· 1. REAL NUMBERS 1-12 2. POLYNOMIALS 13-32 3. PAIR OF LINEAR EQUATIONS IN TWO VARIABLES 33-46 4....

9 789388 395335

1. REAL NUMBERS 1-12

2. POLYNOMIALS 13-32

3. PAIR OF LINEAR EQUATIONS IN TWO VARIABLES 33-46

4. QUADRATIC EQUATIONS 47-61

5. ARITHMETIC PROGRESSIONS 62-79

6. CO-ORDINATE GEOMETRY 80-102

7. TRIANGLES 103-153

8. CIRCLES 154-195

9. TRIGONOMETRIC RATIOS 196-207

10. TRIGONOMETRIC IDENTITIES 208-228

11. HEIGHTS AND DISTANCES 229-252

12. AREAS RELATED TO CIRCLES 253-274

13. SURFACE AREAS AND VOLUMES 275-310

14. STATISTICS 311-328

15. PROBABILITY 329-342

CONTENTS

1Arundeep Handbook of MCQ’s MATHEMATICS

Points to Remember :1. Real Numbers— Numbers, rational and

irrational together, are called real numbers

such as 0, 3, –4, 32

, 3 etc. These number

can be represented on a number line also.2. Divisibility— A non-zero integer ‘a’ is said

to divide an integer ‘b’ there exists an integerc such that b = ac. The integer a is calleddividend, b is called divisor and c is calledquotient. a | b means b is divisible by ‘a’ anda | b means b is not divisible by a.

3. Properties of divisibility—

(i) + divides every non-zero integer i.e. + a1

for every non-zero integer a.(ii) 0 is divided by every non-zero integer a

i.e. 0a

for every non zero integer a.

(iii) 0 does not divide any integer.(iv) If a is a non-zero integer and b is any integer,

then a|b Þ a|–b, –a|b and –a|–b.(v) If a and b are non-zero integers, then a|b

and b|aÞ a = +b.4. Euclid’s Division Lemma— Let a and b be

any two positive integers, then there existsunique integers q and r, such thata = bq + r, 0 < r < b.If b|a, then r = 0, otherwise r satisfies thestronger inequality 0 < r < b.Remarks—

(i) The above Lemma is nothing but arestatement of the long division process wehave been doing all these years and that theintegers q and r are called the quotient andremainder respectively.

(ii) The above Lemma has been stated for

Chapter — 1REAL NUMBERS

positive integers only. But it can be extendedto all integers as stated below :Let a and b be any two integers with b ¹ 0,then there exists unique integers q and r suchthat a = bq + r where 0 < r < |b|.

5. H.C.F. or G.C.D.— The largest or greatestdivisor among the common divisors of twoor more integers is called the GreatestCommon Divisor (G.C.D.) or HighestCommon Factor (H.C.F.) of the givenintegers.

6. Some results of Theorems—Theorem 1. If a and b are positive integerssuch that a = bq + r, then every commondivisor of a and b is a common divisor of band r and vice-versa.Theorem 2. Linear combination— H.C.F.(say a) of two positive integers a and b canbe expressed as a linear combination of aand b i.e. d = xa + yb, for some integers xand y.This representation is not unique, becaused = xa + yb= xa + yb + ab – ab= xa + ab + yb – ab= (x + b) a + (y – a) bNote— To represent the H.C.F. as a linearcombination of the given two numbers, westart from the last but one stop andsuccessively eliminate the previousremainder.

7. How to find H.C.F. and L.C.M. of giventwo numbers a and b

(i) Factorise each of the given positive integersand express them as a product of powers ofprime in ascending order of magnitude ofprimes.

(ii) To find the H.C.F. identify the commonprime factors and find the smallest (least)


exponent of these common factors. Nowraise these common prime factors to theirsmallest exponents and multiply them to getthe H.C.F.

(iii) To find the L.C.M., list all prime factors(once only) occuring in the primefactorization of the given positive integers.For each of these factors, find the greatestexponent and raise each prime factor to thegreatest exponent and multiply them to getthe L.C.M.Note— Product of two positive integers aand b= the product of their H.C.F. and L.C.M.i.e. a × b = H.C.F. × L.C.M.

8. Rational numbers and Irrationalnumbers— The numbers which can be

written in the form of qp

where p and q are

integers and q ¹ 0, are called rational

numbers. For example 65

, 32

, 71

, 15

etc.

Those numbers which cannot be written in

qp

form, are called irrational numbers. For

example 2 , 3 , 5 etc.p is also an irrational number.

9. Terminating and non-terminatingrepeating decimal expansions of arational number— Every rational numbercan be express in decimals either terminatingor non-terminating repeating decimals.

(a) Terminating Decimals— A rationalnumber whose denomina tor can befactorise in the form of 2m × 5n where mand n a re non-negative integers areterminating decimals.

(b) Non-terminating repeating decimals— Ifin a rational number, the denominator is notfactorise in the form of 2m × 5n, where mand n are non-negative integers, the rational

number has non-terminating repeatingdecimals.

MULTIPLE CHOICE QUESTIONS

1. The exponent of 2 in the primefactorisation of 144, is

(a) 4 (b) 5(c) 6 (d) 3

Solution—

2 1442 722 362 183 93 3

1

144 = 24 × 32

\ Exponent of 2 is 4 (a)2. The LCM of two numbers is 1200. Which

of the following cannot be their HCF ?(a) 600 (b) 500(c) 400 (d) 200

Solution—LCM of two number = 1200

\ Their HCF of these two numbers will be thefactor of 1200

\ 500 cannot be its HCF (b)3. If n = 23 × 34 × 44 × 7, then the number of

consecutive zeroes in n, where n is anatural number, is

(a) 2 (b) 3(c) 4 (d) 7

Solution—Because it has four factors 2, 3, 4 and n

_ n = 23 × 34 × 44 × 7\ It has 4 zeroes (c)4. The sum of the exponents of the prime

factors in the prime factorisation of 196,is

(a) 1 (b) 2(c) 4 (d) 6

Solution—


Prime factors of 196

2 1962 987 497 7

1= 2 × 2 × 7 × 7= 22 × 72

Sum of exponents = 2 + 2 = 4 (c)5. The number of decimal places after which

the decimal expansion of the rational

number 52232 ´ will terminate, is

(a) 1 (b) 2(c) 3 (d) 4

Solution—

Decimal expansion of 52232 ´ = 20

23

= 520523´´

= 100115

= 1.15

\ Number of decimal places = 2 (b)6. If p1 and p2 are two odd prime numbers

such that p1 > p2, then p21 – p2

2 is(a) an even number (b) an odd number(c) an odd prime number(d) a prime number

Solution—p1 and p2 are two odd prime numbers suchthat p1 > p2, then p1 – p2 > 0Now p2

1 – p22 = (p1 + p2) (p1 – p2)

But p1 – p2 and p1 + p2 are both even numberas difference and sum of two odd numbersare even numbers.

\ p21 – p2

2 is an even number (a)7. If two positive integers a and b are

expressible in the form a = pq2 and b =p3q; p, q being prime numbers, then LCM(a, b) is

(a) pq (b) p3q3

(c) p3q2 (d) p2q2

Solution—a and b are two positive integers anda = pq2 and b = p3q, where p and q are primenumbers, thenLCM = p3q2 (c)

8. In Q. No. 7, HCF (a, b) is(a) pq (b) p3q3

(c) p3q2 (d) p2q2

Solution— Let a = pq2 and b = p3q where a and b arepositive integers and p, q are prime numbers,thenHCF = pq (a)

9. If two positive integers m and n areexpressible in the form m = pq3 and n =p3q2, where p, q are prime numbers, thenHCF (m, n) =

(a) pq (b) pq2

(c) p3q3 (d) p2q3

Solution—Given m and n are two positive integers s.t.m = pq3 and n = pq2, where p and q are primenumbers, thenHCF = pq2 (b)

10. If the LCM of a and 18 is 36 and the HCFof a and 18 is 2, then a =

(a) 2 (b) 3(c) 4 (d) 1

Solution—LCM of a and 18 = 36and HCF of a and 18 = 2

_ Product of LCM and HCF = product of numbers\ 36 × 2 = a × 18

Þ a = 18236´

= 4 (c)

11. The HCF of 95 and 152, is(a) 57 (b) 1(c) 19 (d) 38

Solution—HCF of 95 and 152 = 19 (c)


)152( 95 57 )95( 1 57 38 )57( 1 38 19 ) 38 ( 2 38 ×

95 1

12. If HCF (26, 169) = 13, then LCM(26, 169) =

(a) 26 (b) 52(c) 338 (d) 13

Solution—HCF (26, 169) = 13Since product of L.C.M. × H.C.F. = Productof Numbers

\ LCM (26, 169) = 1316926´

= 338 (c)

13. If a = 23 × 3, b = 2 × 3 × 5, c = 3n × 5 andLCM (a, b, c) = 23 × 32 × 5, then n =

(a) 1 (b) 2(c) 3 (d) 4

Solution—a = 23 × 3, b = 2 × 3 × 5, c = 3n × 5 andLCM (a, b, c) = 23 × 32 × 5

\ 3n = 32Þ n = 2 (b)14. The decimal expansion of the rational

number 125014587

will terminate after

(a) one decimal place (b) two decimal place(c) three decimal place(d) four decimal place

