

Sol :

Velocity - is a vector quantity as it involves both magnitude as well as direction. Vector quantity has both magnitude and direction. Eg., 5m/s east has magnitude of 5m/s and also direction towards east.

(vii) Acceleration

Sol :

Acceleration is a vector quantity because it involves both magnitude as well as direction.

Question 4

Which vectors are:

(i) Collinear

(ii) Equal

(iii) Coinitial

(iv) Collinear but not equal.

Sol :

(i) Collinear vectors are

$\vec{x}, \vec{z}$ and $\vec{b}$

$\vec{y}, \vec{c}$

$\vec{a}, \vec{d}$

(ii) Equal vectors are

$\vec{y}$ and $\vec{c}$

$\vec{x}$ and $\vec{b}$

$\vec{a}$ and $\vec{d}$

(iii) Coinitial vector are $\vec{a}, \vec{y}$ and $\vec{z}$

(iv) Collinear but not equal

$\vec{b}$ and $\vec{z}$

$\vec{x}$ and $\vec{z}$

Question 5

Answer the following as true or false:

(i) a and b are collinear.

(ii) Two collinear vectors are always equal in magnitude.

(iii). Zero vector is unique.

(iv) Two vectors having same magnitude are collinear.

(v) Two collinear vectors having the same magnitude are equal.

Sol :

(i) $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{a}}$  are collinear. (True)

Two or more vectors that lie on the same line or on a parallel line to this are called collinear vectors.

$\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{a}}$ are collinear.

(ii) Two collinear vectors are always equal in magnitude. (False)

Two or more vectors that lie on the same line or on a parallel line to this are called collinear vectors. Two collinear vectors may point in either same or opposite direction. And they are not necessarily equal in magnitude they can be of different magnitude also.

(iii) Zero vector is unique.(True)

There is only one zero - vector in a vector space. Hence zero vector is unique.

(iv) Two vectors having same magnitude are collinear. (False)

It is not necessary for two vectors having the same magnitude to be parallel to the same line. Hence two vectors having same magnitude need not be collinear.

(v) Two collinear vectors having the same magnitude are equal.(False)

Two vectors are said to be equal if they have the same magnitude and direction, regardless of the positions of their initial points.

ML AGGARWAL CLASS 12 Chapter 1 Relations and Functions Exercise 1.2 (CBSE)

 Exercise 1.2

Question 1

Sol :

ML AGGARWAL CLASS 12 Chapter 1 Relations and Functions Exercise 1.1 (CBSE)

 Exercise 1.1

Question 1

Determine whether each of the following relations are reflexive, symmetric and transitive : 

(i) Relation R in the set A = {1, 2, 3, …, 10} defined by R = {(x, y) : 2x – y = 0}. 

(ii) Relation R in the set Z of all integers defined by 

R = {(x, y) : x – y is an integer}

(iii) Relation R in the set N of all natural numbers defined by 

R = {(x, y) : y = x + 5, x < 4}.

Sol :

Question 2

If the relation R in the set A, where A = {1, 2, 3, 4, 5, 6}, is defined by R = {(x, y) : y is divisible by x}, then express R in the roster form. Also determine whether the relation R is 

(i) reflexive (ii) symmetric (iii) transitive.

Sol :

Question 3

If R is the relation defined on the set of natural numbers N as follows: 

R = {(x, y); x, y ∈ N, 2x + y = 41}, 

find the domain and the range of the relation R. 

Determine whether the relation is reflexive, symmetric and transitive.

Sol :

Question 4

Determine whether each of the following relations in the set A of human beings in a city at a particular time are reflexive, symmetric and transitive :

(i) R = {(x, y) : x and y work at the same place}.

(ii) R = {(x, y) : x and y live in the same locality}.

(iii) R = {(x, y) : x is exactly 5 cm taller than y}. 

(iv) R = {(x, y) : x and y live within 2 kilometres}. 

(v) R = {(x, y) : x is wife of y}.

Sol :

Question 5

Determine whether each of the following relations in the set A of students at a particular time are symmetric and transitive:

(i) R = {(x, y) : x, y ∈ A, x and y are honest} 

(ii) R = {(x, y) : x, y ∈ A, x and y are obedient} 

(iii) R = {(x, y) : x, y ∈ A, x and y are hardworking} 

What are the advantages of students being honest, obedient and hardworking?

Sol :

Question 6

Show that the relation R in the set A = {1, 2, 3} given by R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)} is reflexive but neither symmetric nor transitive.

Sol :

Question 7

Decide in each of the following cases whether the relation is symmetric, transitive and reflexive. Justify your answer by giving examples.

(i) ‘Is less than’ on N 

(ii) ‘Is a factor of’ on N

Sol :

Question 8

Let T be the set of all triangles drawn in a plane with R as a relation in T given by $R = {(T_1, T_2) : T_1 ≅ T_2}$. Show that R is an equivalence relation.

