ncert books for class maths cbsc · Part

. ( ) . . i − MATHEMATICS Fig . . ( + i ) + i ( + i ) . ( – i ) – ( – + i ) . .   −−     . ( – i ) . . i −− Find the multiplicative inverse of each of the complex numbers given in the Exercises to . . – i . i . – i . Express the following expression in the form of a + ib : ) ( . Argand Plane and Polar Representation We already know that corresponding to each ordered pair of real numbers ( x , y ), we get a unique point in the XY- plane and vice-versa with reference to a set of mutually perpendicular lines known as the x -axis and the y -axis. The complex number x + iy which corresponds to the ordered pair ( x , y ) can be represented geometrically as the unique point P( x , y ) in the XY-plane and vice-versa. Some complex numbers such as + i , – + i , + i , + i , – – i and – i which correspond to the ordered pairs ( , ), ( – , ), ( , ), ( , ), ( – , – ), and ( , – ), respectively, have been represented geometrically by the points A, B, C, D, E, and F, respectively in the Fig . . The plane having a complex number assigned to each of its point is called the complex plane or the Argand plane . COMPLEX NUMBERS AND QUADRATIC EQUATIONS Obviously, in the Argand plane, the modulus of the complex number x + iy = is the distance between the point P( x , y ) and the origin O ( , ) (Fig . ). The points on the x -axis corresponds to the complex numbers of the form a + i and the points on the y -axis corresponds to the complex numbers of the form Fig . Fig . + i b . The x -axis and y -axis in the Argand plane are called, respectively, the real axis and the imaginary axis . The representation of a complex number z = x + iy and its conjugate z = x – iy in the Argand plane are, respectively, the points P ( x, y ) and Q ( x, – y ). Geometrically, the point ( x , – y ) is the mirror image of the point ( x , y ) on the real axis (Fig . ). MATHEMATICS . . Polar representation of a complex number Let the point P represent the non- zero complex number z = x + iy . Let the directed line segment OP be of length r and θ be the angle which OP makes with the positive direction of x -axis (Fig . ). We may note that the point P is uniquely determined by the ordered pair of real numbers ( r , θ ), called the polar coordinates of the point P. We consider the origin as the pole and the positive direction of the x axis as the initial line. We have, x = r cos θ , y = r sin θ and therefore, z = r (cos θ + i sin θ ). The latter is said to be the polar form of the complex number . Here z is the modulus of z and θ is called the argument (or amplitude) of z which is denoted by arg z . For any complex number z ≠ , there corresponds only one value of θ in ≤ θ < π . However, any other interval of length π , for example – π < θ ≤ π , can be such an interval.We shall take the value of θ such that – π < θ ≤ π , called principal argument of z and is denoted by arg z, unless specified otherwise. (Figs. . and . ) Fig . Fig . ( ≤θ < π Fig . (– π < θ ≤ π ) COMPLEX NUMBERS AND QUADRATIC EQUATIONS Example Represent the complex number z = + in the polar form. Solution Let = r cos θ , = r sin θ By squaring and adding, we get cos θ sin θ i.e., r = (conventionally, r > ) Therefore, cos θ , sin θ , which gives θ π = Therefore, required polar form is π π cos sin z The complex number z is represented as shown in Fig . . Example Convert the complex number into polar form. Solution The given complex number × – – – – + – – – = – + (Fig . ). Let – = r cos θ , = r sin θ By squaring and adding, we get + = cos θ + sin θ which gives r = , i.e., r = Hence cos θ = − , sin θ = π π θ = π – Thus, the required polar form is π π cos sin Fig . Fig . MATHEMATICS EXERCISE . Find the modulus and the arguments of each of the complex numbers in Exercises to . . z = – – i . z = – + i Convert each of the complex numbers given in Exercises to in the polar form: . – i . – + i . – – i . – . + i . . Quadratic Equations We are already familiar with the quadratic equations and have solved them in the set of real numbers in the cases where discriminant is non-negative, i.e., ≥ , Let us consider the following quadratic equation: ax + bx + c = with real coefficients a , b , c and a ≠ . Also, let us assume that the b – ac < . Now, we know that we can find the square root of negative real numbers in the set of complex numbers. Therefore, the solutions to the above equation are available in the set of complex numbers which are given by x = b b ac b ac b i −± −± A Note At this point of time, some would be interested to know as to how many roots does an equation have? In this regard, the following theorem known as the Fundamental theorem of Algebra is stated below (without proof). “A polynomial equation has at least one root.” As a consequence of this theorem, the following result, which is of immense importance, is arrived at: “A polynomial equation of degree n has n roots.” Example Solve x + = Solution We have, x + = or x = – i.e., x = ± − = ± Example Solve x + x + = Solution Here, b – ac = – × × = – = – COMPLEX NUMBERS AND QUADRATIC EQUATIONS Therefore, the solutions are given by x = −± −± × Example Solve Solution Here, the discriminant of the equation is −× × = – = – Therefore, the solutions are −± −± EXERCISE . Solve each of the following equations: . x + = . x + x + = . x + x + = . – x + x – = . x + x + = . x – x + = . . . . x + + = Miscellaneous Examples Example Find the conjugate of ( )( ) ( )( Solution We have , ( )( ) ( )( = −+ = × = = i Therefore, conjugate of ( )( ) is ( )( MATHEMATICS Example Find the modulus and argument of the complex numbers: (i) − , (ii) i Solution (i) We have, − = −+ × = + i Now, let us put = r cos θ , = r sin θ Squaring and adding, r = i.e., r = so that cos θ = , sin θ = Therefore, π θ Hence, the modulus of − is and the argument is π . (ii) We have ( )( Let = r cos θ , – = r sin θ Proceeding as in part (i) above, we get ; cos θ , sin θ Therefore π θ Hence, the modulus of i + is , argument is π Example If x + iy = ib a ib , prove that x + y = . Solution We have, x + iy = )( )( ib ib ib ib b abi b b ab b b COMPLEX NUMBERS AND QUADRATIC EQUATIONS So that, x – iy = b ab b b Therefore, x + y = ( x + iy ) ( x – iy ) = b a b b b b b = Example Find real θ such that sin θ sin θ is purely real. Solution We have, sin θ sin θ = ( sin θ )( sin θ ) ( sin θ )( sin θ ) + sin θ + sin θ – 4sin θ +4sin θ 4sin θ sin θ 4sin θ 4sin θ We are given the complex number to be real. Therefore 8sin θ 4sin θ = , i.e., sin θ = Thus θ = n π , n ∈ Z . Example Convert the complex number π π cos sin z in the polar form. Solution We have, z = ( ) −+ × Now, put cos , sin θ θ MATHEMATICS Squaring and adding, we obtain × = Hence, r = which gives cos θ , sin θ Therefore, π π π θ (Why?) Hence, the polar form is π π cos sin Miscellaneous Exercise on Chapter . Evaluate:            . For any two complex numbers z and z , prove that Re ( z z ) = Re z Re z – Im z Im z . Reduce     to the standard form . . If ib iy c id prove that ( b c d . Convert the following in the polar form: , – Solve each of the equation in Exercises to . . . . + = COMPLEX NUMBERS AND QUADRATIC EQUATIONS . . If z = – i , z = + i , find – z z z z . If a + ib = , prove that a + b = ( ) . Let z = – i , z = – + i. Find Re z z z  , Im z z . Find the modulus and argument of the complex number . Find the real numbers x and y if ( x – iy ) ( + i ) is the conjugate of – – i . . Find the modulus of + . . If ( x + iy ) = u + iv , then show that ( – u v . If α and β are different complex numbers with β = , then find β α αβ – – . Find the number of non-zero integral solutions of the equation – i . If ( a + ib ) ( c + id ) ( e + if ) ( g + ih ) = A + i B, then show that ( a + b ) ( c + d ) ( e + f ) ( g + h ) = A + B . If m – i , then find the least positive integral value of m . MATHEMATICS Summary ® A number of the form a + ib , where a and b are real numbers, is called a complex number , a is called the real part and b is called the imaginary part of the complex number. ® Let z = a + ib and z = c + id . Then z + z = ( a + c ) + i ( b + d ) z z = ( ac – bd ) + i ( ad + bc ) ® For any non-zero complex number z = a + ib ( a ≠ , b ≠ ), there exists the complex number b b b , denoted by z or z – , called the multiplicative inverse of z such that ( a + ib ) b b b = + i = ® For any integer k , i k = , i k + = i , i k + = – , i k + = – i ® The conjugate of the complex number z = a + ib , denoted by z , is given by z = a – ib. ® The polar form of the complex number z = x + iy is r (cos θ + i sin θ ), where r = (the modulus of z ) and cos θ = x r , sin θ = y r . ( θ is known as the argument of z . The value of θ , such that – π < θ ≤ π , is called the principal argument of z . ® A polynomial equation of n degree has n roots. ® The solutions of the quadratic equation ax + bx + c = , where a , b , c ∈ R, a ≠ , b – ac < , are given by x = b ac b i −± COMPLEX NUMBERS AND QUADRATIC EQUATIONS Historical Note The fact that square root of a negative number does not exist in the real number system was recognised by the Greeks. But the credit goes to the Indian mathematician Mahavira ( ) who first stated this difficulty clearly. “He mentions in his work ‘ Ganitasara Sangraha ’ as in the nature of things a negative (quantity) is not a square (quantity)’, it has, therefore, no square root”. Bhaskara , another Indian mathematician, also writes in his work Bijaganita , written in . “There is no square root of a negative quantity, for it is not a square.” Cardan ( ) considered the problem of solving x + y = , xy = . He obtained x = + and y = – as the solution of it, which was discarded by him by saying that these numbers are ‘useless’. Albert Girard (about ) accepted square root of negative numbers and said that this will enable us to get as many roots as the degree of the polynomial equation. Euler was the first to introduce the symbol i for − and W.R. Hamilton (about ) regarded the complex number a + ib as an ordered pair of real numbers ( a , b ) thus giving it a purely mathematical definition and avoiding use of the so called ‘ imaginary numbers ’. different ways. – MAXWELL v . Introduction In earlier classes, we have studied equations in one variable and two variables and also solved some statement problems by translating them in the form of equations. Now a natural question arises: ‘Is it always possible to translate a statement problem in the form of an equation? For example, the height of all the students in your class is less than cm. Your classroom can occupy atmost tables or chairs or both. Here we get certain statements involving a sign ‘<’ (less than), ‘>’ (greater than), ‘ ≤ ’ (less than or equal) and ≥ (greater than or equal) which are known as inequalities . In this Chapter, we will study linear inequalities in one and two variables. The study of inequalities is very useful in solving problems in the field of science, mathematics, statistics, economics, psychology, etc. . Inequalities Let us consider the following situations: (i) Ravi goes to market with ` to buy rice, which is available in packets of 1kg. The price of one packet of rice is ` . If x denotes the number of packets of rice, which he buys, then the total amount spent by him is ` x . Since, he has to buy rice in packets only, he may not be able to spend the entire amount of ` . (Why?) Hence x < ... ( ) Clearly the statement (i) is not an equation as it does not involve the sign of equality. (ii) Reshma has ` and wants to buy some registers and pens. The cost of one register is ` and that of a pen is ` . In this case, if x denotes the number of registers and y , the number of pens which Reshma buys, then the total amount spent by her is ` ( x + y ) and we have x + y ≤ ... ( ) LINEAR INEQUALITIES LINEAR INEQUALITIES Since in this case the total amount spent may be upto ` . Note that the statement ( ) consists of two statements x + y < ... ( ) and x + y = ... ( ) Statement ( ) is not an equation, i.e., it is an inequality while statement ( ) is an equation. Definition Two real numbers or two algebraic expressions related by the symbol ‘<’, ‘>’, ‘ ≤ ’ or ‘ ≥ ’ form an inequality . Statements such as ( ), ( ) and ( ) above are inequalities. < ; > are the examples of numerical inequalities while x < ; y > ; x ≥ , y ≤ are some examples of literal inequalities . < < (read as is greater than and less than ), < x < (read as x is greater than or equal to and less than ) and < y < are the examples of double inequalities . Some more examples of inequalities are: ax + b < ... ( ) ax + b > ... ( ) ax + b ≤ ... ( ) ax + b ≥ ... ( ) ax + by < c ... ( ) ax + by > c ... ( ) ax + by ≤ c ... ( ) ax + by ≥ c ... ( ) ax + bx + c ≤ ... ( ) ax + bx + c > ... ( ) Inequalities ( ), ( ), ( ), ( ) and ( ) are strict inequalities while inequalities ( ), ( ), ( ), ( ), and ( ) are slack inequalities . Inequalities from ( ) to ( ) are linear inequalities in one variable x when a ≠ , while inequalities from ( ) to ( ) are linear inequalities in two variables x and y when a ≠ , b ≠ . Inequalities ( ) and ( ) are not linear (in fact, these are quadratic inequalities in one variable x when a ≠ ) . In this Chapter, we shall confine ourselves to the study of linear inequalities in one and two variables only. MATHEMATICS . Algebraic Solutions of Linear Inequalities in One Variable and their Graphical Representation Let us consider the inequality ( ) of Section . , viz, x < Note that here x denotes the number of packets of rice. Obviously, x cannot be a negative integer or a fraction. Left hand side (L.H.S.) of this inequality is x and right hand side (RHS) is . Therefore, we have For x = , L.H.S. = ( ) = < (R.H.S.), which is true. For x = , L.H.S. = ( ) = < (R.H.S.), which is true. For x = , L.H.S. = ( ) = < , which is true. For x = , L.H.S. = ( ) = < , which is true. For x = , L.H.S. = ( ) = < , which is true. For x = , L.H.S. = ( ) = < , which is true. For x = , L.H.S. = ( ) = < , which is true. For x = , L.H.S. = ( ) = < , which is false. In the above situation, we find that the values of x , which makes the above inequality a true statement, are , , , , , , . These values of x , which make above inequality a true statement, are called solutions of inequality and the set { , , , , , , } is called its solution set . Thus, any solution of an inequality in one variable is a value of the variable which makes it a true statement. We have found the solutions of the above inequality by trial and error method which is not very efficient. Obviously, this method is time consuming and sometimes not feasible. We must have some better or systematic techniques for solving inequalities. Before that we should go through some more properties of numerical inequalities and follow them as rules while solving the inequalities. You will recall that while solving linear equations, we followed the following rules: Rule Equal numbers may be added to (or subtracted from) both sides of an equation. Rule Both sides of an equation