Solution—

Decimal expansion of 125014587

is terminate

after 4 decimal place (d)

þýü

îíì ´=

´´=

10000814587

81250814587

125014587

Q

15. If p and q are co-prime numbers, then p2

and q2 are(a) coprime (b) not coprime(c) even (d) odd

Solution—p and q are co-prime, thenp2 and q2 will also be coprime (a)Since p2 and q2 have no common factor

16. Which of the following rational numbershave terminating decimal ?

(i) 22516

(ii) 185

(iii) 212

(iv) 2507

(a) (i) and (ii) (b) (ii) and (iii)(c) (i) and (iii) (d) (i) and (iv)

Solution—We know that a rational number hasterminating decimal if the prime factors ofits denominator are in the form 2m × 5n

_ 250 = 2 × 53

\ 2507

has terminating decimals (d)

17. If 3 is the least prime factor of number aand 7 is the least prime factor of numberb, then the least prime factor of a + b, is

(a) 2 (b) 3(c) 5 (d) 10

Solution—3 is the least prime factor of a7 is the least prime factor of b, thenSum of a a and b will be divisible by 2

\ 2 is the least prime factor of a + b (a)

18. 273. is(a) an integer (b) a rational number(c) a natural number


(d) an irrational numberSolution—

Since decimal representation of number isnon-terminating reccuring.

\ 273. is a rational number (b)

19. The smallest number by which 27should be multiplied so as to get a rationalnumber is

(a) 27 (b) 3 3

(c) 3 (d) 3

Solution—

27 = 333 ´´ = 3 3

\ 27 should be multipled by 3

to get a rational number = 27 3 3´

27 3 81= ´ = (c)20. The smallest rational number by which

31

should be multiplied so that its

decimal expansion terminates after oneplace of decimal, is

(a) 103

(b) 101

(c) 3 (d) 1003

Solution—The smallest rational number which should

be multiplied by 31

to get a terminating

decimals = 103

_ 31

× 103

= 101

= 0.1

21. If n is a natural number, then 92n – 42n isalways divisible by

(a) 5 (b) 13(c) both 5 and 13 (d) None of these

Solution—n is natural number, and92n – 42n is the form of a2n – b2n is or(an)2 – (bn)2 which is divisible by (a + b) and(a – b)

or 9 + 4 and 9 – 4or 13 and 5 both (c)

22. If n is any natural number, then 6n – 5n

always ends with(a) 1 (b) 3(c) 5 (d) 7

Solution—Since, n is any natural numberWe know that 6n ends with 6 and 5n endswith 5

\ 6n – 5n will end with 6 – 5 = 1 (a)23. The LCM and HCF of two rational

numbers are equal, then the numbersmust be

(a) prime (b) co-prime(c) composite (d) equal

Solution—_ LCM and HCF of two rational numbers are equal

Then those must be equal (d)24. If the sum of LCM and HCF of two

numbers is 1260 and their LCM is 900more than their HCF, then the productof two numbers is

(a) 203400 (b) 194400(c) 198400 (d) 205400

Solution—Sum of LCM and HCF of two numbers = 1260LCM = 900 more than HCF

\ LCM = 900 + HCFBut LCM = HCF = 1260900 + HCF + HCF = 1260

Þ 2HCF = 1260 – 900 = 360

Þ HCF = 2360

= 180


and LCM = 1260 – 180 = 1080\ Product of two numbers = LCM × HCF =

1080 × 180 = 194400\ Product of numbers = 194400 (b)

25. The remainder when the square of anyprime number greater than 3 is dividedby 6, is

(a) 1 (b) 3(c) 2 (d) 4

Solution—_ The given prime number is greater than 3

Let the prime number be = 6k + 1When k is a natural number

\ (6k + 1)2 = 36k2 + 12k + 1= 6k (6k + 2) + 1

\ Remainder = 1 (a)26. For some integer m, every even integer

is of the form(a) m (b) m + 1(c) 2m (d) 2m + 1

Solution—We know that, even integers are 2, 4, 6, ...So, it can be written in the form of 2mWhere, m = Integer = Z

[Since, integer is represented by Z]or m = ..., –1, 0, 1, 2, 3, ...

\ 2m = ..., –2, 0, 2, 4, 6, ...Alternate MethodLet ‘a’ be a positive integer. On dividing ‘a’by 2, let m be the quotient and r be theremainder. Then, by Euclid’s divisionalgorithm, we havea = 2m + r, where a < r < 2 i.e., r = 0 andr = 1.

Þ a = 2 m or a = 2m + 1When, a = 2m for some integer m, thenclearly a is even. (c)

27. For some integer q, every odd integer isof the form

(a) q (b) q + 1(c) 2q (d) 2q + 1

Solution—

We know that, odd integers are 1, 3, 5, ...So, it can be written in the form of 2q + 1Where, q = integer = Zor q = ..., –1, 0, 1, 2, 3, ...

\ 2q + 1 = ..., –3, –1, 1, 3, 5, ...Alternate MethodLet ‘a’ be given positive integer. On dividing‘a’ by 2, let q be the quotient and r be theremainder. Then, by Euclid’s divisionalgorithm, we havea = 2q + r, where 0 < r < 2

Þ a = 2q + r, where r = 0 or r = 1Þ a = 2q or 2q + 1

When a = 2q + 1 for some integer q, thenclearly a is odd. (d)

28. n2 – 1 is divisible by 8, if n is(a) an integer (b) a natural number(c) an odd integer (d) an even integer

Solution—Let a = n2 – 1Here n can be even or odd.Case I : n = Even i.e., n = 2k, where k is aninteger.

Þ a = (2k)2 – 1Þ a = 4k2 – 1

At k = –1, 4(–1)2 – 1 = 4 – 1 = 3, which isnot divisible by 8.At k = 0, a = 4(0)2 – 1 = 0 – 1 = –1, whichis not divisible by 8.Case II : n = Odd i.e., n = 2k + 1, where kis an odd integer.

Þ a = 2k + 1Þ a = (2k + 1)2 – 1Þ a = 4k2 + 4k + 1 – 1Þ a = 4k2 + 4kÞ a = 4k(k + 1)

At k = –1, a = 4(–1)(–1 + 1) = 0 which isdivisible by 8.At k = 0, a = 4(0)(0 + 1) = 0 which isdivisible by 8.At k = 1, a = 4(1)(1 + 1) = 8 which isdivisible by 8.


Hence, we can conclude from above twocases, if n is odd, then n2 – 1 is divisible by 8.

(c)29. The decimal expansion of the rational

number 52332 ´ will terminate after

(a) one decimal place(b) two decimal places(c) three decimal places(d) more than 3 decimal places

Solution—

52332 ´

Multiply and divide the expansion by 5

2 233 5

2 5´´

= 233 510´

= 1.65

Hence, the decimal expansion of the rational

number 233

2 5´ will terminate after two

decimal places. (b)30. If two positive integers a and b are

written as a = x3y2 and b = xy3; x, y areprime numbers, then HCF (a, b) is

(a) xy (b) xy2

(c) x3y3 (d) x2y2

Solution—Given that, a = x3y2 = x × x × x × y × yand b = xy3 = x × y × y × y

\ HCF of a and b = HCF (x3y2, xy3) = x × y × y = xy2

[Since, HCF is the product of the smallestpower of each common prime factor involvedin the numbers] (b)

31. The least number that is divisible by allthe numbers from 1 to 10 (both inclusive)is

(a) 10 (b) 100(c) 504 (d) 2520

Solution—Factors of 1 to 10 numbers

1 = 12 = 1 × 23 = 1 × 34 = 1 × 2 × 25 = 1 × 56 = 1 × 2 × 37 = 1 × 78 = 1 × 2 × 2 × 29 = 1 × 3 × 310 = 1 × 2 × 5

\ LCM of number 1 to 10 = LCM (1, 2, 3, 4,5, 6, 7, 8, 9, 10)= 1 × 2 × 2 × 2 × 3 × 3 × 5 × 7 = 2520 (d)

32. The largest number which divides 70 and125, leaving remainders 5 and 8,respectively, is

(a) 13 (b) 65(c) 875 (d) 1750

Solution—Since, 5 and 8 are the remainders of 70 and125, respectively. Thus, after subtractingthese remainders from the numbers, we havethe numbers 65 = (70 – 5), 117 = (125 – 8),which is divisible by the required number.Now, required number = HCF of 65, 117

[For the largest number]For this, 117 = 65 × 1 + 52[_ Dividend = divisor × quotient + remainder]

Þ 65 = 52 × 1 + 13Þ 52 = 13 × 4 + 0\ HCF = 13

Hence, 13 is the largest number whichdivides 70 and 125, leaving remainders 5 and8. (a)

