Kvpy Mathematics

Resonance Pre-foundation Career Care Programmes (PCCP) Division

WORKSHOP TAPASYA SHEET MATHEMATICS COURSE : KVPY (STAGE-) I

KVPY

Subject : Mathematics

S. No.

Topics

Page No.

1.

Ratio & Proportion, Mixture & Alligation

1-4

2.

Linear Inequalities

5-7

2.

Quadratic equation & Progressions

8 - 12

3.

Sets

13 - 16

4.

Trigonometry

17 - 22

5.

Lines, Angles & Quadrilaterals

23 - 25

6.

Triangles

26 - 28

7.

Circles

29 - 30

8.

Number System

31 - 34

9.

Mensuration

35 - 38

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RATIO-PROPORTION, PARTNERSHIP AND MIXTURE & ALLIGATION a c  then, the dividendo is b d

Dividendo : If

a–b c –d  . d b

Ratio : The comparison of two quantities a and b of similar kind is represented as a : b is called a ratio also it can

Componendo and Dividendo : If

a . b In the ratio a : b, we call a as the first term or antecedent and b, the second term or consequent.

be represented as

a c  , then the b d

componendo-dividendo is a  b  c  d . a–b c–d VARIATION :

5 e.g. The ratio 5 : 9 represents , with antecedent = 5 9

(i) We say that x is directly proportional to y, if x = ky for some constant k and we write, x  y. (ii) We say that x is inversely proportional to y, if xy = k

and consequent = 9.

 The multiplication or division of each term of a ratio by the same non-zero number does not affect the ratio.

for some constant k and we write, x 

e.g. 4 : 5 = 8 : 10 = 12 : 15 etc. Also, 4 : 6 = 2 : 3.

1 . y

Ex.1 If a : b = 5 : 9 and b : c = 4 : 7, find a : b : c. Proportion : The equality of two ratios is called proportion. If a : b = c : d, we write, a : b : : c : d and we say that a, b, c, d are in proportion. where, a is called first proportional, b is called second proportional, c is called third proportional and d is called fourth proportional.

 Law of Proportion : Product of means = Product of extremes Thus, if a : b : : c : d  (b × c) = (a × d), Here a and d are called extremes, while b and c are called mean terms.

9  9  63 Sol. a : b = 5 : 9 and b : c = 4 : 7 =  4   :  7   = 9 : 4  4  4

 a:b:c=5:9:

Ex.2 Find out : (i) the fourth proportional to 4, 9, 12; (ii) the third proportional to 16 and 36; (iii) the mean proportional between 0.08 and 0.18. Sol. (i) Let the fourth proportional to 4, 9, 12 be x. Then, 4 : 9 : : 12 : x  4 × x = 9 × 12 

 Mean proportional of two given numbers a and b is

63 = 20 : 36 : 63. 4

x=

9  12 = 27. 4

 Fourth proportional to 4, 9, 12 is 27.

ab .

(ii) Let the third proportional to 16 and 36 is x. Then, 16 : 36 : : 36 : x  16 × x = 36 × 36

Some other ratios : Compounded Ratio : The compounded ratio of the ratios (a : b), (c : d), (e : f) is (ace : bdf). Duplicate ratio : The duplicate ratio of (a : b) is (a2 : b2).

36  36 = 81. 16  Third proportional to 16 and 36 is 81.

Sub-duplicate ratio : The sub-duplicate ratio of (a : b)

(iii) Mean proportional between 0.08 and 0.18

is ( a :



x=

b ). = 3

Triplicate ratio : The triplicate ratio of (a : b) is (a : b ). Sub-triplicate ratio : The sub-triplicate ratio of (a : b) is

ab c d .  b d

8 18 144 12    100 100 100  100 100

= 0.12 Ex.3 If x : y = 3 : 4, find (4x + 5y) : (5x – 2y).

1  1  a 3 : b 3 .    

Componendo : If

0.08  0.18 

3

Sol. a c  then, the componendo is b d

x 3  y 4 x 3 4   5  4   5  y 4 x  5y 4     (3  5)  32 .    3 5 x – 2y 7 x   7   5  – 2  5  – 2  4 4 y      

PAGE # 11

Ex.4 Divide Rs. 1162 among A, B, C in the ratio 35 : 28 : 20. Sol. Sum of ratio terms = (35 + 28 + 20) = 83. 35   A’s share = Rs. 1162   = Rs. 490; 83   28   B’s share = Rs. 1162   = Rs. 392; 83   20   C’s share = Rs. 1162   = Rs. 280. 83  

Ex.5 A bag contains 50 p, 25 p and 10 p coins in the ratio 5 : 9 : 4, amounting to Rs. 206. Find the number of

which y units are taken out and replaced by water. After n operations, the quantity of pure liquid n   y  =  x 1 – x   units.     Ex.7 The cost of Type 1 rice is Rs.15 per kg and Type 2 rice is Rs.20 per kg. If both type-1 and type-2 are mixed in ratio of 2 : 3 , then find the price per kg of the mixed variety of rice. Sol. Let the price of the mixed variety be Rs. x per kg. By the rule of alligation, we have :

Cost of 1 kg of Type 1 rice

coins of each type. Sol. Let the number of 50 p, 25 p and 10 p coins be 5x, 9x

Rs. 15

and 4x respectively.





Rs. 20 Mean price Rs. x (x – 15)

(20 – x)

5x 9 x 4x   = 206 2 4 10 50x + 45x + 8x = 4120

Then,  

Cost of 1 kg of Type 2 rice



103x = 4120 x = 40. Number of 50 p coins = (5 × 40) = 200; Number of 25 p coins = (9 × 40) = 360; Number of 10 p coins = (4 × 40) = 160.