may be multiplied (or divided) by the same non-zero number. In the case of solving inequalities, we again follow the same rules except with a difference that in Rule , the sign of inequality is reversed (i.e., ‘<‘ becomes ‘>’, ≤ ’ becomes ‘ ≥ ’ and so on) whenever we multiply (or divide) both sides of an inequality by a negative number. It is evident from the facts that > while – < – , – < – while (– ) (– ) > (– ) (– ) , i.e., > . LINEAR INEQUALITIES Thus, we state the following rules for solving an inequality: Rule Equal numbers may be added to (or subtracted from) both sides of an inequality without affecting the sign of inequality. Rule Both sides of an inequality can be multiplied (or divided) by the same positive number. But when both sides are multiplied or divided by a negative number, then the sign of inequality is reversed. Now, let us consider some examples. Example Solve x < when (i) x is a natural number, (ii) x is an integer. Solution We are given x < or x < (Rule ), i.e., x < / . When x is a natural number, in this case the following values of x make the statement true. , , , , , . The solution set of the inequality is { , , , , , }. When x is an integer , the solutions of the given inequality are ..., – , – , – , , , , , , , The solution set of the inequality is {...,– , – ,– , , , , , , , } Example Solve x – < x + when x is an integer, x is a real number. Solution We have, x – < x + or x – + < x + + (Rule ) or x < x + or x – x < x + – x (Rule ) or x < or x < (Rule ) When x is an integer, the solutions of the given inequality are ..., – , – , – , – , , When x is a real number , the solutions of the inequality are given by x < , i.e., all real numbers x which are less than . Therefore, the solution set of the inequality is x ∈ (– ∞ , ). We have considered solutions of inequalities in the set of natural numbers, set of integers and in the set of real numbers. Henceforth, unless stated otherwise, we shall solve the inequalities in this Chapter in the set of real numbers. MATHEMATICS Example Solve x + < x + . Solution We have, x + < x + or x – x < x + – x or – x < or x > – i.e., all the real numbers which are greater than – , are the solutions of the given inequality. Hence, the solution set is (– , ∞). Example Solve – x – ≤ Solution We have – x – ≤ or ( – x ) ≤ x – . or – x ≤ x – or – x ≤ – , i.e., x ≥ Thus, all real numbers x which are greater than or equal to are the solutions of the given inequality, i.e., x ∈ [ , ∞ ). Example Solve x + < x + . Show the graph of the solutions on number line. Solution We have x + < x + or x < or x < The graphical representation of the solutions are given in Fig . . Fig . Example Solve ≥ − . Show the graph of the solutions on number line. Solution We have ≥ or ≥ or ( x – ) ≥ ( x – ) LINEAR INEQUALITIES or x – ≥ x – or x ≥ or x ≥ The graphical representation of solutions is given in Fig . . Fig . Example The marks obtained by a student of Class XI in first and second terminal examination are and , respectively. Find the minimum marks he should get in the annual examination to have an average of at least marks. Solution Let x be the marks obtained by student in the annual examination. Then ≥ or + x ≥ or x ≥ Thus, the student must obtain a minimum of marks to get an average of at least marks. Example Find all pairs of consecutive odd natural numbers, both of which are larger than , such that their sum is less than . Solution Let x be the smaller of the two consecutive odd natural number, so that the other one is x + . Then, we should have x > ... ( ) and x + ( x + ) < ... ( ) Solving ( ), we get x + < i.e., x < ... ( ) From ( ) and ( ), we get < x < Since x is an odd number, x can take the values , , , and . So, the required possible pairs will be ( , ), ( , ), ( , ), ( , ) MATHEMATICS EXERCISE . . Solve x < , when x is a natural number. x is an integer. . Solve – x > , when x is a natural number. x is an integer. . Solve x – < , when x is an integer. x is a real number. . Solve x + > , when x is an integer. x is a real number. Solve the inequalities in Exercises to for real x . . x + < x + . x – > x – . ( x – ) ≤ ( x – ) . ( – x ) ≥ ( – x ) . x + < . > . ( ) ( ≤ . ) ≥ . ( x + ) – < ( x – ) . – ( x + ) > x – ( x – ) . ( ) ( ) < . ( ) ( ) ( ≥ Solve the inequalities in Exercises to and show the graph of the solution in each case on number line . x – < x + . x – > x – . ( – x ) < ( x + ) . ( – ) ( – ) – ≥ . Ravi obtained and marks in first two unit test. Find the minimum marks he should get in the third test to have an average of at least marks. . To receive Grade ‘A’ in a course, one must obtain an average of marks or more in five examinations (each of marks). If Sunita’s marks in first four examinations are , , and , find minimum marks that Sunita must obtain in fifth examination to get grade ‘A’ in the course. . Find all pairs of consecutive odd positive integers both of which are smaller than such that their sum is more than . . Find all pairs of consecutive