33. If the HCF of 65 and 117 is expressible inthe form 65m – 117, then the value of mis

(a) 4 (b) 2(c) 1 (d) 3

Solution—By Euclid’s division algorithm,


b = aq + r, 0 < r < a[_ dividend = divisor × quotient + remainder]

Þ 117 = 65 × 1 + 52Þ 65 = 52 × 1 + 13Þ 52 = 13 × 4 + 0\ HCF (65, 117) = 13 ...(i)

Also, given that HCF (65, 117) = 65m – 117...(ii)

From equations (i) and (ii),65m – 117 = 13

Þ 65m = 130Þ m = 2 (b)

34. The decimal expansion of the rational

number 125014587

will terminate after:

(a) one decimal place(b) two decimal places(c) three decimal places(d) four decimal places

Solution—

Rational number = 125014587

= 41 5214587´

2 12505 6255 1255 255 5

1

Now 1 41425872 5´

= 351014587´

× 3

3

)2()2(

= 100010814587

´´

= 10000116696

= 11.6696

Hence, given rational number will terminateafter four decimal places. (d)

35. Euclid’s division lemma states thatfor two positive integers a and b, thereexist unique integers q and r such that

a = bq + r, where r must satisfy(a) 1 < r < b (b) 0 < r < b(c) 0 < r < b (d) 0 < r < b

Solution—According to Euclid’s Division lemma, for apositive pair of integers there exists uniqueintegers q and r, such thata = bq + r, where 0 < r < b (c)

36. Which of the following is a pair of co-primes.

(a) (14, 35) (b) (18, 25)(c) (31, 93) (d) (32, 62)

Solution—We know that HCF of two co-primenumber is 1HCF of 14, 35 is 7HCF of 18, 25 is 1HCF of 31, 93 is 31HCF of 32, 60 is 4

\ Required co-prime number is (18, 25) (b)37. If a = (22 × 33 × 54) and b = (23 × 32 × 5)

then HCF (a, b) = ?(a) 90 (b) 180(c) 360 (d) 540

Solution—a = (22 × 33 × 54), b = (23 × 32 × 5)HCF = 22 × 32 × 5 = 2 × 2 × 3 × 3 × 5

= 180 (b)38. HCF of (23 × 32 × 5), (22 × 33 × 52) and

(24 × 3 × 53 × 7) is(a) 30 (b) 48(c) 60 (d) 105

Solution— HCF of 23 × 32 × 5, 22 × 33 × 52 &24 × 3 × 53 × 7HCF = 22 × 3 × 5

= 2 × 2 × 3 × 5 = 60 (c)39. LCM of (23 × 3 × 5) and (24 × 5 × 7) is(a) 40 (b) 560(c) 1120 (d) 1680

Solution—LCM of 23 × 3 × 5, 24 × 5 × 7= 24 × 3 × 5 × 7


= 2 × 2 × 2 × 2 × 3 × 5 × 7= 1680 (d)

40. The HCF of two numbers is 27 and theirLCM is 162. If one of the number is 54,what is the other number:

(a) 36 (b) 45(c) 9 (d) 81

Solution— HCF of two numbers = 27LCM = 162One number = 54

\ Second number = numberOneLCMHCF´

= 5416227´

= 81 (d)

41. The product of two numbers is 1600 andtheir HCF is 5. The LCM of the numbersis

(a) 8000 (b) 1600(c) 320 (d) 1605

Solution—Product of two numbers = 1600HCF = 5

\ LCM = HCFnumberstwoofProduct

= 51600

= 320 (c)

42. What is the largest number that divideseach one of 1152 and 1664 exactly:

(a) 32 (b) 64(c) 128 (d) 256

Solution—Largest number that divides each oneof 1152 and 1664

\ HCF of 1152 and 1664 = 128 (c)

1152)1664(1 1152 512)1152(2 1024 128)512(4 512 ×

43. What is the largest number that divides70 and 125, leaving remainders 5 and 8respectively

(a) 13 (b) 9(c) 3 (d) 585

Solution— Largest number that divides 70 and 125leaving remainders as 5 and 8 respectively.

\ Required number = 70 – 5 = 65and 125 – 8 = 117

\ HCF of 65, 117 = 13 (a)

65)117(1 65 52)65(1 52 13)52(4 52 ×

44. What is the largest number that divides245 and 1029, leaving remainder 5 in eachcase?

(a) 15 (b) 16(c) 9 (d) 5

Solution—Largest number that divides 245 and1029 leaving remainder as 5 in each case.

\ Required number = 245 – 5 = 240and 1029 – 5 = 1024Now, HCF of 240 and 1020 = 16 (b)

240)1024(4 960 54)240(4 192 48)64(1 48 16)48(3 48 ×

45. The simplest form of 11681095

is

(a) 2617

(b) 2625


(c) 1613

(d) 1615

Solution—Simplest form of 11681095

HCF of 1095 and 1168 = 73Dividing each by 73,

11681095

= 731168731095¸¸

= 1615

(d)

1095)1168(1 1095 73)1095(15 73 365 365 ×

46. Euclid's division lemma states that forany positive integer a and b, there existunique integers q and r such that a = bq+ r, where r must satisfy:

(a) 1 < r < b (b) 0 < r < b(c) 0 < r < b (d) 0 < r < b

Solution— In a = bq + rr must satisfy i.e. 0 < r < b (c)

47. A number when divided by 143 leaves 31as remainder. What will be the remainderwhen the same number is divided by 13?

(a) 0 (b) 1(c) 3 (d) 5

Solution—Let the given number when divided by143 gives q as quotient and 31 as remainder.

\ Number = 143q + 31= (13 × 11)q + 31= 13 × 11q + 13 × 2 + 5= 13(11q + 2) + 5

The number where divided by 73, gives 5 asremainder. (d)

48. Which of the following is an irrationalnumber?

(a) 722

(b) 3.1416

(c) 1416.3 (d) 3.141141114...Solution—3.141141114... is an irrational number

because it has non-terminating non-repeatingdecimal representation. (d)

49. p is(a) an integer(b) a rational number(c) an irrational number(d) none of these

Solution—p is an irrational number. (c)_ it has non-terminating non-repeating decimal

representation. (d)

50. 35.2 is(a) an integer(b) a rational number(c) an irrational number(d) none of these

Solution— 35.2 is a rational number as it is non-terminating repeating decimal. (b)

51. 2.13113111311113... is(a) an integer(b) a rational number(c) an irrational number(d) none of these

Solution—2.13113111311113... is an irrationalnumber.It is non-terminating non-repeating decimal.

(c)52. The number 3.24636363... is(a) an integer(b) a rational number(c) an irrational number(d) none of these

Solution— 3.24636363...

= 6324.3_ It is non-terminating repeating decimal.\ It is a rational number. (b)


53. Which of the following rational numbersis expressible as a terminating decimal?

(a) 165124

(b) 30131

(c) 6252027

(d) 4621625

Solution— 6252027

= 452027

is a rational because it

has terminating decimal as q = 54 which isin form of 2m × 5n. (c)

54. The decimal expansion of the rational

number 52372 ´ will terminate after


Solution—Decimal expansion of 52372 ´

= 5437´ = 20

37 = 520

537´´

= 100185

= 1.85

It is upto 2 decimal places. (b)55. The decimal expansion of the number

125014753

will terminate after


Solution—Decimal expansion of 125014753

=

41 5214753´

2 12505 6255 1255 255 5

1

_ q is in form of 2m × 5n

\ It is terminating and 81250814753´´

= 10000118024

= 11.8024

Which is upto 4 places of decimal. (d)56. The number 1.732 is(a) an irrational number(b) a rational number(c) an integer(d) a whole number

Solution—1.732 is a rational number.As it is terminating decimal. (b)

57. a and b are two positive integers suchthat the least prime factor of a is 3 andthe least prime factor of b is 5. Then, theleast prime factor of (a + b) is

(a) 2 (b) 3(c) 5 (d) 8

Solution—_ Least prime factor of a positiveinteger a is 3 and b is 5

\ 2 is neither a factor of a nor of b\ a and b are odd

Then (a + b) = even(Sum of two odd numbers is even)

\ (a + b) is divisible by 2Which is the least prime factor. (a)

58. 2 is

(a) a rational number(b) an irrational number(c) a terminating decimal


(d) a nonterminating repeating decimal

Solution— 2 is an irrational number.. (b)

_ it has non-terminating non-repeating decimalrepresentation. (d)

59. 21

is

(a) a fraction(b) a rational number(c) an irrational number(d) none of these

Solution— 21

=22

12´´

= 21

2

It is an irrational number. (c)

60. (2 + 2 ) is

(a) an integer(b) a rational number(c) an irrational number(d) none of these

Solution—2 + 2 is an irrational number as sumof a rational and an irrational is an irrational.(c)

61. What is the least number that is divisibleby all the natural numbers from 1 to 10(both inclusive)?

(a) 100 (b) 1260(c) 2520 (d) 5040

Sol. LCM of 1 to 10= 2 × 2 × 2 × 3 × 3 × 5 × 7 = 2520 (c)2 1, 2, 3, 4, 5, 6, 7, 8, 9, 102 1, 1, 3, 2, 5, 3, 7, 4, 9, 53 1, 1, 3, 1, 5, 3, 7, 2, 9, 55 1, 1, 1, 1, 5, 1, 7, 2, 3, 5

1, 1, 1, 1, 1, 1, 7, 2, 3, 1


Points to Remember :1. Polynomial— Let x be a variable, n be a

positive integer and a1, a2, a3 ..... an beconstants (real numbers),then f (x) = an xn + an – 1 xn – 1 + an – 2 xn – 2 + .....+ a1x + a0 is called a polynomial in variable x.an xn, an – 1 xn – 1, ..... a1x, a0 are known as theterms of the polynomial and an, an – 1 ..... a1,a0 are their co-efficients.