Ex.6 If a man goes from a place A to another place B 100 m apart in 4 hours at a certain speed. With the same speed going from B to C 400 m apart, what time will he take ? Sol. d = st, where d is distance in m, s is speed in m/sec.,

(20 – x ) 2 = ( x – 15 ) 3

 60 – 3x = 2x – 30  5x = 90  x = 18. So, price of the mixture is Rs.18 per kg. Ex.8 A milk vendor has 2 cans of milk. The first contains 25% water and the rest milk. The second contains 50% water. How much milk should he mix from each of the containers so as to get 12 litres of milk such that the ratio of water to milk is 3 : 5 ? Sol. Let cost of 1 litre milk be Rs.1.

t is time in seconds. Speed is same  d  t New distance is 4 times, now the time will be 4 times

Milk in 1 litre mixture in 1st can =

the time it takes from A to B .So, the time taken from B to C is 4 × 4 = 16 hours.

C.P. of 1 litre mixture in 1st can = Rs. Milk in 1 litre mixture in 2nd can =

3 litre, 4 3 . 4

1 litre, 2

Alligation : It is the rule that enables us to find the ratio in which two or more ingredients at the given price must be mixed to produce a mixture of a desired price. Mean Price : The cost price of a unit quantity of mixture is called the mean price. Rule of Alligation : If two ingredients are mixed, then, Quantity of cheaper C.P. of dearer  – Mean price   Quantity of dearer Mean price  – C.P. of cheaper 

We can also represent this thing as under C.P. of a unit quantity of cheaper

C.P. of a unit quantity of dearer

(c)

(d)

Milk in 1 litre of final mixture =

Suppose a container contains x units of liquid from

5 litre, 8

5 . 8 By the rule of alligation, we have :

x 3/4  5 / 8 y = 5 / 8  1/ 2 x 1/ 8 y = 1/ 8 3/4

(m – c)

1 . 2

Mean price = Rs.

C.P. of 1 litre mixture in 1st can

Mean price (m) (d – m)

C.P. of 1 litre mixture in 2nd can = Rs.

C.P. of 1 litre mixture in 2nd can

Mean price 5/8

1/8  We will mix 6 from each can.

1/2

1/8

PAGE # 22

Ex.9 Tea worth Rs.126 per kg and Rs.135 per kg are mixed with a third variety in the ratio 1 : 1 : 2. If the mixture is worth Rs. 153 per kg, then find the price of the third variety per kg. Sol. Since first and second varieties are mixed in equal  126  135   proportions, so their average price = Rs. 2   = Rs.130.50 So, the mixture is formed by mixing two varieties, one at Rs. 130.50 per kg and the other at say, Rs. x per kg in the ratio 2 : 2, i.e., 1 : 1. We have to find x. By the rule of alligation, we have : Cost of 1 kg tea of 1st kind

130.50

Cost of 1 kg tea of 2nd kind

Rs. x Mean price Rs. 153

1=

x  153 22.5

Ex.12 A and B invested Rs. 3600 and Rs. 4800 respectively to open a shop. At the end of the year B’s profit was Rs. 1208. Find A’s profit.

Profit of A 3  Profit of B 4

 Profit of A =

153 + 22.5 = x x = Rs.175.50

 Profit of A =

Ex.10 A jar full of whisky contains 40% alcohol . A part of this whisky is replaced by another containing 19% alcohol and now the percentage of alcohol was found to be 26%. Find the quantity of whisky replaced. Sol. By the rule of alligation, we have : Strength of first jar Strength of 2nd jar

40%

Distribution of Profit/Loss when unequal capital is invested for equal interval of time : When partners invest different amounts of money, for equal interval of time, then profit/loss is divided in the ratio of their investment.

Sol. Profit sharing ratio = 3600 : 4800 = 3 : 4

22.50

(x – 153)

When two or more persons jointly start a business with an objective to earn money. This is called partnership. These persons are called partners and the money invested in the business is known as capital.

Mean strength 26%

7

19%

14

So, ratio of 1st and 2nd quantities = 7 : 14 = 1 : 2. 2  Required quantity replaced = . 3 Ex.11 A vessel is filled with liquid, 3 parts of which are water and 5 parts syrup. How much of the mixture must be drown off and replaced with water so that the mixture may be half water and half syrup ? Sol. Suppose the vessel initially contains 8 litres of liquid. Let x litres of this liquid be replaced with water. 3x    x  litres. Quantity of water in new mixture =  3 – 8  

3 Profit of B 4 3 × 1208 = Rs. 906 4

Distribution of P/L when equal capital is invested for different intervals of time : Ex.13 Govind & Murari started a business with equal capitals. Govind terminated the partnership after 7 months. At the end of the year, they earned a profit of Rs. 7600. Find the profit of each of them. Sol. Govind invested for 7 month, Murari invested for 12 month. Since investment is same for both (Let it be Rs. x)  Profit sharing ratio = 7x : 12x = 7 : 12 7 × 7600 = 2800 7  12 12 Murari’s profit = × 7600 = 4800. 7  12

 Govind’s profit =

Ex.14 Ramesh started a business by investing Rs. 25000. 3 months later Mahesh joined the business by investing Rs. 25000. At the end of the year Ramesh got Rs. 1000 more than Mahesh out of the profit. Find the total profit. Sol. Ramesh invested for 12 month, Mahesh invested for 9 month.  Profit sharing ratio = 12x : 9x = 12 : 9 = 4 : 3. Let Capital be Rs P. 4 P 7 3 Profit of Ramesh = P 7

5x    litres. Quantity of syrup in new mixture =  5 – 8  

Profit of Ramesh =

5x  3x     x  = 5 – 3 –  8  8     5x + 24 = 40 – 5x 8  10x = 16  x = . 5

3 4 P = P + 1000 7 7



 8 1 1 So, part of the mixture replaced =    = . 5 5 8

4 3 P – P = 1000 7 7 P = 1000  P = Rs.7000. 7

PAGE # 33

Distribution of P/L when capital and time both are

Rs. 75000 and Rs. 90000 respectively. It was decided to pay Tanoj a monthly salary of Rs. 1875 as he was

unequal : Ex.15 Suresh & Ramesh entered into a partnership by investing Rs.14000 and Rs. 18000 respectively suresh with drew his money after 4 months. If the total profit at the end is Rs. 12240, find the profit of each. Sol. Profit sharing ratio = 14000 × 4 : 18000 × 12 = 7 : 27 Suresh’s profit =