Sol :

Question 9

Show that the relation R in the set A of all books in a library of a school, given by

R = {(x, y) : x and y have same number of pages}, is an equivalence relation. 

Sol :

Question 10

Show that the relation R in the set A of points in a plane, given by R = {(P, Q) : points P and Q have equal distances from the origin}, is an equivalence relation. Also show that the set of all points related to a point P (different from origin) is the circle passing through P with origin as its centre.

Sol :

Question 11

Show that the relation R in the set A of all triangles, given by $R = {(T_1, T_2) : T_1 ~ T_2}$, is an equivalence relation. Consider three triangles $T_1$ with sides 3, 4, 5; $T_2$ with sides 5, 12, 13 and $T_3$ with sides 6, 8, 10. Which triangles among $T_1$, $T_2$ and $T3_$ are related?

Sol :

Question 12

Show that the relation R in the set A of all polygons, given by $R = {(P_1, P_2) : P_1$ and $P_2$ have same number of sides}, is an equivalence relation. What is the set of all elements in A related to the right triangle T with sides 3, 4 and 5?

Sol :

Question 13

Show that the relation R defined in the set L of all straight lines drawn in the XY-plane, given by $R = {(L_1, L_2) : L_1 \text{ is parallel to }L_2}$, is an equivalence relation. Find the set of all straight lines related to the line y = 2x + 4.

Sol :

Question 14

Show that the relation R on the set I of all integers defined by R = {(a, b) : a – b is divisible by 3, a, b ∈ I} is an equivalence relation

Sol :

Question 15

Let I be the set of all integers and R be the relation on I defined by R = {(a, b) : a – b is divisible by 5}. Prove that R is an equivalence relation. Find the set of all elements of I related to 1.

Sol :

Question 16

Give examples of relations which are

(i) symmetric but neither reflexive nor transitive.

(ii) transitive but neither reflexive nor symmetric.

(iii) symmetric and transitive but not reflexive.

(iv) symmetric and reflexive but not transitive.

(v) reflexive and transitive but not symmetric. 

Sol :

Question 17

Give an example of a relation R on A = {a, b, c} which is 

(i) neither reflexive nor symmetric but transitive. 

(ii) neither symmetric nor transitive but reflexive. 

(iii) neither transitive nor reflexive but symmetric.

Sol :

Question 18

Show that the relation R in the set A = {1, 2, 3, 4, 5}, given by R = {(a, b) : |a – b| is even}, is an equivalence relation. Also show that all the elements of {1, 3, 5} are related to each other and all the elements of {2, 4} are related to each other but no element of {1, 3, 5} is related to any element of {2, 4}.

Sol :

Question 19

Show that the relation S in the set A = {x ∈ Z : 0 ≤ x ≤ 12} given by

S = {(a, b) : a, b ∈ A, | a – b | is divisible by 4} is an equivalence relation.

Find the set of elements related to 1

Sol :

Question 20

Show that the relation R in the set A = {x ∈ W, 0 ≤ x ≤ 17} given by

(i) R = {(a, b) : |a – b| is a multiple of 5} 

(ii) R = {(a, b) : a = b} 

are equivalence relations. Find the set of all elements related to 2 in each case.

Sol :

Question 21

Consider the division of set A = {1, 2, 3, 4, 5, 6, 7, 8} by subsets {1, 6}, {2, 7}, {3, 8}, {4} and {5}. Show that the relation R in A, given by R = {(a, b) : a and b lie in the same subset of the division of A}, is an equivalence relation. Find the set of all elements of A related to 6 and the set of all elements related to 5.

Sol :

Very Short Answer Type Question

Question 22

If A = {– 1, 1, 3}, then what is the number of relations on A?

Sol :

Question 23

Let A be any non-empty set. State true or false : 

(i) Identity relation on A is reflexive. 

(ii) Every reflexive relation on A is identity relation on A. 

(iii) Identity relation on A is symmetric. 

(iv) Identity relation on A is an equivalence relation. 

(v) Universal relation on A is an equivalence relation

Sol :

Question 24

State the reason for the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} not to be transitive.

Sol :

Question 25

If A = {0, 1, 2, …, 9} and the relation R on A is defined by R = {(x, y) : x, y ∈ A, y = 2x + 1}, then determine whether the relation R is 

(i) reflexive (ii) symmetric (iii) transitive

Sol :

Question 26

Let R be the relation on the set N given by R = {(a, b) : a = b – 2, b > 6}, then determine whether 

(i) (2, 4) ∈ R

(ii) (6, 8) ∈ R

(iii) (9, 7) ∈ R.

Sol :

Question 27

If the relation R on the set N of natural numbers is defined by

R = {(a, b) : a, b ∈ N, a = 2b – 1, b > 4}, then determine whether

(i) (6, 11) ∈ R

(ii) (13, 7) ∈ R

(iii) (5, 3) ∈ R

Sol :

Question 28

If P be the set of people living in Delhi and R be the relation on P defined by

R = {(a, b) : a, b ∈ P, a lives within 4 km of b}, then determine whether R is transitive.