2. Degree of Polynomial— The exponent ofthe highest degree term in a polynomial isknown as its degree.

3. Constant Polynomial— A polynomial ofdegree zero (0) is called a constantpolynomial.

4. Linear Polynomial— A polynomial ofdegree 1 is called the linear polynomial. Alinear polynomial may be monomial or abinomial having one term or two termsrespectively e.g., ax or ax + b.

5. Quadratic Polynomial— A polynomial ofdegree 2 is called a quadratic polynomial e.g.,f (x) = ax2 + bx + c where a ¹ 0.A quadratic polynomial can be monomial,binomial or trinomial having one term, twoterms or three terms respectively.

6. Cubic Polynomial— A polynomial of degree3 is called a cubic polynomiale.g., f (x) = ax3 + bx2 + cx + d where a ¹ 0,b, c, d are real numbers.

7. Bi-quadratic Polynomial— A polynomial ofdegree 4 is called a bi-quadratic polynomiale.g., f (x) = ax4 + bx3 + cx2 + dx + e.Where a ¹ 0, b, c, d, e are real numbers.

8. Value of a Polynomial— If f (x) is apolynomial and a is any real number thenthe real number obtained by replacing x bya in f (x), is called the value of f (x) at x = aand is denoted by f (a).

Chapter — 2POLYNOMIALS

9. Zero of a Polynomial— A real number a isa zero of a polynomial f (x) if f (a) = 0.Finding a zero of a polynomial f (x) meanssolving the polynomial equation f (x) = 0. Apolynomial of degree one, has one zero,having degree of two, has two zeros, havingdegree of 3, has three zeros.

10. Graph of Polynomials—(i) Graph of a Linear Polynomial is a line, so

f (x) = ax + b is called a linear polynomial.(ii) Graph of a quadratic polynomial

f (x) = ax2 + bx + c is a parabola in shapewhich opens upward or downwardaccording to as a > 0 or a < 0.

(iii) Graph of a cubic polynomial does not havea fixed shape. It may cut the axe at the mostthrice.

11. Geometrically the zeros of a polynomialf (x)— Geometrically the zeros of apolynomial f (x) are the x-coordinates of thepoints where the graph y = f (x) intersectsx-axis.

12. (i) Relationship of zeros and co-efficientsof a quadratical polynomialf (x) = ax2 + bx + c, Let a and b be thezeros of the polynomial f (x),

then a + b = ab-

= – 2oftcoefficienoftcoefficien

xx

and ab = ac

= 2oftcoefficientermconstant

x

(ii) In cubic polynomial f (x) = ax3 + bx2 + cx + d,Let a, b and g are the zeros of thepolynomial f (x)

then a + b + g =ab-

= – 3

2

oftcoefficienoftcoefficien

xx


ab + b g + g a = ac

= 3oftcoefficienoftcoefficien

xx

and ab g = ad-

= – 3oftcoefficientermconstant

x

13. Division Algorithm— If f (x) and g (x) areany two polynomials with g (x) ¹ 0, thenwe can always find polynomials q (x) andr (x) such that f (x) = q (x) g (x) + r (x)where r (x) = 0 or degree of r (x) < degreeof g (x).If r (x) = 0, then polynomial g (x) is a factorof the polynomial f (x).Therefore Dividend = Quotient × Divisior +Remainder which is similar to Euclid’sDivision Algorithm.

MULTIPLE CHOICE QUESTIONS

Mark the correct alternative in each ofthe following :

1. If a, b are the zeros of the polynomial

f (x) = x2 + x + 1, then a1

+ b1

=

(a) 1 (b) –1(c) 0 (d) None of these

Solution—f (x) = x2 + x + 1Here a = 1, b = 1, c = 1

_ a and b are its zeros

\ a + b = = ab-

= 11-

= –1

ab = ac

= 11

= 1

\ a1

+ b1

= abab +

= 11-

= –1, i.e., (b)


p (x) = 4x2 + 3x + 7, then a1

+ b1

is

equal to

(a) 37

(b) – 37

(c) 73

(d) – 73

Solution—_ a and b are the zeros of p (x) = 4x2 + 3x + 7

Here a = 4, b = 3, c = 7

\ a + b = ab-

= 43-

, ab = ac

= 47

a1

+ b1

= abab +

= 7443

´´-

= 73-

, i.e., (d)

3. If one zero of the polynomialf (x) = (k2 + 4) x2 + 13x + 4k is reciprocalof the other, then k =

(a) 2 (b) –2(c) 1 (d) –1

Solution—f (x) = (k2 + 4) x2 + 13x + 4kHere a = k2 + 4, b = 13, c = 4k

_ One zero is reciprocal of the otherLet first zero = a

Then second zero = a1

\ a + a1

= ab-

= 113

2 +-

k

and a × a1

= ac

= 442 +k

k

Þ 442 +k

k = 1 Þ k2 + 4 = 4k

Þ k2 – 4k + 4 = 0Þ (k – 2)2 = 0 Þ k – 2 = 0\ k = 2, i.e., (a)4. If the sum of the zeros of the polynomial

f (x) = 2x3 – 3kx2 + 4x – 5 is 6, then valueof k is

(a) 2 (b) 4


(c) –2 (d) –4Solution—

f (x) = 2x3 – 3kx2 + 4x – 5Here a = 2, b = –3k, c = 4, d = –5

Sum of zeros = ab-

= –( )

23k-

\ 23k

= 6 Þ k = 326´

= 4, i.e., (b)

5. If a and b are the zeros of the polynomialf (x) = x2 + px + q, then a polynomial

having a1

and b1

is its zeros is

(a) x2 + qx + p (b) x2 – px + q(c) qx2 + px + 1 (d) px2 + qx + 1

Solution—f (x) = x2 + px + q

_ a and b are the zeros of f (x)

Then a + b = 1p-

= –p

ab = 1q

= q

\ Sum of zeros of the new polynomial

= a1

+ b1

= abab +

= qp-

and product of zeros = a1

× b1

= ab1

= q1

\ Polynomial will be given byx2 – (sum of zeros) x + product of zeros

= x2 – ÷÷ø

öççè

æ -qp

x + q1

=q1

(qx2 + px + 1),

(c)6. If a, b are the zeros of polynomial

f (x) = x2 – p(x + 1) – c, then (a + 1)(b + 1) =(a) c – 1 (b) 1 – c

(c) c (d) 1 + cSolution—

f (x) = x2 – p (x + 1) – c= x2 – px – p – c = x2 – px – (p + c)Here A = 1, B = –p, C = –(p + c)

\ a + b = – ÷øöç

èæ -

1p

= p

ab =( )

1cp +-

Now (a + 1) (b + 1) = ab + a + b + 1= ab + (a + b) + 1= –(p + c) + {(p) + 1}= –p – c + p + 1= –c + 1 i.e. 1 – c, (b)

7. If a, b are the zeros of the polynomialf (x) = x2 – p (x + 1) – c such that (a + 1)(b + 1) = 0, then c =

(a) 1 (b) 0(c) –1 (d) 2

Solution—f (x) = x2 – p (x + 1) – c= x2 – px – p – c= x2 – px – (p + c)

_ a and b are the zeros\ a + b = –(–p) = p

ab = – ÷øöç

èæ +

1cp

= –(p + c)

Now (a + 1) (b + 1) = 0Þ ab + (a + b) + 1 = 0Þ –(p + c) + p + 1 = 0Þ –p – c + p + 1 = 0Þ –c + 1 = 0Þ c = 1, i.e., (a)8. If f (x) = ax2 + bx + c has no real zeros

and a + b + c < 0, then(a) c = 0 (b) c > 0


(c) c < 0 (d) None of theseSolution—

f (x) = ax2 + bx + c_ Zeros are not real

\ b2 – 4ac < 0 ....(i)but a + b + c < 0b < –(a + c)Squaring both sidesb2 < (a + c)2

Þ (a + c)2 – 4ac < 0 {From (i)}Þ (a – c)2 < 0Þ a – c < 0Þ a < c i.e., (d)9. If the diagram in figure shows the graph

of the polynomial f (x) = ax2 + bx + c,then

(a) a > 0, b < 0 and c > 0(b) a < 0, b < 0 and c < 0(c) a < 0, b > 0 and c > 0(d) a < 0, b > 0 and c < 0