Ex.18 Tanoj & Manoj started a business by investing

7 × 12240 = 2520 34

27 Ramesh’s profit = × 12240 = 9720. 34

the active partner. At the end of the year if the total profit is Rs. 39000, find the profit of each. Sol. Profit sharing ratio = 75000 : 90000 = 5 : 6 Total profit = Rs. 39000 Salary of Tanoj = 12 × 1875 = Rs. 22500 Profit left = Rs.39000 – Rs. 22500 = Rs.16500. Tanoj’s profit =

 Total profit of Tanoj = 22500 + 7,500 = Rs. 30,000

Ex.16 David started a business establishment by investing Manoj’s profit =

Rs.15000. After 4 months william entered into a partnership by investing a certain amount. At the end

5 × 16500 = 7500. 11

6 × 16500 = Rs. 9,000 11

Change in invested capital :

of the year; the profit was shared in the ratio 9 : 8. Find how much money was invested by william. Sol. Let William invested Rs. x. Profit sharing ratio = 15000 × 12 : 8x = 1,80,000 : 8x Also profit ratio = 9 : 8

months Rajeev invested an additional capital of Rs. 4000, Sanjeev withdrew Rs. 4000 after 9 months. At the end of the year of their profit was Rs. 45800. Find the profit of each.

 ATP 180000 : 8x = 9 : 8



Ex.19 Rajeev & Sanjeev entered into a partnership and invested Rs. 36000 and Rs. 40000 respectively. After 8

Sol. Rajeev’s capital = 36000 × 8 + (36000 + 4000) × 4 = Rs. 448000 for 1 month

180000  8 =x 98

Sanjeev’s capital = 40000 × 9 + (40000 – 4000) × 3 = Rs. 468000 for 1 month

 x = Rs. 20,000

Profit sharing ratio = 448000 : 468000 = 112 : 117 Working and Sleeping partner : Rajeev’s profit = Active Partner : A partner who manages the business is known as active or working partner. Sleeping Partner : A partner who only invests the money is known as sleeping partner.

Sanjeev’s profit =

112 × 45800 = Rs. 22400 229 117 × 45800 = Rs. 23400. 229

Ex.20 A, B and C start a business each investing Rs. 20000.

Ex.17 Nitesh & Jitesh invested Rs.15000 and Rs.18000 respectively in a business. If the total profit at the end of the year is Rs. 8800 and Nitesh, being an active partner, gets an additional 12.5% of the profit, find the total profit of Nitesh. Sol. Profit sharing ratio = 15000 : 18000 = 5 : 6 Total profit = 8800 Nitesh gets 12.5% of the profit =

12.5 × 8800 1000

After 5 months A withdrew Rs. 5000, B withdrew Rs. 4000 and C invests Rs. 6000 more. At the end of the year, a total profit of Rs. 69900 was recorded. Find the share of each. Sol. Ratio of the capitals of A, B and C = 20000 × 5 + 15000 × 7 : 20000 x 5 + 16000 × 7 : 20000 × 5 + 26000 × 7 = 205000 : 212000 : 282000 = 205 : 212 : 282. 

A’s share = Rs. (69900 ×

205 ) = Rs. 20500 ; 699

B's share = Rs. (69900 ×

215 ) = Rs. 21200 ; 699

C’s share = Rs. (69900 x

282 ) = Rs. 28200. 699

= Rs. 1100 Net profit = 8800 – 1100 = Rs. 7700 Nitesh share in profit =

5 × 7700 = 3500 56

Total profit of Nitesh = 3500 + 1100 = Rs. 4600.

PAGE # 44

I N E Q UAT I O N S

A statement involving variable (s) and the sign of inequality viz, >,<  or  is called an inequation. An inequation may contain one or more variables. Also, it may be linear or quadratic or cubic etc. (i) 3x – 2 < 0

(ii) 2x2 + 3x + 4 > 0

(iii) 2x + 5y  4

Let a be a non-zero real number and x be a variable. Then inequations of the form ax + b < 0, ax + b  0, ax + b > 0 and ax + b  0 are known as linear inequations in one variable x. For example, 9x – 15 > 0, 5x – 4  0, 3x + 2 < 0 and 2x – 3  0 are linear inequations in one variable.

(a) Properties of inequalities

(a) Solving linear inequations in one variable

(i) If ‘a’ is a positive no. i.e. a > 0 then for x < y

Rule 1: Same number may be added to (or subtracted from) both side of an inequation without changing the sign of inequality.



x y  & ax < ay.. a a

(ii) If ‘a’ is –ve i.e. a < 0 then for x < y



x y  & ax > ay.. a a

(iii) If ‘a’ is a +ve no. i.e. a > 0 then for x > y



Rule 3 : Any term of an inequation may be taken to the other side with its sign changed without affecting the sign of inequality.

x y  & ax > ay.. a a

(iv) If ‘a’ is a –ve no. i.e. a < 0 then for x > y



x y  & ax < ay.. a a

(i) Closed interval : Let a and b be two given real numbers such that a < b. Then the set of all real numbers x such that a  x  b is called closed interval and is denoted by [a, b] may be graphed as :

a

b



(ii) Open interval : If a and b are two real numbers such that a < b, then the set of all real numbers x satisfying a < x < b is called an open interval and is denoted by (a, b) or ]a, b[ and may be graphed as :

a

Rule 2 : Both sides of an inequation can be multiplied (or divided) by the same positive real number without changing the sign of inequality. However, the sign of inequality is reversed when both sides of an inequation are multiplied or divided by a negative number.

b

(iii) Semi-closed or semi-open interval : If a and b are two real numbers such that a < b, then the sets (a, b] = {x  R : a < x  b} and [a, b) = {x  R : a  x < b} are known as semi-open or semi-closed intervals. (a, b]

Ex. 1 Solve the inequality ax > a. Sol. This inequality has the parameter a that needs to be investigated further. If a > 0, then x > 1 If a < 0, then x < 0 Ex.2 Solve : 24 x < 100 when (i) x is a natural number Sol. We are given 24 x < 100

(ii) x is an integer.

