01. Real Numbers

Exercise 1.1


Question 1:

Use Euclid’s division algorithm to find the HCF of:
(i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 255

Answer:
(i) 135 and 225
Since 225 > 135, we apply the division lemma to 225 and 135 to obtain
225 = 135 × 1 + 90
Since remainder 90 ≠ 0, we apply the division lemma to 135 and 90 to obtain
135 = 90 × 1 + 45
We consider the new divisor 90 and new remainder 45, and apply the division lemma to obtain
90 = 2 × 45 + 0
Since the remainder is zero, the process stops.
Since the divisor at this stage is 45,
Therefore, the HCF of 135 and 225 is 45.
(ii)196 and 38220
Since 38220 > 196, we apply the division lemma to 38220 and 196 to obtain
38220 = 196 × 195 + 0
Since the remainder is zero, the process stops.
Since the divisor at this stage is 196,
Therefore, HCF of 196 and 38220 is 196.
(iii)867 and 255
Since 867 > 255, we apply the division lemma to 867 and 255 to obtain
867 = 255 × 3 + 102
Since remainder 102 ≠ 0, we apply the division lemma to 255 and 102 to obtain
255 = 102 × 2 + 51
We consider the new divisor 102 and new remainder 51, and apply the division lemma to obtain
102 = 51 × 2 + 0
Since the remainder is zero, the process stops.
Since the divisor at this stage is 51,
Therefore, HCF of 867 and 255 is 51.

Question 2:

Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is
some integer.
Answer:

Let a be any positive integer and b = 6. Then, by Euclid’s algorithm,
a = 6q + rfor some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r < 6.
Therefore, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5
Also, 6q + 1 = 2 × 3q + 1 = 2k₁ + 1, where k₁ is a positive integer
6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = 2k₂ + 1, where k₂ is an integer
6q + 5 = (6q + 4) + 1 = 2 (3q + 2) + 1 = 2k₃ + 1, where k₃ is an integer
Clearly, 6q + 1, 6q + 3, 6q + 5 are of the form 2k + 1, where k is an integer.
Therefore, 6q + 1, 6q + 3, 6q + 5 are not exactly divisible by 2. Hence, these expressions of numbers are odd numbers.
And therefore, any odd integer can be expressed in the form 6q + 1, or 6q + 3,
or 6q + 5

Question 3:
An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?

Answer:

HCF (616, 32) will give the maximum number of columns in which they can march.
We can use Euclid’s algorithm to find the HCF.
616 = 32 × 19 + 8
32 = 8 × 4 + 0
The HCF (616, 32) is 8.
Therefore, they can march in 8 columns each.

Question 4:

Use Euclid’s division lemma to show that the square of any positive integer is either of form 3m or 3m + 1 for some integer m.
[Hint: Let x be any positive integer then it is of the form 3q, 3q + 1 or 3q + 2. Now square each of these and show that they can be rewritten in the form 3m or 3m + 1.]

Answer:
Let a be any positive integer and b = 3.
Then a = 3q + r for some integer q ≥ 0
And r = 0, 1, 2 because 0 ≤ r < 3
Therefore, a = 3q or 3q + 1 or 3q + 2
Or,

Where k₁, k₂, and k₃ are some positive integers
Hence, it can be said that the square of any positive integer is either of the form 3m or 3m + 1.

Question 5:
Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.
Answer:

Let a be any positive integer and b=3
a=3q+r, where q ≥ 0 and 0 ≤  r <  3
 a=3q or 3q+1 or 3q+2
Therefore, every number can be represented as these three forms.
There are three cases.
Case 1: When a=3q,
a³ = (3q)³
a³ = 27q³
a³ = 9(3q³) = 9m

Case 2: When a = 3q + 1,
a³ = (3q +1)³
a³ = 27q³ + 27q² + 9q + 1
a³ = 9(3q³ + 3q² + q) + 1
a³ = 9m + 1
Where m is an integer such that m = (3q³ + 3q² + q)
Case 3: When a = 3q + 2,
a³ = (3q +2)³
a³ = 27q³ + 54q² + 36q + 8
a³ = 9(3q³ + 6q² + 4q) + 8
a³ = 9m + 8
Where m is an integer such that m = (3q³ + 6q² + 4q)
Therefore, the cube of any positive integer is of the form 9m, 9m + 1,
or 9m + 8.