even positive integers, both of which are larger than such that their sum is less than . LINEAR INEQUALITIES Fig . Fig . . The longest side of a triangle is times the shortest side and the third side is cm shorter than the longest side. If the perimeter of the triangle is at least cm, find the minimum length of the shortest side. . A man wants to cut three lengths from a single piece of board of length 91cm. The second length is to be 3cm longer than the shortest and the third length is to be twice as long as the shortest. What are the possible lengths of the shortest board if the third piece is to be at least 5cm longer than the second? [ Hint : If x is the length of the shortest board, then x , ( x + ) and x are the lengths of the second and third piece, respectively. Thus, x + ( x + ) + x ≤ and x ≥ ( x + ) + ]. . Graphical Solution of Linear Inequalities in Two Variables In earlier section, we have seen that a graph of an inequality in one variable is a visual representation and is a convenient way to represent the solutions of the inequality. Now, we will discuss graph of a linear inequality in two variables. We know that a line divides the Cartesian plane into two parts. Each part is known as a half plane. A vertical line will divide the plane in left and right half planes and a non-vertical line will divide the plane into lower and upper half planes (Figs. . and . ). A point in the Cartesian plane will either lie on a line or will lie in either of the half planes I or II. We shall now examine the relationship, if any, of the points in the plane and the inequalities ax + by < c or ax + by > c . Let us consider the line ax + by = c, a ≠ , b ≠ ... ( ) MATHEMATICS Fig . There are three possibilities namely: ax + by = c ax + by > c (iii) ax + by < c . In case (i), clearly, all points ( x , y ) satisfying (i) lie on the line it represents and conversely. Consider case (ii), let us first assume that b > . Consider a point P ( α , β ) on the line ax + by = c, b > , so that a α + b β = c. Take an arbitrary point Q ( α , γ ) in the half plane II (Fig . ). Now, from Fig . , we interpret, γ > β (Why?) or b γ > b β or a α + b γ > a α + b β (Why?) or a α + b γ > c i.e., Q( α, γ ) satisfies the inequality ax + by > c . Thus, all the points lying in the half plane II above the line ax + by = c satisfies the inequality ax + by > c . Conversely, let ( α , β ) be a point on line ax + by = c and an arbitrary point Q( α , γ ) satisfying ax + by > c so that a α + b γ > c ⇒ a α + b γ > a α + b β (Why?) ⇒ γ > β (as b > ) This means that the point ( α , γ ) lies in the half plane II. Thus, any point in the half plane II satisfies ax + by > c , and conversely any point satisfying the inequality ax + by > c lies in half plane II. In case b < , we can similarly prove that any point satisfying ax + by > c lies in the half plane I, and conversely. Hence, we deduce that all points satisfying ax + by > c lies in one of the half planes II or I according as b > or b < , and conversely. Thus, graph of the inequality ax + by > c will be one of the half plane (called solution region ) and represented by shading in the corresponding half plane. Note The region containing all the solutions of an inequality is called the solution region. . In order to identify the half plane represented by an inequality, it is just sufficient to take any point ( a , b ) (not online) and check whether it satisfies the inequality or not. If it satisfies, then the inequality represents the half plane and shade the region A LINEAR INEQUALITIES Fig . which contains the point, otherwise, the inequality represents that half plane which does not contain the point within it. For convenience, the point ( , ) is preferred. . If an inequality is of the type ax + by ≥ c or ax + by ≤ c , then the points on the line ax + by = c are also included in the solution region. So draw a dark line in the solution region. . If an inequality is of the form ax + by > c or ax + by < c , then the points on the line ax + by = c are not to be included in the solution region. So draw a broken or dotted line in the solution region. In Section . , we obtained the following linear inequalities in two variables x and y: x + 20y ≤ ... ( ) while translating the word problem of purchasing of registers and pens by Reshma. Let us now solve this inequality keeping in mind that x and y can be only whole numbers , since the number of articles cannot be a fraction or a negative number. In this case, we find the pairs of values of x and y , which make the statement ( ) true. In fact, the set of such pairs will be the solution set of the inequality ( ). To start with, let x = . Then L.H.S. of ( ) is x + y = ( ) + y = y . Thus, we have y ≤ or y ≤ ... ( ) For x = , the corresponding values of y can be , , , , , , only. In this case, the solutions of ( ) are ( , ), ( , ), ( , ), ( , ), ( , ), ( , ) and ( , ). Similarly, other solutions of ( ), when x = , and are: ( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , ) This