24 x 100 < 24 24 25  x< 6 (i) When x is a natural number, the following values of x make the statement true. 1, 2, 3, 4. (ii) When x is an integer, the solutions of the given equations are ....,– 3, – 2, – 1, 0, 1, 2, 3, 4. The solution set of the equation is : { ...,– 3,– 2,– 1, 0,1, 2, 3, 4}.



Ex.3 Solve & graph the solution set of 3x + 6  9 and – 5x > –15, x  R. Sol. 3x + 6  9 and –5x > –15  3x  3  – x > – 3  x1  x 5. x –1

Sol. We have,

Sol. –4 –3 –2 –1

    

0

1

2

3



– 2 < 2x – 4 2x – 4 > –2 and –2x + 5  13 2x > 2 and –2x  13 – 5 x>1 and –2x  8 x>1 and –x  4 x>1 and x  –4



 x > 1 or x  – 4 or x  (–  , – 4]  (1,  ). Ex.5 Solve the following equation : 2(2x + 3) – 10 < 6 (x – 2) Sol. We have,  2(2x + 3) – 10  6 (x – 2)  4x + 6 – 10  6x – 12  4x – 4  6x – 12  4x – 6x  – 12 + 4 [Transposing – 4 to RHS and 6x to LHS]  – 2x  – 8 

–2 x –8  –2 –2

 x4  x  [4, ) Hence, the solution set of the given inequation is [4, ) which can be graphed on real line as shown in Figure.

2x  4 5 x –1 2x  4 –50 x –1



2x  4 – 5( x – 1) 0 x –1



2x  4 – 5x  5 0 x –1



–3 x  9 0 x –1



3x – 9 0 x –1



3( x – 3) 0 x –1



x–3 0 x –1



1 15 4

3

Hence, the solution set of the given inequations is (1, 3].

Ex. 8 Solve : – 5 

Sol. – 5 × 4 

25 x – 10 – 21x  9 x > 15 4

+

–

2 – 3x < 9. 4

2 – 3x ×49×4 4 [Multiplying throughout by 4]

– 20  2 – 3x  36

4x – 1 x > 15 4

– 20 – 2  – 3x  36 – 2

[Subtracting 2 throughout]

– 22  – 3x  34

4 (4x – 1) > 15 x [Multiplying both sides by 60 i.e. Lcm of 15 and 4] 16x – 4 > 15x 16x – 15x > 4 [Transposing 15 x to LHS and – 4 to RHS] x>4 x  (4, ) Hence, the solution set of the given inequation is (4, ). This can be graphed on the real number line as shown in figure.

–22 –3 x 34   –3 –3 –3

[Dividing throughout by – 3]

–34 22 x 3 3 –34 22 x 3 3

x  [– 34/3, 22/3] Hence, the interval [– 34/3, 22/3] is the solution set of

0

4

the given system of inequations.

PAGE # 66

Ex.9 Solve 5x - 3 < 3x + 1 when

(a) Inequations involving absolute value

(i) x is an integer, Sol. We have, 5x – 3 < 3x + 1

    

(ii) x is a real number. Result 1. If a is a positive real number, then (i)

5x – 3 + 3 < 3x + 1 + 3

| x | < a  – a < x < a i.e. x  (– a, a)

5x < 3x + 4

a

–a 5x – 3 x < 3x + 4 – 3x

(ii) | x |  a  – a  x  a i.e. x  [–a, a]

2x < 4 x a  x < – a or x > a

which are less than 2. Therefore, the solution set of the inequality is x  (-  , 2).

–a

a

(ii) | x |  a  x  – a or x  a

–a

a

The function f(x) defined by

Result 3. Let r be a positive real number and a be a

x, when x  0 f(x) = x   – x, when x  0 

fixed real number. Then,

is called the modulus function. It is also called the absolute value function.

(i)

| x – a | < r a – r < x < a + r i.e. x  (a – r, a + r)

(ii) | x – a |  r a – r  x  a + r i.e. x  [a – r, a + r] (iii) | x – a | > r x < a – r, or x > a + r

y

(iv) | x – a |  r x  a – r, or x  a + r f(x )=

x'

)= f(x

–x

Ex.12 Find x from 1  x  2 and represent it on number line.

x

x

o

Sol. 1  x  x  1  x > 1 or x < –1 

x  ( ,  1)  (1,  )

...(i)

also x  2  x < 2 or x > – 2

y' The distance between two real numbers x and y is

 x  (–2, 2)

defined as x – y .

Combining the two results, we get

x – 4, where x – 4  0  x  4 Sol. x – 4   – ( x – 4 ), where x – 4  0  x  4 and – (x – 4) = 7  x–4=7 x = 11 and –x+4=7

Ex. 11



–x=3



x=–3

...(ii)

1  x  2  {– 2 < x < –1}  {1 < x < 2}

Ex.10 Solve : x – 4 = 7

Sol.

 x lies between – 2 & 2

i.e. x  (– 2, –1)  (1, 2)

Ex.13

 x  –3

 Ans.  x  11

Evaluate 3  – 2 – 3 – 3 – – 7 3  –2–3 –3– –7

–1 0 1

–2

2

Find x satisfying x – 5  3 .

Sol. as x – a  r  a – r  x  a  r i.e. x  [a – r, a  r ]

 x – 5  3  5 – 3  x  5  3 i.e. 2  x  8 i.e. x  [2, 8] 2

8

= 3 + 5 – 3 – {–(–7)} = 3 + {–(–5)} – 3 – 7 = 3 + 5 – 10 = 8 – 10 = –2.

PAGE # 77

QUADRATIC EQUATION & PROGRESSION In this case both the roots are irrational and distinct. [See remarks also] If P(x) is quadratic expression in variable x, then P(x) = 0 is known as a quadratic equation.