Solve these Question :

1.  Using Euclid’s division algorithm prove that : 847, 2160 are co-primes (relatively primes).
2. Using Euclid’s division Lemma, find the H.C.F. of 92690, 7378 and 7161.
3.  Using Euclid’s division algorithm find HCF of 255, 1309 and 1326.
4.  Find the largest number that divides 628, 3129 and 15630 to leave remainders 3,4 and 5 respectively.
5. Find the greatest number which divides the number 90,115 and 140 completely, when these numbers are increased by 1,2 and 3 respectively.
6.    Show that the square of an odd positive integer can be of the form 6q+1 or 6q+3 for some integer q.
7.  Show that the square of any positive integer can be in the form of 6q, 6q+1, 6q+3, 6q+4.

Exercise 1.2 

Question 1:

Express each number as product of its prime factors:
(i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429

Answer:
(i)  140 = 2×2×5×7 = 2²×5×7
(ii) 156 = 2×2×3×13 =2²×3×13
(iii) 3825 = 3×3×5×5×17 = 3²×5²×17
(iv) 5005 = 5×7×11×13
(v) 7429 = 17×19×23

 Question 2:

Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.
(i) 26 and 91 (ii) 510 and 92 (iii) 336 and 54

Answer:
(i) 26 and 91
    26= 2×13
    91= 7×13
    HCF = 13
    LCM = 2×7×13 = 182
    Product of the two Number = 26×91=2366
    HCF × LCM = 13×182 = 2366
Hence, product of two numbers = HCF × LCM

(ii) 510 and 92
      510 = 2×3×5×17
      92 = 2×2×23
      HCF = 2
      LCM = 2×2×3×5×17×23 = 23460
      Product of the two Number = 510 × 92 = 46920
      HCF × LCM = 2×23460 = 46920
Hence, product of two numbers = HCF × LCM

(iii) 336 and 54
       336 = 2×2×2×2×3×7
       54 = 2×3×3×3
       HCF = 2×3 = 6
       LCM = 2×2×2×2×3×3×3×7 = 3024
       Product of the two Number = 336 × 54 = 18144
      HCF × LCM = 6 × 3024 = 18144
Hence, product of two numbers = HCF × LCM

Question 3:

Find the LCM and HCF of the following integers by applying the prime factorisation method.
(i) 12, 15 and 21 (ii) 17, 23 and 29 (iii) 8, 9 and 25
Answer:
(i) 12, 15 and 21
     12 = 2×2×3
     15 = 3×5
     21 = 3×7
     HCF = 3
     LCM = 2×2×3×5×7 = 420

(ii) 17, 23 and 29
      17 = 1×17
      23 = 1×23
      29 = 1×29
      HCF = 1
     LCM = 17×23×29 = 11339

(iii) 8, 9 and 25
       8 =2×2×2
       9 = 3×3
       25 = 5×5
       HCF = 1
     LCM = 2×2×2×3×3×5×5= 1800

Question 4:
Given that HCF (306, 657) = 9, find LCM (306, 657).
Answer:

Question 5:
Check whether 6ⁿ can end with the digit 0 for any natural number n.

Answer:

If any number ends with the digit 0, it should be divisible by 10 or in other words, it will also be divisible by 2 and 5 as 10 = 2 × 5
Prime factorisation of 6ⁿ = (2 ×3)ⁿ
It can be observed that 5 is not in the prime factorisation of 6ⁿ.
Hence, for any value of n, 6ⁿ will not be divisible by 5.
Therefore, 6ⁿ cannot end with the digit 0 for any natural number n.

Question 6 :
Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.

Answer:

Numbers are of two types - prime and composite. Prime numbers can be divided by 1 and only itself, whereas composite numbers have factors other than 1 and itself.
It can be observed that
7 × 11 × 13 + 13 = 13 × (7 × 11 + 1) = 13 × (77 + 1)
= 13 × 78
= 13 ×13 × 6
The given expression has 6 and 13 as its factors. Therefore, it is a composite number.
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5 ×(7 × 6 × 4 × 3 × 2 × 1 + 1)
= 5 × (1008 + 1)
= 5 ×1009
1009 cannot be factorised further. Therefore, the given expression has 5 and 1009 as its factors. Hence, it is a composite number.

Question 7:
There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?