Solution—_ Curve ax2 + bx + c intersects x-axis at two

points and curve is upward.\ a > 0, b < 0 and c > 0 (a)

10. Figure shows the graph of the polynomialf (x) = ax2 + bx + c for which

(a) a < 0, b > 0 and c > 0(b) a < 0, b < 0 and c > 0(c) a < 0, b < 0 and c < 0

(d) a > 0, b > 0 and c < 0

Solution—_ Curve ax2 + bx + c intersects x-axis at two

points and curve is downward.\ a < 0, b < 0 and c > 0 (b)

11. If the product of zeros of the polynomialf (x) = ax3 – 6x2 + 11x – 6 is 4, then a =

(a) 23

(b) – 23

(c) 32

(d) – 32

Solution—f (x) = ax3 – 6x2 + 11x – 6Given, product of zeros = 4

But product of zeros = ac-

\( )a

6-- = 4 Þ 4a = 6 Þ a = 4

6 = 2

3,

i.e., (a)12. If zeros of the polynomial f (x) = x3 –

3px2 + qx – r are in AP, then(a) 2p3 = pq – r (b) 2p3 = pq + r(c) p3 = pq – r (d) None of these

Solution—f (x) = x3 – 3px2 + qx – rHere a = 1, b = –3p, c = q, d = –r

_ Zeros are in AP

Y

Xf (x) = ax + bx + c

OX

Y¢

–2

ba

, –D4a

Y

XOX¢

Y¢

–2

ba

, –D4a


Let the zeros be a – d, a, a + d

Then sum of zeros = ab-

= – ÷øöç

èæ -

13p

= 3p

Þ a – d + a + a + d = 3pÞ 3a = 3p Þ a = p

Also, (a – d) a + a (a + d) + (a + d) (a – d)

= ac

= 1q

= q

Þ a2 – ad + a2 + ad + a2 – d2 = qÞ 3a2 – d2 = qÞ 3p2 – d2 = q Þ d2 = 3p2 – q

and (a – d) × a × (a + d) =d

a-

=( )1

r--= r

Þ a (a2 – d2) = rÞ p (p2 – 3p2 + q) = r Þ p3 – 3p3 + pq = rÞ –2p3 + pq = rÞ –2p3 = –pq + rÞ 2p3 = pq – r, i.e., (a)

13. If the product of two zeros of thepolynomial f (x) = 2x3 + 6x2 – 4x + 9 is 3,then its third zero is

(a) 23

(b) – 23

(c) 29

(d) – 29

Solution—f (x) = 2x3 + 6x2 – 4x + 9Product of 2 zeros = 3Here a = 2, b = 6, c = –4, d = 9Let a, b and g are its zeros

Product of zeros = ad-

Þ abg = ad-

= 29-

Þ 3g = 29-

(_ Product of two zeros = 3)

Þ g = 329

´-

= 23-

, i.e., (b)

14. If the polynomial f (x) = ax3 + bx – c isdivisible by the polynomial g (x) = x2 + bx + c,then ab =

(a) 1 (b) c1

(c) –1 (d) – c1

Solution—f (x) = ax3 + bx – c = ax3 + ox2 + bx – cg (x) = x2 + bx + cDividing f (x) by g (x)

) + (

+ ( ) c

a c

b ac x

3x bxax abx acx

abx

3

2

+ +

+

- -

(

-

-

- -- - -

- -

2

2 2

2

abx ab x abc

b ac ab x abc c+ + +

) +

x bx c + +2ax ab-

_ f (x) is completely divisible by g (x)\ r (x) = 0

(b – ac + ab2) x + abc – c = 0b – ac + ab2 = 0 and abc – c = 0

Þ abc = c Þ ab = 1, i.e., (a)15. In Q. No. 14, ac =(a) b (b) 2b(c) 2b2 (d) –2b

Solution—In the previous questionsRemainder = 0

\ (b – ac + ab2) = 0b + ab2 = ac

Þ ac = b (1 + ab) = b (1 + 1) = 2b, i.e., (b)16. If one root of the polynomial f (x) = 5x2 +

13x + k is reciprocal of the other, thenthe value of k is

(a) 0 (b) 5

(c) 61

(d) 6


Solution—f (x) = 5x2 + 13x + k

_ One root is reciprocal of the other

Then roots are a and a1

Now product of roots = ac

Þ a × a1

= 5k

Þ 5k

= 1

Þ k = 5, i.e., (b)17. If a, b, g are the zeros of the polynomial

f (x) = ax3 + bx2 + cx + d, then a1

+ b1

+ g1

=

(a) – db

(b) dc

(c) – dc

(d) – ac

Solution—f (x) = ax3 + bx2 + cx + dand a, b, g are its zeros

\ a + b + g = ab-

ab + bg + ga = ac

abg = ad-

\ a1

+ b1

+ g1

= abgabgabg ++

=

ad

ac

- = ac

× da

- = dc-

, i.e., (c)

18. If a, b, g are the zeros of the polynomial f(x) = ax3 + bx2 + cx + d, then a2 + b2 + g2=

(a) 2

2

aacb -

(b)a

acb 22 -

(c) 2

2 2b

acb +(d) 2

2 2a

acb -

Solution—f (x) = a3 + bx2 + cx + d

_ a, b, g are its zeros

\ a + b + g = ab-

ab + bg + ga = ac

and abg = ad-

Nowa2 + b2 + g2 = (a + b + g)2 –2 (ab + bg + ga)

=2

÷øöç

èæ -

ab

– 2 ac

= 2

2

ab

– ac2

= 2

2 2a

acb -, i.e., (d)

19. If a, b, g are the zeros of the polynomial f (x)

= x3 – px2 + qx – r, then ab1

+ bg1

+ ga1

=

(a) pr

(b) rp

(c) – rp

(d) – pr

Solution—f (x) = x3 – px2 + qx – rHere a = 1, b = –p, c = q, d = –r

_ a, b and g are its zeros

\ a + b + g = ab-

= – ÷øöç

èæ -

1p

= p

ab + bg + ga = ac

= 1q

= q

abg = ad-

=( )1

r- - = r


Now ab1

+ bg1

+ ga1

= abgbag ++

= abggba ++

= rp

, i.e., (b)


f (x) = ax2 + bx + c, then 21

a + 21

b =

(a) 2

2 2a

acb -(b) 2

2 2c

acb -

(c) 2

2 2a

acb +(d) 2

2 2c

acb +

Solution—f (x) = ax2 + bx + ca and b are its zeros

\ a + b = ab-

, ab =ca

Now 21

a + 21

b = 22

22

baab +

= 22

22

baba +

=( )

22

2 2ba

abba -+ = 2

2

2

÷øöç

èæ

-÷øöç

èæ -

ac

ac

ab

=

2

2

2

2 2

ac

ac

ab -

=

2

2

2

2 2

aca

acb -

= 2

2 2a

acb - × 2

2

ca

= 2

2 2c

acb -, i.e., (b)

21. If two of the zeros of the cubic polynomialax3 + bx2 + cx + d are each equal to zero,then the third zero is

(a) ad-

(b) ac

(c) ab-

(d) ab

Solution—Two of the zeros of the cubic polynomialax3 + bx2 + cx + d are each equal to zeroLet a, b and g are its zeros, then

a + b + g = ab-

Þ 0 + 0 + a = ab-

Þ a = ab-

\ Third zero will be ab-

(c)

22. If two zeros of x3 + x2 – 5x – 5 are 5 ,

and – 5 then its third zero is

(a) 1 (b) –1(c) 2 (d) –2

Solution—

Two zeros of x3 + x2 – 5x – 5 are 5 and – 5Here a = 1, b = 1, c = –5, d = –5Let a, b and g are are three zeros, then

a + b + g = ab-

Þ 5 + ( )5- + g = 11-

Þ g = –1 (b)23. The product of the zeros of x3 + 4x2 + x – 6

is(a) –4 (b) 4(c) 6 (d) –6

Solution—Quadratic polynomial is x3 + 4x2 + x – 6Here a = 1, b = 4, c = 1, d = –6

\ Product of zeros = abg = ad-

=( )1

6-- = 1

6 = 6 (c)


24. What should be added to the polynomialx2 – 5x + 4, so that 3 is the zero of theresulting polynomial ?

(a) 1 (b) 2(c) 4 (d) 5

Solution—_ 3 is the zero of the polynomial

f (x) = x2 – 5x + 4\ x – 3 is a factor of f (x)

Now f (3) = (3)2 – 5 × 3 + 4 = 9 – 15 + 4= 13 – 15 = –2

\ –2 is to be subtractingor 2 is added (b)

25. What should be subtracted to thepolynomial x2 – 16x + 30, so that 15 isthe zero of the resulting polynomial ?

(a) 30 (b) 14(c) 15 (d) 16

Solution—_ 15 is the zero of polynomial f (x) = x2 – 16x + 30

Then f (15) = 0\ f (15) = (15)2 – 16 × 15 + 30

= 225 – 240 + 30= 255 – 240 = 15

\ 15 is to be subtracted (c)26. A quadratic polynomial, the sum of whose

zeroes is 0 and one zero is 3, is(a) x2 – 9 (b) x2 + 9(c) x2 + 3 (d) x2 – 3

Solution—In a quadratic polynomialLet a and b be its zerosand a + b = 0and one zero = 3