Case-2 When b2 – 4ac = 0, (D = 0) In this case both the roots are real and equal to –

General form of a Quadratic Equation : The general form of a quadratic equation is ax2 + bx + c = 0, where a, b, c are real numbers and a  0.

b . 2a

Case-3 When b2 – 4ac < 0, (D < 0) In this case b2 – 4ac < 0, then 4ac – b2 > 0 

The value of x which satisfies the given quadratic equation is known as its root. The roots of the given equation are known as its solution. For quadratic equation ax2 + bx + c = 0, the roots are

=

and  = or

 b  b 2  4ac  b  b 2  4ac and . 2a 2a

=

 b   ( 4ac  b 2 ) 2a  b   ( 4ac  b 2 ) 2a

 b  i 4ac  b 2 2a

 b  i 4ac  b 2 [  1 = i ] 2a i.e. in this case both the roots are imaginary and distinct. and  =

Consider the quadratic equation, a x2 + b x + c = 0 having  and  as its roots and b2  4ac is called discriminant of roots of quadratic equation. It is denoted by D or .



REMARKS :



If a, b, c  Q and b2 – 4ac is positive (D > 0) but not a perfect square, then the roots are irrational and they

Roots of the given quadratic equation may be (i) Real and unequal (ii) Real and equal (iii) Imaginary and unequal. Let the roots of the quadratic equation ax2 + bx + c = 0 (where a  0, b, c  R) be  and  then =

 b  b 2  4ac 2a

and b2 – 4ac is positive but not a perfect square, then the roots may not occur in conjugate pairs. 

If b2 – 4ac is negative (D < 0), then the roots are complex conjugate of each other. In fact, complex roots of an equation with real coefficients always occur in conjugate pairs like 2 + 3i and 2 – 3i. However, this may not be true in case of equations with complex coefficients. For example, x2 – 2ix – 1 = 0 has both roots equal to i.



If a and c are of the same sign and b has a sign opposite to that of a as well as c, then both the roots are positive, the sum as well as the product of roots is positive

... (i)

 b  b 2  4ac ... (ii) 2a The nature of roots depends upon the value of expression ‘b2 – 4ac’ with in the square root sign. This is known as discriminant of the given quadratic equation. and

always occur in conjugate pairs like 2 + 3 and 2 – 3 . However, if a, b, c are irrational numbers

=

Consider the Following Cases :

(D  0).

Case-1 When b2 – 4ac > 0, (D > 0) In this case roots of the given equation are real and distinct and are as follows  b  b 2  4ac  b  b 2  4ac = and  = 2a 2a

(i) When a( 0), b, c  Q and b2 – 4ac is a perfect square In this case both the roots are rational and distinct. (ii) When a( 0), b, c  Q and b2 – 4ac is not a perfect square



If a, b, c are of the same sign then both the roots are negative, the sum of the roots is negative but the product of roots is positive (D  0). (a) Relation Between Roots & Coefficients : (i) The solutions of quadratic equation a x2 + b x + c = 0 are given by x=

 b  b2  4 a c

2a (ii) The expression b2  4 a c  D is called discriminant of the quadratic equation a x2 + b x + c = 0. If ,  are the roots of the quadratic equation a x2 + b x + c = 0, then PAGE # 88

coefficien t of x

(a) Sum of the roots = –

Step-(iv) Add square of half of the coefficient of x.i.e.

coefficien t of x 2

2

 b    on both sides to obtain  2a 

b  +  =  a

(b) Product of the roots =

 b  x2 + 2   x +  2a 

constant term

coefficient of x 2 c  = a (iii) A quadratic equation whose roots are  and  is (x ) (x ) = 0 i.e. x2  (sum of roots) x + (product of roots) = 0.

2

b  b 2  4ac  x   = . 2a  4a 2  Step-(vi) Take square root of both sides to get

x+

(i)  –  = ± (  )2  4

term

b 2  4ac 4a 2

b on RHS. 2a

(c) By Using Quadratic Formula :

3

= ± (   ) 2  4 2 ± 3 (  )2  4 (vi) 4 + 4 = (2 +  2)2 – 222= {(+)2 –2 }2 – 222 (vii) 4 –  4 = ( + ) ( – )(2 +  2)



= ± (+)

b =± 2a

Step (vii) Obtain the values of x by shifting the constant

(  )2  4

(iii) 2 +  2 = ( + )2 – 2 (iv) 3 +  3 = ( + )3 – 3(+) (v) 3 –  3 = ( – )3 + 3(–)



2

Step-(v) Write L.H.S. as the perfect square of a binomial expression and simplify R.H.S. to get

(b) Symmetric functions of roots of a quadratic equation

(ii) 2 –  2 = ± ( + )

2

 b   b    =   – c 2 a    2a  a

Solve the quadratic equation in general form viz. ax2 + bx + c = 0. Step (i) By comparison with general quadratic equation, find the value of a, b and c.

(  )2  4 {(+)2 –2 }

Step (ii) Find the discriminant of the quadratic equation. D = b2 – 4ac Step (iii) Now find the roots of the equation by given equation

(a) By Factorisation : ALGORITHM :

x=

Step (i) Factorise the constant term of the given quadratic equation. Step (ii) Express the coefficient of middle term as the sum or difference of the factors obtained in step 1. Clearly, the product of these two factors will be equal to the product of the coefficient of x2 and constant term. Step (iii) Split the middle term in two parts obtained in step 2. Step (iv) Factorise the quadratic equation obtained in step 3. (b) By the Method of Completion of Square : ALGORITHM : Step-(i) Obtain the quadratic equation. Let the quadratic equation be ax2 + bx + c = 0, a  0. Step-(ii) Make the coefficient of x2 unity, if it is not unity. i.e., obtain x2 +

b c x+ = 0. a a



b D b D , 2a 2a

REMARK : If b2 – 4ac < 0, i.e., negative, then

b 2 – 4ac is not real

and therefore, the equation does not have any real roots. C o mm on Roo ts : Consider two quadratic equations, a1 x2 + b1 x + c1 = 0 & a 2 x2 + b 2 x + c 2 = 0. (i) If two quadratic equations have both roots common, then the equations are identical and their coefficients are in proportion. i.e. a1 b c = 1 = 1 . a2 b2 c2 (ii) If only one root is common, then the common root '  ' will be : =

c 1 a 2  c 2 a1 b1 c 2  b 2 c 1 = a1 b 2  a 2 b1 c 1 a 2  c 2 a1

Hence the condition for one common root is:

c

Step-(iii) Shift the constant term on R.H.S. to get a b c x2 + x = – a a

 c 1 a 2  c 2 a1 2 = a1 b 2  a 2 b1  b1 c 2  b 2 c 1 

PAGE # 99

SEQUENCE A sequence is an arrangement of numbers in a definite order according to some rule. e.g. (i) 2, 5, 8, 11, ... (ii) 4, 1, – 2, – 5, ... (iii) 3, –9, 27, – 81, ...