Sol :

Question 29

Is the relation R on the set R of real numbers defined by

R = {(a, b) : a, b ∈ R, 1 + ab ≥ 0} transitive? Justify your answer.

Sol :

Question 30

Is the relation R on the set Q of rational numbers defined by

$R = {(x, y) : x, y ∈ Q, x < y^2}$, symmetric? Justify your answer

Sol :

Question 31

If A = {1, 3, 7} and R be the relation ‘is greater than’ on the set A. Write R as a set of ordered pairs. Is this relation an equivalence relation?

Sol :

Question 32

If the relation R on the set A = {1, 2, 3} is defined by R = {(1, 1), (2, 2), (3, 3), (2 1), (3, 2)}, then determine whether the relation R is 

(i) reflexive (ii) symmetric (iii) transitive

Sol :

Question 33

If R be the relation in the set {1, 2, 3, 4} given by R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}, then determine whether 

(i) R is reflexive and symmetric but not transitive. 

(ii) R is reflexive and transitive but not symmetric. 

(iii) R is symmetric and transitive but not reflexive

Sol :

ML AGGARWAL CLASS 12 Chapter 6 Applications of derivative Exercise 6.2 (CBSE)

applied math 6.1 ml aggarwal class 12 

 Exercise 6.1

Question 1

Solve for x

(i) x(x-2)(x-5)(x+3)>0

Sol :

Here, x=0,2,5,3

Let's check sign's of intervals 

-∞ and -3

Take any value between -∞ and -3

let say -10

and put it in x(x-2)(x-5)(x+3)

⇒-10(-10-2)(-10-5)(-10+3)

⇒-10(-12)(-15)(-7)

⇒12600

which means overall sign is (+ve) in interval -∞ and -3

In the same way, we going to find sign's for intervals

-3 and 0 (-ve)

0 and 2 (+ve)

2 and 5 (-ve)

5 and ∞ (+ve)

Since, the function is greater than 0, it include all positive intervals

So, x∈(-∞,-3)∪(0,2)∪(5,∞)

(ii) $x^{4}-5 x^{2}+4 \geqslant 0$

Sol :

Let $x^2=t$

$t^{2}-5 t+4 \geqslant 0$

$t^{2}-4 t-t+4 \geqslant 0$

(t-4)(t-1)≥0

$\left(x^{2}-4\right)\left(x^{2}-1\right) \geqslant 0$

(x-2)(x+2)(x-1)(x+1)≥0

So, x=2,-2,1-1

Since, the function is greater than 0, it include all positive intervals

x∈(-∞,-2]∪[-1,1]∪[2,∞)

Question 2

Find all real values of n :

(i) $x^{3}(x-1)(x-2)>0$

Sol :

Here , x=0,1,2

Since, the function is greater than 0, it include all positive intervals

Solution set : x∈(1,0)∪(2,∞)

(ii) $x^{2}(x-1)(x-2) \leq 0$

Sol :

Here, x=0,1,2

For (-∞,0) from inequality it is +ve

For (0,1) from inequality it is +ve

For (1,2) from inequality it is -ve

For (2,∞) from inequality it is +ve

Since, the function is less than 0, it include all negative intervals. Also, we include zero as well.

Solution set : x∈[1,2]∪{0}

Question 3

Solve for x

(i) $\frac{1}{x-2} \leq 1$

Sol :

$\frac{1}{x-2} \leq 1$ , x≠2

Multiplying both side by $(x-2)^{2}$

$(x-2)^{2} \times \frac{1}{(x-2)} \leq 1 \times(x-2)^{2}$

$(x-2) \leqslant x^{2}-4 x+4$

$0 \leq x^{2}-5 x+6$

$x^{2}-2x-3x+4 \geq 0$

(x-2)(x-3)≥0

Here, x=2,3

For (-∞,2) from inequality it is +ve

For (2,3) from inequality it is -ve

For (3,∞) from inequality it is +ve

Since, the function is greater than 0, it include all positive intervals

Solution set : x∈(-∞,2)∪[3,∞)

(ii) $\frac{(x+1)(x-3)}{(x+2)} \geqslant 0$

Sol :

$\frac{(x+1)(x-3)}{(x+2)} \geqslant 0$, x≠-2

Multiplying both side by (x+2)

(x+1)(x-3)(x+2)≥0

Here, x=-1,+3,-2

For (-∞,-2) from inequality it is -ve

For (-2,-1) from inequality it is +ve

For (-1,3) from inequality it is -ve

For (3,∞) from inequality it is +ve

Since, the function is greater than 0, it include all positive intervals

Solution set : x∈(-2,-1]∪[3,∞)

Question 1