\ 3 + b = 0 Þ b = –3\ Second zero = –3\ Quadratic polynomial will be given by

(x – 3) (x + 3) i.e. x2 – 9 (a)27. If two zeroes of the polynomial

x3 + x2 – 9x – 9 are 3 and –3, then itsthird zero is

(a) –1 (b) 1(c) –9 (d) 9

Solution—Two zeroes of the polynomial x3 + x2 – 9x – 9are 3 and –3Here a = 1, b = 1, c = –9, d = –9

\ Sum of zeros = a + b + g = ab-

Þ 3 + (–3) + g = 11-

Þ g = –1\ Third zero = –1 (a)

28. If 5 and – 5 are two zeroes of thepolynomial x3 + 3x2 – 5x – 15, then itsthird zero is

(a) 3 (b) –3(c) 5 (d) –5

Solution—

5 and – 5 are two zeroes of polynomial

x3 + 3x2 – 5x – 15Here a = 1, b = 3, c = –5, d = –15

\ Sum of zeros = ab-

Þ a + b + g = 13-

Þ 5 + ( )5- + g = –3

g = –3\ Third zero = –3 (b)

29. If x + 2 is a factor x2 + ax + 2b and a + b =4, then

(a) a = 1, b = 3 (b) a = 3, b = 1(c) a = –1, b = 5 (d) a = 5, b = –1

Solution—x + 2 is a factor of x2 + ax + 2band a + b = 4

_ x + 2 is one of the factor\ x = –2 is its one zero\ f (–2) = 0


Þ (–2)2 + a (–2) + 2b = 0Þ 4 – 2a + 2b = 0Þ 2a – 2b = 4 ...(1)

a – b = 2 ...(2)But a + b = 4On adding eqn. (1) (2) ; we get,

2a = 6 Þ a = 26

= 3

and a + b = 4 Þ 3 + b = 4Þ b = 4 – 3 = 1\ a = 3, b = 1 (b)

30. The polynomial which when divided by–x2 + x – 1 gives a quotient x – 2 andremainder 3, is

(a) x3 – 3x2 + 3x – 5 (b) –x3 – 3x2 – 3x – 5(c) –x3 + 3x2 – 3x + 5 (d) x3 – 3x2 – 3x + 5

Solution—Divisor = –x2 + x – 1, Quotient = x – 2 andRemainder = 3, ThereforePolynomial = Divisor × Quotient + Remainder= (–x2 + x – 1) (x – 2) + 3= –x3 + x2 – x + 2x2 – 2x + 2 + 3= –x3 + 3x2 – 3x + 5 (c)

31. The number of polynomials having zeroes–2 and 5 is

(a) 1 (b) 2(c) 3 (d) more than 3

Solution—Let p(x) = ax2 + bx + c be the requiredpolynomial whose zeroes are –2 and 5.

\ Sum of zeroes = ab-

Þ ab-

= –2 + 5 = 13

= 1)3(--

...(i)

and product of zeroes = ac

Þ ac

= –2 × 5 = 110-

...(ii)

From equations (i) and (ii),

a = 1, b = –3 and c = –10\ p(x) = ax2 + bx + c = 1· x2 – 3x – 10

= x2 – 3x – 10But we know that, if we multiply/divide anypolynomial by any arbitrary constant. Then,the zeroes of polynomial never change.

\ p(x) = kx2 – 3kx – 10k[where, k is a real number]

Þ p(x) =kx2

– k3

x – k10

[where, k is a non-zero real number]Hence, the required number of polynomialsare infinite i.e., more than 3. (d)

32. If one of the zeroes of the quadraticpolynomial (k – 1)x2 + kx + 1 is –3, thenthe value of k is

(a) 34

(b) 34-

(c) 32

(d) 32-

Solution—Given that, one of the zeroes of the quadraticpolynomial say p(x) = (k – 1)x2 + kx + 1 is–3, then p(–3) = 0

Þ (k – 1)(–3)2 + k(–3) + 1 = 0Þ 9(k – 1) – 3k + 1 = 0Þ 9k – 9 – 3k + 1 = 0Þ 6k – 8 = 0

\ k = 34

(a)

33. The zeroes of the quadratic polynomialx2 + 99x + 127 are

(a) both positive (b) both negative(c) both equal(d) one positive and one negative

Solution—Let given quadratic polynomial bep(x) = x2 + 99x + 127One comparing p(x) with ax2 + bx + c, wegeta = 1, b = 99 and c = 127


We know that, x =a

acbb2

42 -±-

=12

12714)99(99 2

´´´-±-

=2

508980199 -±-

=2

929399 ±- = 2

4.9699 ±-

= 24.9699 +-

, 24.9699 --

= 26.2-

, 24.195-

= –1.3, –97.7Hence, both zeroes of the given quadraticpolynomial p(x) are negative.Alternate MethodWe know that,In quadratic polynomial,

ifþýü

<<<>>>

0,0or00,0or0

cbacba

, then both zeroes

are negative.In given polynomial, we see thata = 1 > 0, b = 99 > 0 and c = 127 > 0which satisfy the above condition,So, both zeroes to the given quadraticpolynomial are negative. (b)

34. If the zeroes of the quadratic polynomialx2 + (a + 1) x + b are 2 and –3, then

(a) a = –7, b = –1 (b) a = 5, b = –1(c) a = 2, b = –6 (d) a = 0, b = –6

Solution—Let p(x) = x2 + (a + 1) x + bGiven that, 2 and –3 are the zeroes of thequadratic polynomial p(x).

\ p(2) = 0 and p(–3) = 0Þ 22 + (a + 1) (2) + b = 0

Þ 4 + 2a + 2 + b = 0Þ 2a + b = –6 ...(i)

and (–3)2 + (a + 1) (–3) + b = 0Þ 9 – 3a – 3 + b = 0Þ 3a – b = 6 ...(ii)

On adding equations (i) and (ii), we get5a = 0 Þ a = 0Put the value of a in equation (i), we get2 × 0 + b = –6 Þ b = –6So, the required values are a = 0 and b = –6

(d)35. Given that one of the zeroes of the cubic

polynomial ax3 + bx2 + cx + d is zero, theproduct of the other two zeroes is

(a) ac- (b) a

c

(c) 0 (d) ab-

Solution—Let p(x) = ax3 + bx2 + cx + dGiven that, one of the zeroes of the cubicpolynomial p(x) is zero.Let a, b and g are the zeroes of cubicpolynomial p(x), where a = 0.We know that,

Sum of product of two zeroes at a time = ac

Þ ab + bg + ga = ac

Þ 0 × b + bg + g × 0 = ac

[_ a = 0, given]

Þ 0 + bg + 0 = ac

Þ bg = ac

Hence, product of other two zeroes = ac

(b)36. The zeroes of the quadratic polynomial


x2 + ax + a, a ¹ 0,(a) cannot both be positive(b) cannot both be negative(c) area always unequal(d) are always equal

Solution—Let p(x) = x2 + ax + a, a ¹ 0On comparing p(x) with ax2 + bx + c, we geta = 1, b = a and c = a

Now, x =a

acbb2

42 -±-

[By quadratic formula]

=12

42

´-±- aaa

=2

)4( -±- aaa, a ¹ 0

+ – +0 4

Here, we see thata(a – 4) > 0

Þ a Î (–¥, 0) È (4, ¥)Now, we know thatIn quadratic polynomial ax2 + bx + cIf a > 0, b > 0, c > 0 or a < 0, b < 0, c < 0Then the polynomial has always all negativezeroesand if a > 0, c < 0 or a < 0, c > 0, then thepolynomial has always zeroes of oppositesign.Case I : If a Î (–¥, 0) i.e. a < 0

Þ a = 1 > 0, b, c = a < 0So, both zeroes are of opposite sign.Case II : If a Î (4, ¥) i.e., a > 4

Þ a = 1 > 0, b, c > 4So, both zeroes are negative.Hence, in any case zeroes of the givenquadratic polynomial cannot both the

positive. (a)37. If one of the zeroes of the cubic

polynomial x3 + ax2 + bx + c is –1, thenthe product of other two zeroes is

(a) b – a + 1 (b) b – a – 1(c) a – b + 1 (d) a – b – 1

Solution—Let p(x) = x3 + ax2 + bx + cLet a, b and g be the zeroes of the givencubic polynomial p(x).