No. of Terms

Terms

Common Difference

For 3 terms

a – d, a, a + d

d

For 4 terms

a – 3d, a – d, a + d, a + 3d

2d

For 5 terms

a – 2d, a – d, a, a + d, a + 2d

d

For 6 terms

a – 5d, a – 3d, a – d, a + d, a + 3d, a + 5d

2d

Types of Sequence On the basis of the number of terms there are two types of sequence :

SUM OF n TERMS OF AN A.P. Let A.P. be a, a + d, a + 2d, a + 3d,............., a + (n – 1)d Then, Sn = a + (a + d) +...+ {a + (n – 2) d} + {a + (n – 1) d} ..(i) also Sn= {a + (n – 1) d} + {a + (n – 2) d} +....+ (a + d) + a ..(ii) Add (i) & (ii)

(i) Finite sequences : A sequence is said to be finite if it has finite number of terms. (ii) Infinite sequences : A sequence is said to be infinite if it has infinite number of terms.

 2Sn = 2a + (n – 1)d + 2a + (n – 1)d +....+ 2a + (n – 1)d

PROGRESSIONS Those sequence whose terms follow certain patterns are called progressions. Generally there are three types of progressions.



2Sn = n [2a + (n – 1) d]



Sn 

n 2a ( n  1) d  2

Sn =

n n [a + a + (n – 1)d] = [a +  ] 2 2

Sn 

n a    where, is the last term. 2

(i) Arithmetic Progression (A.P.) (ii) Geometric Progression (G.P.) (iii) Harmonic Progression (H.P.)



ARITHMETIC PROGRESSION A sequence is called an A.P., if the difference of a term and the previous term is always same. i.e. d = tn + 1 – tn = Constant for all n  N. The constant difference, generally denoted by ‘d’ is called the common difference.

GENERAL FORM OF AN A.P. If we denote the starting number i.e. the 1st number by ‘a’ and a fixed number to be added is ‘d’ then a, a + d, a + 2d, a + 3d, a + 4d,........... forms an A.P. th

n FORM OF AN A.P. Let A.P. be a, a + d, a + 2d, a + 3d,........... Then, First term (a1) = a + 0.d Second term (a2) = a + 1.d Third term (a3) = a + 2.d . . . . . . nth term (an) = a + (n – 1) d  an = a + (n – 1) d is called the nth term. th



rth term of an A.P. when sum of first r terms is Sr is given by, tr = Sr – Sr – 1.

PROPERTIES OF A.P. (i) For any real number a and b, the sequence whose nth term is an = an + b is always an A.P. with common difference ‘a’ (i.e. coefficient of term containing n). (ii) If a constant term is added to or subtracted from each term of an A.P. then the resulting sequence is also an A.P. with the same common difference. (iii) If each term of a given A.P. is multiplied or divided by a non-zero constant K, then the resulting sequence is also an A.P. with common difference d respectively. Where d is the common K difference of the given A.P.

Kd or

(iv) In a finite A.P. the sum of the terms equidistant from the beginning and end is always same and is equal to the sum of 1st and last term.

m TERM OF AN A.P. FROM THE END Let ‘a’ be the 1st term and ‘d’ be the common difference of an A.P. having n terms. Then mth term from the end is (n – m + 1)th term from beginning or {n – (m – 1)}th term from beginning.

SELECTION OF TERMS IN AN A.P. Sometimes we require certain number of terms in A.P. The following ways of selecting terms are generally very convenient.

(v) If three numbers a, b, c are in A.P. , then 2b = a + c. Arithmetic Mean (Mean or Average) (A.M.) If three terms are in A.P. then the middle term is called the A.M. between the other two, so if a, b, c are in A.P., b is A.M. of a & c. A.M. for any n number a 1, a 2,..., a n is; A=

a1  a 2  a 3  .....  a n . n

PAGE # 1010

n - Arithmetic Numbers :

Means

Between

Two

If a, b are any two given numbers & a, A1, A2,...., An, b are in A.P. then A1, A2,... An are the n-A.M.’s between a & b. Total terms are n + 2. ba Last term b = a + (n+2–1)d.Now, d = . n 1

A1 = a + An = a +

(iii) Any four consecutive terms of a G.P. can be taken a a as 3 , , ar, ar3. r r (iv) If each term of a G.P. be multiplied or divided or raised to power by the some nonzero quantity, the resulting sequence is also a G.P.

ba 2(b  a ) , A2 = a + ,. .. ... ... .. ., n1 n1

n (b  a) . n 1

If a, b, c are in G.P., b is the G.M. between a & c. b² = ac, therefore b = a c ; a > 0, c > 0.

 NOTE : Sum of all n-A.M.’s inserted between a & b is equal to n times the single A.M. between a & b. n

i.e.



r 1

Ar = nA where A is the single A.M. between a & b.

n-Geometric Means Between a, b : If a, b are two given numbers & a, G1, G2,....., Gn, b are in G.P.. Then, G1, G2, G3,...., Gn are n-G.M.s between a & b. G1 = a(b/a)1/n+1, G2 = a(b/a)2/n+1,......, Gn = a(b/a)n/n+1

 NOTE : G.P. is a sequence of numbers whose first term is non zero & each of the succeeding terms is equal to the preceding term multiplied by a constant. Thus in a G.P. the ratio of successive terms is constant. This constant factor is called the common ratio of the series & is obtained by dividing any term by that which immediately precedes it. 1 1 1 1 , , , ...are in G.P. .P. 3 9 27 81

Example : 2, 4, 8, 16 ... &

(i) Therefore a, ar, ar2, ar3, ar4,...... is a G.P. with ‘a’ as the first term and ‘r’ as common ratio. nth term = a rn1 (ii) Sum of the first n terms.



n



a r  1   r 1 Sn =  a 1 rn   1  r





n

 G =  ab  = G r

n

where G is the single G.M.

r 1

between a & b.