is shown in Fig . . Let us now extend the domain of x and y from whole numbers to real numbers, and see what will be the solutions of ( ) in this case. You will see that the graphical method of solution will be very convenient in this case. For this purpose, let us consider the (corresponding) equation and draw its graph. x + y = ... ( ) In order to draw the graph of the inequality ( ), we take one point say ( , ), in half plane I and check whether values of x and y satisfy the inequality or not. MATHEMATICS We observe that x = , y = satisfy the inequality. Thus, we say that the half plane I is the graph (Fig . ) of the inequality. Since the points on the line also satisfy the inequality ( ) above, the line is also a part of the graph. Thus, the graph of the given inequality is half plane I including the line itself. Clearly half plane II is not the part of the graph. Hence, solutions of inequality ( ) will consist of all the points of its graph (half plane I including the line). We shall now consider some examples to explain the above procedure for solving a linear inequality involving two variables. Example Solve x + y > graphically. Solution Graph of x + y = is given as dotted line in the Fig . . This line divides the xy -plane in two half planes I and II. We select a point (not on the line), say ( , ), which lies in one of the half planes (Fig . ) and determine if this point satisfies the given inequality, we note that ( ) + ( ) > or > , which is false. Hence, half plane I is not the solution region of the given inequality. Clearly, any point on the line does not satisfy the given strict inequality. In other words, the shaded half plane II excluding the points on the line is the solution region of the inequality. Example Solve x – ≥ graphically in two dimensional plane. Solution Graph of x – = is given in the Fig . . We select a point, say ( , ) and substituting it in given inequality, we see that: ( ) – ≥ or – ≥ which is false. Thus, the solution region is the shaded region on the right hand side of the line x = . Fig . Fig . Fig . LINEAR INEQUALITIES Fig . Fig . Example Solve y < graphically. Solution Graph of y = is given in the Fig . . Let us select a point, ( , ) in lower half plane I and putting y = in the given inequality, we see that × < or < which is true. Thus, the solution region is the shaded region below the line y = . Hence, every point below the line (excluding all the points on the line) determines the solution of the given inequality. EXERCISE . Solve the following inequalities graphically in two-dimensional plane: . x + y < . x + y ≥ . x + y ≤ . y + ≥ x . x – y ≤ . x – y > . – x + y ≥ – . y – x < . y < – . x > – . . Solution of System of Linear Inequalities in Two Variables In previous Section, you have learnt how to solve linear inequality in one or two variables graphically. We will now illustrate the method for solving a system of linear inequalities in two variables graphically through some examples. Example Solve the following system of linear inequalities graphically. x + y ≥ ... ( ) x – y ≤ ... ( ) Solution The graph of linear equation x + y = is drawn in Fig . . We note that solution of inequality ( ) is represented by the shaded region above the line x + y = , including the points on the line. On the same set of axes, we draw the graph of the equation x – y = as shown in Fig . . Then we note that inequality ( ) represents the shaded region above MATHEMATICS the line x – y = , including the points on the line. Clearly, the double shaded region, common to the above two shaded regions is the required solution region of the given system of inequalities. Example Solve the following system of inequalities graphically x + y ≤ ... ( ) x ≥ ... ( ) y ≥ ... ( ) Solution We first draw the graph of the line x + y = , x = and y = Then we note that the inequality ( ) represents shaded region below the line x + y = and inequality ( ) represents the shaded region right of line x = but inequality ( ) represents the shaded region above the line y = . Hence, shaded region (Fig . ) including all the point on the lines are also the solution of the given system of the linear inequalities. In many practical situations involving system of inequalities the variable x and y often represent quantities that cannot have negative values, for example, number of units produced, number of articles purchased, number of hours worked, etc. Clearly, in such cases, x ≥ , y ≥ and the solution region lies only in the first quadrant. Example Solve the following system of inequalities x + y ≤ ... ( ) x ≥ ... ( ) y ≥ ... ( ) Solution We draw the graph of the line x + y = The inequality x + y ≤ represents the shaded region below the line, including the points on the line x + y = (Fig . ). Fig . Fig . LINEAR INEQUALITIES Since x ≥ , y ≥ , every point in the shaded region in the first quadrant, including the points on the line and the axes, represents the solution of the given system of inequalities. Example