\ a = –1 [given]and p(–1) = 0

Þ (–1)3 + a(–1)2 + b(–1) + c = 0Þ –1 + a – b + c = 0Þ c = 1 – a + b ...(i)

We know that,Product of all zeroes

= (–1)3· 3oftCoefficientermConstant

x = 1c-

abg = –cÞ (–1)bg = –c [_ a = –1]Þ bg = cÞ bg = 1 – a + b [from Eq. (i)]

Hence, product of the other two roots is1 – a + b.Alternate MethodSince, –1 is one of the zeroes of the cubicpolynomial f(x) = x2 + ax2 + bx + ci.e., (x + 1) is a factor of f(x).Now, using division algorithm,

)1()1(

)1(

)1()1()1(

)1()1(

)1(

1

2

2

2

23

23

+-+-+

-+-

+-+-++-

-+-

+-

+

++++abxax

abc

abxabcxab

xaxa

bxxa

xx

cbxaxxx


\ x3 + ax2 + bx + c = (x + 1) × {x2 + (a – 1)x +(b – a + 1)} + (c – b + a – 1)

Þ x3 + ax2 + bx + (b – a + 1) = (x + 1) {x2 + (a– 1)x + (b – a + 1)}Let a and b be the other two zeroes of thegiven polynomial, thenProduct of zeroes

= (–1)a·b = 3oftCoefficientermConstant

x-

Þ –a·b = 1)1( +-- ab

Þ ab = –a + b + 1Hence, the required product of other tworoots is (–a + b + 1). (a)

38. Given that two of the zeroes of the cubicpolynomial ax3 + bx2 + cx + d are 0, thethird zero is

(a) ab- (b) a

b

(c) ac

(d) ad-

Solution—Two of the zeroes of the cubic polynomialax3 + bx2 + cx + d = 0, 0Let the third zero be aThen, use the relation between zeroes andcoefficient of polynomial, we have

a + 0 + 0 = ab-

Þ a = ab- (a)

39. If one zero of the quadratic polynomialx2 + 3x + k is 2, then the value of k is

(a) 10 (b) –10(c) 5 (d) –5

Solution—Let the given quadratic polynomial beP(x) = x2 + 3x + kIt is given that one of its zeros is 2

\ P(2) = 0Þ (2)2 + 3(2) + k = 0Þ 4 + 6 + k = 0Þ k + 10 = 0Þ k = –10 (b)

40. If the zeroes of the quadratic polynomialax2 + bx + c, c ¹ 0 are equal, then

(a) c and a have opposite signs(b) c and b have opposite signs(c) c and a have the same sign(d) c and b have the same sign

Solution—The zeroes of the given quadratic polynomialax2 + bx + c, c ¹ 0 are equal. If coefficientof x2 and constant term have the same signi.e., c and a have the same sign. While bi.e., coefficient of x can be positive/negativebut not zero.e.g.,

(i) x2 + 4x + 4 = 0Þ (x + 2)2 = 0Þ x = –2, –2

(ii) x2 – 4x + 4 = 0Þ (x – 2)2 = 0Þ x = 2, 2

Alternate MethodGiven that, the zeroes of the quadraticpolynomial ax2 + bx + c, where c ¹ 0, areequal i.e., discriminant (D) = 0

Þ b2 – 4ac = 0 Þ b2 = 4ac

Þ ac =c

b2

Þ ac > 0Which is only possible when a and c havethe same signs. (c)

41. If one of the zeroes of a quadraticpolynomial of the form x2 + ax + b is thenegative of the other, then it

(a) has no linear term and constant term isnegative.


(b) has no linear term and the constant termis positive.

(c) can have a linear term but the constantterm is negative.

(d) can have a linear term but the constantterm is positive.

Solution—Let p(x) = x2 + ax + b

Now, product of zeroes = 2oftCoefficientermConstant

x

Let a and b be the zeroes of p(x)

\ Product of zeroes (a·b) = 1b

Þ ab = b ...(i)Given that, one of the zeroes of a quadraticpolynomial p(x) is negative of the other.

\ ab < 0So, b < 0 [from Eq. (i)]Hence, b should be negative

Put a = 0, then, p(x) = x2 + b = 0Þ x2 = –b

Þ x = + b- [_ b < 0]Hence, if one of the zeroes of quadraticpolynomial p(x) is the negative of the other,then it has no linear term i.e., a = 0 and theconstant term is negative i.e., b < 0.Alternate MethodLet f(x) = x2 + ax + band by given condition the zeroes are a and–a.

\ Sum of the zeroes = a – a = aÞ a = 0\ f(x) = x2 + b, which cannot be linear

and product of zeroes = a · (–a) = bÞ –a2 = b

which is possible when, b < 0.Hence, it has no linear term and the constantterm is negative. (a)

42. Which of the following is not the graph of a quadratic polynomial?

Y

XO

X¢

Y¢

Y

XO

X¢

Y¢

Y

XO

X¢

Y¢(a) (b)

(d)(c)


Solution—For any quadratic polynomial ax2 + bx + c, a ¹ 0, the graph of the corresponding equation y = ax2

+ bx + c has one of the two shapes either open upwards like È or open downwards like Çdepending on whether a > 0 or a < 0. These curves are called parabolas. So, option (d) cannot bepossible.Also, the curve of a quadratic polynomial crosses the X-axis on at most two points but in option(d) the curve crosses the X-axis on the three points, so it does not represent the quadraticpolynomial. (d)

43. Which of the following is a polynomial?

(a) x2 – 5x + 4 x + 3 (b) x3/2 – x + x1/2 + 1

(c) x + x1

(d) 2 x2 – 3 3 x + 6

Solution— 2 x2 – 3 3 x + 6 is polynomial, others are not polynomial. Since in all other expressions,the power of x in one of the terms be not a non-negative integers. (d)

44. Which of the following is not a polynomial?

(a) 3 x2 – 2 3 x + 5 (b) 9x2 – 4x + 2

(c) 23

x3 + 6x2 – 21

x – 8 (d) x + x3

Solution—x + x3

is not a polynomial since the power of x in3x i.e. 3x–1 be –1 which is not a non-

negative integer and others are polynomial. (d)45. The zeros of the polynomial x2 – 2x – 3 are(a) –3, 1 (b) –3, –1 (c) 3, –1 (d) 3, 1

Solution—Let f(x) = x2 – 2x – 3= x2 – 3x + x – 3 = x(x – 3) + 1(x – 3) = (x – 3)(x + 1)

If x – 3 = 0, then x = 3and if x + 1 = 0, then x = –1

\ Zeros are 3, –1 (c)

46. The zeros of the polynomial x2 – 2 x – 12 are

(a) 2 , 2- (b) 3 2 , –2 2

(c) –3 2 , 2 2 (d) 3 2 , 2 2

Solution—Let f(x) = x2 – 2 x – 12


= x2 – 3 2 x + 2 2 x – 12

ïþ

ïýü

ïî

ïíì

+-=-

´-=-

22232

222312Q

= x(x – 3 2 ) + 2 2 (x – 3 2 )

= (x – 3 2 )(x + 2 2 )

If x – 3 2 = 0, then x = 3 2

If x + 2 2 = 0, then x = –2 2

Hence, zeros are 3 2 , –2 2 (b)

47. The zeros of the polynomial 4x2 + 5 2 x– 3 are

(a) –3 2 , 2 (b) –3 2 ,22

(c)2

23-,

42

(d) none of these

Solution—Polynomial f(x) = 4x2 + 5 2 x – 3

= 4x2 + 6 2 x – 2 x – 3

þýü

îíì

´-=-\-=-´

1121212)3(4Q

= 2 2 x( 2 x + 3) – 1( 2 x + 3)

= ( 2 x + 3)(2 2 x – 1)

\ Either 2 x + 3 = 0,

then x = 23-

=22

23´

- =

223-

or 2 2 x – 1 = 0,

then x = 221

=222

12´

´ =

42

\ Zeros are :2

23- and

42

(c)

48. The zeros of the polynomial x2 + 61

x – 2

are

(a) –3, 4 (b) 23-

, 34

(c) 34-

, 23

(d) none of these

Solution—Polynomial is x2 + 61

x – 2

= x2 + 23

x – 34

x – 2

ïïþ

ïïý

ü

ïïî

ïïí

ì

-=

÷øöç

èæ -´=-

34

23

61

34

232 xxQ

= x ÷øöç

èæ +

23x – 3

4 ÷øöç

èæ +

23x

= ÷øöç

èæ +

23x ÷

øöç

èæ -

34x

Either x + 23

= 0, then x = 23-

or x – 34

= 0, then x = 34

Hence, zeros are 23-

, 34

(b)

49. The zeros of the polynomial

7x2 – 311x

– 32

are

(a) 32

, 71-

(b) 72

, 31-


(c) 32-

, 71

(d) none of these

Solution—Polynomial = 7x2 – 311

x – 32

= 7x2 – 314

x + x – 32

2 1473 3

14 14 13 311 14 13 3

ì ü- -æ ö´ =ç ÷ï ïè øï ïï ï- -\ = ´í ýï ï

- -ï ï= ´ï ïî þ

Q

= 7x ÷øöç

èæ -

32x + 1 ÷

øöç

èæ -

32x

= ÷øöç

èæ -

32x (7x + 1)

Either x – 32

= 0, then x = 32

If 7x + 1 = 0, then x = 71-

Hence, zeros are 32

, 71-

(a)