A sequence is said to be H.P. if the reciprocals of its terms are in A.P. If the sequence a 1, a 2, a 3,...., a n is an H.P. then 1/a 1, 1/a 2,...., 1/a n is an A.P. Here we do not have the formula for the sum of the n terms of a H.P. For H.P. whose first term is a and second term is b, the n th term is tn

ab . b  (n  1)(a  b )

, r 1 If a, b, c are in H.P.  b =

, r 1

n  rn  0 if r < 1 therefore, a (| r |  1) . S = 1– r

(i) If a, b, c are in G.P. then b 2 = ac, in general if a 1, a 2, a 3, a 4,......... a n – 1 , a n are in G.P., then a 1a n = a 2a n – 1 = a 3 a n – 2 = .............. (ii) Any three consecutive terms of a G.P. can be a , a , ar.. r

n

i.e.

tn =

(iii) Sum of an infinite G.P. when r < 1. When

taken as

The product of n G.M.s between a & b is equal to the nth power of the single G.M. between a & b

2ac a ab or = . ac c bc

 NOTE : (i)

If a, b, c are in A.P. 

ab a = bc a

ab a (ii) If a, b, c are in G.P.  b  c = b

If a, b, c are in H.P., b is the H.M. between a & c, then 2ac . ac If a 1, a2 , ........ an are ‘n’ non-zero numbers then H.M. H of these numbers is given by :

b=

1 1  1  1  .......  1    = an  . H n  a1 a 2

PAGE # 1111

Relation bet ween means :



If A, G, H are respectively A.M., G.M., H.M. between a & b both being unequal & positive then, G² = AH i.e. A, G, H are in G.P. A.M.  G.M.  H.M. Let a1, a2, a3, .......an be n positive real numbers, then we define their a1  a 2  a 3  .......  a n , n G.M. = (a1 a2 a3 .........an)1/n and

A.M. =

n H.M. = 1 1 1 .   .......  a1 a 2 an

It can be shown that A.M.  G.M.  H.M. and equality holds at either places iff a1 = a2 = a3 = ..............= an.



PAGE # 1212

SETS Ex. 5 Write the set A = {0,1,4,9,16,........} in set builder form. Sol. A = {x2 : x Z). A well defined collection of objects is known as sets.

1

builder form. Sol. W e observe that the elements of set X are the reciprocals of the squares of all natural numbers. So,

For example : The collection of all states in the Indian union is a set but collection of good cricket players of India is not a set, since the term “good player is vague and it is not well defined.

16

,

1

say a belongs to A. If a does not belong to A, then a  A is written.

9

,

1

Ex.6 Write the set X = {1,

4

,

1

If a is an element of a set A, then we write a  A and

25

, ........} in the set

 1  the set X in set builder form is X =  2 ; n N .  n 

Some letters are reserved for the sets as listed below : N : For the set of Natural numbers. Z : For the set of Integers. Z+ :For the set of all positive Integers. Q : For the set of all Rational numbers. +

Q : For the set of all positive Rational numbers.

A set is said to be empty or null or void set if it has no element and it is denoted by or { }. Ex. 7 Write {x  N : 5 < x < 6} in roster form. Sol. A = { }.

R : For the set of all Real numbers.

(b) Singleton Set :

R+ : For the set of all Positive real numbers.

A set consisting of a single element is called a singleton set.

C : For the set of all Complex numbers.

A set is often described in the following two ways : (a) Roster Method : In this method a set is described by listing elements, separated by commas, within braces { }. Ex.1 Write the set of vowels of English alphabet in roster form. Sol. A = {a, e, i, o, u}. Ex.2 Write the set of even natural numbers in roster form. Sol. B = {2,4,6,.....}. Ex.3 Write the set of all prime numbers less than 11 in roster form. Sol. C = {2,3,5,7}. Ex.4 Write the set A = {x  z, x2 < 20} in the roster from. Sol. We observe that the squares of integers 0, ± 1, ± 2, ± 3, ± 4 are less than 20. Therefore, the set A in roster form i s A = { – 4, – 3, – 2, –1, 0, 1, 2, 3, 4}. •

(a) Empty Set :

NOTE : The order in which the element are written in a set makes no difference. (b) Set Builder Method : In this method, a set is described by a characterizing property P(x) of its elements x. In such a case the set is described by {x : P(x) holds } or, {x | P(x) holds,} which is read as ‘the set of all x such that P(x) holds’. The symbol ‘|’ or ‘:’ is read as ‘such that’.

Ex. 8 Write the set {x : x  N and x2 = 9} in roster form. Sol. Let B is the set. So B is a singleton set equal to {3}. (c) Finite Set : A set is called a finite set if it is either void set or its element can be listed (counted labelled) by natural numbers 1, 2, 3 ....... and the process of listing terminates at a certain natural number n (say). For example : Set of all persons on the earth is a finite set. (d) Infinite Set : A set whose elements cannot be listed by natural numbers 1, 2, 3,...... for any natural number n is called an infinite set. For example : Set of all points in a plane is an infinite set. Ex. 9 Which of the following sets are finite and which are infinite ? (a) Set of concentric circle in a plane. (b) Set of letters of English alphabets. (c) { x  N, x > 5 } (d) { x  R, 0 < x < 1 } (e) { x  N, x < 200} Sol. (a) Infinite set (b) Finite set (c) Infinite set (d) Infinite set (e) Finite set (e) Cardinal Number of a Finite Set : The number n in the above definition is called the cardinal number or order of a finite set A and is denoted by n(A).