Solve the following system of inequalities graphically x + y ≤ ... ( ) x + y ≤ ... ( ) x > ... ( ) y > ... ( ) Solution We draw the graphs of the lines x + y = and x + y = . The inequality ( ) and ( ) represent the region below the two lines, including the point on the respective lines. Since x ≥ , y ≥ , every point in the shaded region in the first quadrant represent a solution of the given system of inequalities (Fig . ). EXERCISE . Solve the following system of inequalities graphically: . x ≥ , y ≥ . x + y ≤ , x ≥ , y ≥ . x + y ≥ , x + y < . x + y ≥ , x – y < . x – y > , x – y < – . x + y ≤ , x + y ≥ . x + y ≥ , x + y ≥ . x + y ≤ , y > x , x ≥ . x + y ≤ , x ≥ , y ≥ . x + y ≤ , x + y ≤ , x ≥ , y ≥ . x + y ≥ , x + y ≤ , x – y ≤ . x – y ≤ , x + y ≥ , x ≥ , y ≥ . x + y ≤ , y ≥ x , x ≥ , x , y ≥ . x + y ≤ , x + y ≤ , x ≤ , y ≥ , x ≥ . x + y ≤ , x + y ≥ , x – y ≤ , x ≥ , y ≥ Fig . MATHEMATICS Miscellaneous Examples Example Solve – ≤ x – < . Solution In this case, we have two inequalities, – ≤ x – and x – < , which we will solve simultaneously. We have – ≤ x – < or – ≤ x < or – ≤ x < Example Solve – ≤ – x ≤ . Solution We have – ≤ – x ≤ or – ≤ – x ≤ or – ≤ – x ≤ or ≥ x ≥ – which can be written as – ≤ x ≤ Example Solve the system of inequalities: x – < + x ... ( ) – x ≤ ... ( ) and represent the solutions on the number line. Solution From inequality ( ), we have x – < + x or x < ... ( ) Also, from inequality ( ), we have – x ≤ or – x ≤ – i.e., x ≥ ... ( ) If we draw the graph of inequalities ( ) and ( ) on the number line, we see that the values of x , which are common to both, are shown by bold line in Fig . . Fig . Thus, solution of the system are real numbers x lying between and including , i.e., ≤ x < LINEAR INEQUALITIES Example In an experiment, a solution of hydrochloric acid is to be kept between ° and ° Celsius. What is the range of temperature in degree Fahrenheit if conversion formula is given by C = (F – ), where C and F represent temperature in degree Celsius and degree Fahrenheit, respectively. Solution It is given that < C < . Putting C = (F – ), we get < (F – ) < , or × ( ) < (F – ) < × ( ) or < (F – ) < or < F < . Thus, the required range of temperature is between ° F and ° F. Example A manufacturer has litres of a % solution of acid. How many litres of a % acid solution must be added to it so that acid content in the resulting mixture will be more than % but less than %? Solution Let x litres of % acid solution is required to be added. Then Total mixture = ( x + ) litres Therefore % x + % of > % of ( x + ) and % x + % of < % of ( x + ) or x + ( ) > ( x + ) and x + ( ) < ( x + ) or x + > x + and x + < x + 10800 or x > and x < or x > and x < , i.e. < x < MATHEMATICS Thus, the number of litres of the % solution of acid will have to be more than litres but less than litres. Miscellaneous Exercise on Chapter Solve the inequalities in Exercises to . . ≤ x – ≤ . ≤ – ( x – ) < . – ≤ ≤ . ( x < ≤ . < − ≤ . ( x ≤ ≤ Solve the inequalities in Exercises to and represent the solution graphically on number line. . x + > – , x – < . ( x – ) < x + , ( x + ) > – x . x – > ( x – ) , – x > – x . ( x – ) – ( x + ) ≤ , x + ≤ x + . . A solution is to be kept between ° F and ° F. What is the range in temperature in degree Celsius (C) if the Celsius / Fahrenheit (F) conversion formula is given by F = C + ? . A solution of % boric acid is to be diluted by adding a % boric acid solution to it. The resulting mixture is to be more than % but less than % boric acid. If we have litres of the % solution, how many litres of the % solution will have to be added? . How many litres of water will have to be added to litres of the % solution of acid so that the resulting mixture will contain more than % but less than % acid content? . IQ of a person is given by the formula IQ = MA CA × , where MA is mental age and CA is chronological age. If ≤ IQ ≤ for a group of years old children, find the range of their mental age. LINEAR INEQUALITIES Summary ® Two real numbers or two algebraic expressions related by the symbols <, >, ≤ or ≥ form an inequality. ® Equal numbers may be added to (or subtracted from ) both sides of an inequality. ® Both sides of an inequality can be multiplied (or divided ) by the same positive number. But when both sides are multiplied (or divided) by a negative number, then the inequality is reversed. ® The values of x , which make an inequality a true statement, are called solutions of the inequality . ® To represent x < a (or x > a ) on a number line, put a circle on the number a and dark line to the left (or right) of the number a . ® To represent x ≤ a (or x ≥ a ) on a number line, put a dark circle on the number a and dark the line to the left (or right) of the number x . ® If an inequality is having ≤ or ≥ symbol, then the points on the line are also included in the solutions of the inequality and the graph of the inequality lies left (below) or right (above) of the graph of the equality represented by dark