50. The sum and the product of the zeros ofa quadratic polynomial are 3 and –10respectively. The quadratic polynomial is

(a) x2 – 3x + 10 (b) x2 + 3x – 10(c) x2 – 3x – 10 (d) x2 + 3x + 10

Solution—Sum of zeros = 3Product of zeros = –10

\ Polynomial : x2 – (Sum of zeros) x +Product of zeros

= x2 – 3x – 10 (c)51. A quadratic polynomial whose zeros are

5 and –3, is

(a) x2 + 2x – 15 (b) x2 – 2x + 15(c) x2 – 2x – 15 (d) none of these

Solution—Zeros are 5 and –3\ Sum of zeros = 5 – 3 = 2

Product of zeros = 5 × (–3) = –15\ Polynomial : x2 – (Sum of zeros)x + Product

of zeros= x2 – 2x – 15 (c)

52. A quadratic polynomial whose zeros are

53

and 21-

, is

(a) 10x2 + x + 3 (b) 10x2 + x – 3(c) 10x2 – x + 3 (d) 10x2 – x – 3

Sol. Zeros are 53

and 21-

Sum of zeros = 53

– 21

= 1056 -

= 101

and product of zeros = 53

× ÷øöç

èæ -

21

= 103-

Now polynomial is given byx2 – (Sum of zeros)x + Product of zeros

= x2 – 101

x – 103

= 101

[10x2 – x – 3] (d)

53. The zeros of the quadratic polynomialx2 + 88x + 125 are

(a) both positive(b) both negative(c) one positive and one negative(d) both equal

Solution—Let f(x) = x2 + 88x + 125

Here, sum of roots = ab-

= –88

and product = ac

= 125


_ Product is positive,\ Both zeros can be both positive or both

negative._ Sum is negative.

\ Both zeros are negative. (b)54. If a and b are the zeros of x2 + 5x + 8

then the value of (a + b) is(a) 5 (b) –5(c) 8 (d) –8

Solution—_ a and b are the zeros of x2 + 5x + 8Then sum of zeros (a + b)

= ab-

= 15-

= –5 (b)

55. If a and b are the zeros of 2x2 + 5x – 9then the value of ab is

(a) 25-

(b) 25

(c) 29-

(d) 29

Solution—_ a and b are the zeros of 2x2 + 5x – 9

\ Product of zeros (ab) = ac

= 29-

(c)

56. If one zero of the quadratic polynomialkx2 + 3x + k is 2 then the value of k is

(a) 65

(b) 65-

(c) 56

(d) 56-

Solution—_ 2 is a zero of kx2 + 3x + k\ It will satisfy the quadratic equation

kx2 + 3x + k = 0Þ k(2)2 + 3 × 2 + k = 0Þ 4k + 6 + k = 0 Þ 5k = –6

Þ k = 56-

(d)

57. If one zero of the quadratic polynomial(k – 1)x2 + kx + 1 is –4 then the value of

k is

(a) 45-

(b) 45

(c) 34-

(d) 34

Solution—_ –4 is a zero of (k – 1)x2 + 4x + 1\ –4 will satisfy the equation

(k – 1)x2 + kx + 1 = 0Þ (k – 1)(–4)2 + k(–4) + 1 = 0Þ 16k – 16 – 4k + 1 = 0Þ 12k – 15 = 0 Þ 12k = 15

Þ k = 1215

= 45

(b)

58. If –2 and 3 are the zeros of the quadraticpolynomial x2 + (a + 1)x + b then

(a) a = –2, b = 6 (b) a = 2, b = –6(c) a = –2, b = –6 (d) a = 2, b = 6

Sol. _ –2 and 3 are the zeros of x2 + (a + 1)x + b

Sum of roots = ab-

Þ 11+a

= –(–2 + 3)

Þ a + 1 = –1 Þ a = –1 – 1 = –2and product of roots = b = –2 × 3 = –6

\ a = –2, b = –6 (c)59. If one zero of 3x2 + 8x + k be the

reciprocal of the other then k = ?(a) 3 (b) –3

(c) 31

(d) 31-

Solution—Polynomial is 3x2 + 8k + k

Let one zero be a, then second zero is = α1

\ Product of zeros = a × α1

= 1 = ac

\ 1 = 3k

Þ k = 3 (a)


60. If the sum of the zeros of the quadraticpolynomial kx2 + 2x + 3k is equal to theproduct of its zeros then k = ?

(a) 31

(b) 31-

(c) 32

(d) 32-

Solution—Polynomial : kx2 + 2x + 3k

Sum of zeros = ab-

= k2-

and product of zeros ac

= kk3

= 3

But sum of zeros = Product of zeros

\ k2-

= 3 Þ k = 32-

(d)


x2 + 6x + 2 then ÷÷ø

öççè

æ+β1

α1

= ?

(a) 3 (b) –3(c) 12 (d) –12

Solution—a and b are the zeros of thepolynomial x2 + 6x + 2

\ a + b = ab-

= 16-

= –6

and ab = ac

= 12

= 2

Now, α1

+ β1

=αβαβ +

=αββα +

= 26-

= –3

(b)62. If a, b, g are the zeros of the polynomial

x3 – 6x2 – x + 30 then (ab + bg + ga) = ?(a) –1 (b) 1(c) –5 (d) 30

Solution—_ a, b, g are the zeros ofx3 – 6x2 – x + 30

Then ab + bg + ga = ac

= 11-

= –1 (a)

63. If a, b, g are the zeros of the polynomial2x3 + x2 – 13x + 6 then abg = ?

(a) –3 (b) 3

(c) 21-

(d) 213-

Solution—_ a, b, g are the zeros of 2x3 + x2 – 13x + 6,

then abg = ad-

= 26-

= –3 (a)

64. If a, b, g be the zeros of the polynomialp(x) such that (a + b + g) = 3,(ab + bg + ga) = –10 andabg = –24 then p(x) = ?

(a) x3 + 3x2 – 10x + 24(b) x3 + 3x2 + 10x – 24(c) x3 – 3x2 – 10x + 24(d) None of these

Solution—a, b, g are the zeros of p(x) such that

a + b + g = ab-

= 3,

ab + bg + ga = ac

= –10

and abg = ad-

= –24

\ x3 – (3)x2 + (–10)x + 24= x3 – 3x2 – 10x + 24 (c)

65. If two of the zeros of the cubic polynomialax3 + bx2 + cx + d are 0 then the thirdzero is

(a) ab-

(b) ab

(c) ac

(d) ad-


Solution—Let a, b, g are the zeros of thepolynomial ax3 + bx2 + cx + d

Then a + b + g = ab-

But two zeros are 0

\ 0 + 0 + g = ab-

Þ g = ab-

(a)

66. If one of the zeros of the cubic polynomialax3 + bx2 + cx + d is 0 then the product ofthe other two zeros is

(a) ac-

(b) ac

(c) 0 (d) ab-

Solution—If one zero of cubic polynomialax3 + bx + c + d = 0Let a be zero, then

ab + bg + ga = ac

0 + bg + 0 = ac

Þ bg = ac

\ Product of the other two zeros = ac

(b)

67. If one of the zeros of the cubic polynomialx3 + ax2 + bx + c is –1 then the product ofthe other two zeros is

(a) a – b – 1 (b) b – a – 1(c) 1 – a + b (d) 1 + a – b

Solution—One of the zeros of the polynomialx3 + ax2 + bx + c is –1 ...(i)Let a, b, g be the zeros and a = –1, thenSubstituting the value of x in (i)and (–1)3 + a(–1)2 + b(–1) + c = 0–1 + a – b + c = 0

Þ c = 1 – a + b

and abg = –c(–1) (bg) = –c Þ bg = c

Þ bg = +c= (1 – a + b)= 1 – a + b (c)

68. If a, b be the zeros of the polynomial2x2 + 5x + k such that a2 + b2 + ab =

421

then k = ?

(a) 3 (b) –3(c) –2 (d) 2

Solution—a, b be the zeros of 2x2 + 5x + k

Then a + b = ab-

= 25-

and ab = ac

= 2k

and a2 + b2 + ab = 421

Þ (a + b)2 – ab = 421

Þ2

25

÷øöç

èæ -

– 2k

= 421

Þ 425

– 2k

= 421

Þ 425

– 421

= 2k

Þ 44

= 2k

Þ 2k

= 1 Þ k = 2 (d)

69. On dividing a polynomial p(x) by anonzero polynomial q(x), let g(x) be thequotient and r(x) be the remainder thenp(x) = q(x) · g(x) + r(x), where

(a) r(x) = 0 always(b) deg r(x) < deg g(x) always


(c) either r(x) = 0 or deg r(x) < deg g(x)(d) r(x) = g(x)

Solution—p(x) is divided by q(x), thenp(x) = q(x) × g(x) + r(x)

\ Either r(x) = 0Degree of r(x) < deg of g(x) (c)

70. Which of the following is a truestatement?

(a) x2 + 5x – 3 is a linear polynomial.

(b) x2 + 4x – 1 is a binomial.(c) x + 1 is a monomial.(d) 5x3 is a monomial.

Solution—(a) is not a linear polynomial.(b) is trinominal not binomial.(c) is not a monomial.(d) 5x2 is monomial is true. (d)

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