PAGE # 1313

(f) Equivalents Set : Two finite sets A and B are equivalent if their cardinal numbers are same. i.e. n(A) = n(B). For example : A = {1,2,3} and B = {a,b,c} are equivalent sets. (g) Equal Set : Two sets A and B are said to be equal if every element of A is a member of B, and every element of B is member of A. NOTE : Equal sets are equivalents but equivalent sets need not be equal. Ex. 10 Are the following sets equal ? A = { x : x is a letter in the word reap } B = { x : x is a letter in the word paper }. Sol. A = { r, e, a, p} B = { p, a, e, r } So, A and B are equal sets.

Ex.13 Let A = {1,2,3,4}, B = {1,2,3} and C = {2,4}. Find sets X satisfying each pair of conditions. (i) X  B and X

Sol. We have (i) 

X  B and X  C X is subset of B but X is not a subset of C

 

X  P (B) but X  P (C) X = {1}, {3}, {1,2}, {1,3}, {2,3}, {1,2,3}.

(ii) We have, X  B, X  B and X  C  X is a subset of B other than B itself and X is not a subset of C  

X  P(B), X  P(C) and X  B X = {1}, {3}, {1,2}, {1,3}, {2,3}.

(iii) We have, X  A, X  B and X  C

(h) Subset : Let A and B be two sets. If every element of A is an element of B, then A is called a subset of B. If A is a

C

(ii) X  B, X  B and X  C (iii) X  A, X  B and X  C

  

X   P(A), X  P(B) and X  X (C) X is a subset of A, B and C X = , {2}

subset of B, we write A  B, which is read as “A is a subset of B” or “A is contained in B”. Thus, A  B if a  A  a  B. The symbol “” stands for “implies”. If A is not a subset of B, we write A  B. NOTE : Every set is a subset of itself and the empty set is subset of every set. These two subsets are called improper subsets. A subsets A of a set B is called a

Diagram drawn to represent sets are called Venn-Euler diagram or simply Venn diagram. In Venn-diagram the universal set U is represented by points within a rectangle and its subsets are represented by points in closed curves (usually circles) within the rectangle.

proper subset of B if A  B and we write A  B. SOME RESULTS ON SUBSET : (i) Every set is a subset of itself (ii) The empty set is a subset of every set. (iii) The total number of subsets of a finite set containing n element is 2n.

(a) Union of Sets : Let A and B be two sets. The union of A and B is the set of all those elements which belong either to A or to B or to both A and B. Thus, A  B = { x : x  A or x  B}.

(i) Universal Set : A set that contains all sets in a given context is called the Universal Set. Ex.11 If A = {1,2,3}, B = {2,4,5,6} and C = {1,3,5,7}, then find the universal set. Sol. U = {1, 2, 3, 4, 5, 6, 7} can be taken as the universal set. ( j) Power Set : Let A be a set. Then the collection or family of all subsets of A is called the power set of A and is denoted by P(A).

Ex.14 If A {1, 2, 3} and B = {1, 3, 5, 7}, then find A  B. Sol. A  B = {1, 2, 3, 5, 7}. (b) Intersection of Sets : Let A and B be two sets. The intersection of A and B is the set of all those elements that belong to both A and B.

Ex.12 Let A = {1,2,3}. Then find the power set of A. Sol. Subset of A are : , {1}, {2}, {3}, {1,2}, {1,3}, {2,3} and {1,2,3}. Hence, total number of subset are = 23 = 8. Hence, P(A) ={ {1}, {2}, {3}, {1,2}, {1,3}, {2,3}, {1,2,3} }. Thus, A  B = {x : x  A and x  B}. PAGE # 1414

Ex.15 If A = { 1, 2, 3, 4, 5 } and B = { 1, 3, 9, 12 }, then find A  Sol. A = { 1, 3 }. (c) Disjoint Sets : Two sets A and B are said to be disjoint, if A B = . If A  B  , then A and B are said to be intersecting or overlapping sets. (d) Difference of Sets : Let A and B be two sets. The difference of A and B, written as A – B, is the set of all those elements of A which do not belong to B. u

Ex.18 Let U = { 1, 2, 3, 4, 5, 6, 7, 8, 9 }, A = { 1, 2, 3, 4 }, B = { 2, 4, 6, 8 } and C = { 3, 4, 5, 6 }. Find (i) Ac (ii) ( A  C )c (iii) ( A  B )c (iv) (B – C)c c Sol. (i) A = U – A = { 1, 2, 3, 4, 5, 6, 7, 8, 9 } – { 1, 2, 3, 4 } = { 5, 6, 7, 8, 9 }. (ii) ( A  C )c = U – ( A  C ) = { 1, 2, 3, 4, 5, 6, 7, 8, 9 } – { 3, 4 } = { 1, 2, 5, 6, 7, 8, 9 }. (iii) ( A  B )c = U – ( A  B ) = { 1, 2, 3, 4, 5, 6, 7, 8, 9 } – { 1, 2, 3, 4, 6, 8 } = { 5, 7, 9 }. (iv) (B – C)c = U – ( B – C ) = { 1, 2, 3, 4, 5, 6, 7, 8, 9 } – { 2, 8 } = { 1, 3, 4, 5, 6, 7, 9 }.

A–B A

B

Thus, A – B = {x : x  A and x  B} or, A – B = {x  A : x  B}. Clearly, x A – B  x  A and x  B. Similarly, the difference B – A is the set of all those elements of B that do not belong to A i.e. B – A = {x  B : x  A}. u

A

B

Ex.16 If A = { 2, 3, 4, 5, 6, 7 } and B = { 3, 5, 7, 9, 11, 13 }, then find A – B and B – A. Sol. A – B = { 2, 4, 6 } and B – A = { 9, 11, 13 }. (e) Symmetric Difference of Two Sets : Let A and B be two sets. The symmetric difference of sets A and B is the set (A – B)  (B – A) and is denoted by A  B.

Ex.17 If A = { x  R : 0 < x < 3 }, B = { x  R : 1  x  5 }, then find A  B. Sol. A – B = { x  R : 0 < x < 1 }, B – A = { x  R : 3  x  5 } AB= {xR:0