line that satisfies an arbitrary point in that part. ® If an inequality is having < or > symbol, then the points on the line are not included in the solutions of the inequality and the graph of the inequality lies to the left (below) or right (above) of the graph of the corresponding equality represented by dotted line that satisfies an arbitrary point in that part. ® The solution region of a system of inequalities is the region which satisfies all the given inequalities in the system simultaneously. v Every body of discovery is mathematical in form because there is no other guidance we can have – DARWIN v . Introduction Suppose you have a suitcase with a number lock. The number lock has wheels each labelled with digits from to . The lock can be opened if specific digits are arranged in a particular sequence with no repetition. Some how, you have forgotten this specific sequence of digits. You remember only the first digit which is . In order to open the lock, how many sequences of -digits you may have to check with? To answer this question, you may, immediately, start listing all possible arrangements of remaining digits taken at a time. But, this method will be tedious, because the number of possible sequences may be large. Here, in this Chapter, we shall learn some basic counting techniques which will enable us to answer this question without actually listing -digit arrangements. In fact, these techniques will be useful in determining the number of different ways of arranging and selecting objects without actually listing them. As a first step, we shall examine a principle which is most fundamental to the learning of these techniques. . Fundamental Principle of Counting Let us consider the following problem. Mohan has pants and shirts. How many different pairs of a pant and a shirt, can he dress up with? There are ways in which a pant can be chosen, because there are pants available. Similarly, a shirt can be chosen in ways. For every choice of a pant, there are choices of a shirt. Therefore, there are × = pairs of a pant and a shirt. PERMUTATIONS AND COMBINATIONS PERMUTATIONS AND COMBINATIONS Let us name the three pants as P , P , P and the two shirts as S , S . Then, these six possibilities can be illustrated in the Fig. . . Let us consider another problem of the same type. Sabnam has school bags, tiffin boxes and water bottles. In how many ways can she carry these items (choosing one each). A school bag can be chosen in different ways. After a school bag is chosen, a tiffin box can be chosen in different ways. Hence, there are × = pairs of school bag and a tiffin box. For each of these pairs a water bottle can be chosen in different ways. Hence, there are × = different ways in which, Sabnam can carry these items to school. If we name the school bags as B , B , the three tiffin boxes as T , T , T and the two water bottles as W , W , these possibilities can be illustrated in the Fig. . . Fig . Fig . MATHEMATICS In fact, the problems of the above types are solved by applying the following principle known as the fundamental principle of counting , or, simply, the multiplication principle , which states that “If an event can occur in m different ways, following which another event can occur in n different ways, then the total number of occurrence of the events in the given order is m×n.” The above principle can be generalised for any finite number of events. For example, for events, the principle is as follows: ‘If an event can occur in m different ways, following which another event can occur in n different ways, following which a third event can occur in p different ways, then the total number of occurrence to ‘the events in the given order is m × n × p .” In the first problem, the required number of ways of wearing a pant and a shirt was the number of different ways of the occurence of the following events in succession: the event of choosing a pant the event of choosing a shirt. In the second problem, the required number of ways was the number of different ways of the occurence of the following events in succession: the event of choosing a school bag the event of choosing a tiffin box (iii) the event of choosing a water bottle. Here, in both the cases, the events in each problem could occur in various possible orders. But, we have to choose any one of the possible orders and count the number of different ways of the occurence of the events in this chosen order. Example Find the number of letter words, with or without meaning, which can be formed out of the letters of the word ROSE, where the repetition of the letters is not allowed. Solution There are as many words as there are ways of filling in vacant places by the letters, keeping in mind that the repetition is not allowed. The first place can be filled in different ways by anyone of the letters R,O,S,E. Following which, the second place can be filled in by anyone of the remaining letters in different ways, following which the third place can be fille