Answer:

It can be observed that Ravi takes lesser time than Sonia for completing 1 round of the circular path. As they are going in the same direction, they will meet again at the same time when Ravi will have completed 1 round of that circular path with respect to Sonia. And the total time taken for completing this 1 round of circular path will be the LCM of time taken by Sonia and Ravi for completing 1 round of circular path respectively i.e., LCM of 18 minutes and 12 minutes.
18 = 2 × 3 × 3
And, 12 = 2 × 2 × 3
LCM of 12 and 18 = 2 × 2 × 3 × 3 = 36
Therefore, Ravi and Sonia will meet together at the starting pointafter 36 minutes.

Solve these Question :

Exercise 1.3

Question 1:

Question 2:

Question 3:

Solve these Question :

Exercise 1.4

Question 1:

Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:

Answer:
(i)
3125 = 5⁵
The denominator is of the form 5^m
Hence the decimal expansion of  13/3125 is terminating.

(ii)
8 = 2³

The denominator is of the form 2^m
Hence the decimal expansion of  17/8 is terminating.

(iii)
455 = 5×7×13
Since the denominator is not in the form 2^m×5ⁿ

Hence the decimal expansion  is not terminating.

(iv)

1600= 2⁶ × 5²

the denominator is of the form 2^m×5ⁿ

Hence the decimal expansion of  15/1600 is terminating.

(v)

343 = 7³

Since the denominator is not in the form 2^m×5ⁿ

Hence the decimal expansion   is not terminating.

(vi)

2³×5²

the denominator is of the form 2^m×5ⁿ

Hence the decimal expansion   is terminating.

(vii)

2²×5⁷×7⁵

Since the denominator is not in the form 2^m×5ⁿ

Hence the decimal expansion   is not terminating.

Question 2:
Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions.
Answer:

Question 3:

The following real numbers have decimal expansions as given below. In each case,
decide whether they are rational or not. If they are rational, and of the form , p/q
what can you say about the prime factors of q?
(i) 43.123456789 (ii) 0.120120012000120000. . . (iii) 43.123456789
Answer :

02. Polynomials

Exercise 2.1

Question 1:

The graphs of y = p(x) are given in following figure, for some polynomials p(x). Find the number of zeroes of p(x), in each case.

Answer:
(i) The number of zeroes is 0 as the graph does not cut the x-axis at any point.

(ii) The number of zeroes is 1 as the graph intersects the x-axis at only 1 point.
(iii) The number of zeroes is 3 as the graph intersects the x-axis at 3 points.
(iv) The number of zeroes is 2 as the graph intersects the x-axis at 2 points.
(v) The number of zeroes is 4 as the graph intersects the x-axis at 4 points.
(vi) The number of zeroes is 3 as the graph intersects the x-axis at 3 points.

 Exercise 2.2

Question 1:

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
i) x² – 2x – 8                       (ii) 4s² – 4s + 1                         (iii) 6x² – 3 – 7x

(iv) 4u² + 8u                 (v) t² – 15                          (vi) 3x² – x – 4

Answer:

Question 2:
Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

Answer :

Exercise 2.3

Question 1:

Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:

Answer:

Quotient = x - 3
Remainder = 7x - 9

Quotient = x² + x - 3
Remainder = 8



Quotient = -x² - 2
Remainder = -5x + 10  

Concept insight: While dividing one polynomial by another, first arrange the polynomial  in descending powers of the variable. In the process of division, be careful about the signs of the coefficients of the terms of the polynomials. After performing division, one can check his/her answer obtained by the division algorithm which is as below:

Dividend = Divisor x Quotient + Remainder
Also, remember that the quotient obtained is a polynomial only.

Question 2:

Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:

Answer:
i) The polynomial 2t⁴ + 3t³ - 2t² - 9t - 12  can be divided by the polynomial t² - 3 = t² + 0.t - 3 as follows:

(















Since the remainder is 0, t² - 3 is a factor of 2t⁴ + 3t³ - 2t² - 9t - 12 .

(ii) The polynomial 3x⁴ + 5x³ - 7x² + 2x + 2 can be divided by the polynomial x² + 3x + 1 as follows:

Since the remainder is 0, x² + 3x + 1 is a factor of 3x⁴ + 5x³ - 7x² + 2x + 2

(iii) The polynomial x⁵ - 4x³ + x² + 3x + 1 can be divided by the  polynomial x³ - 3x + 1 as follows:



Since the remainder is not equal to 0, x³ - 3x + 1 is not a factor of x⁵ - 4x³ + x² + 3x + 1.

Question 3:

Answer:
Let p(x) = 3x⁴ + 6x³ - 2x² - 10x -5



Now, x² + 2x + 1 = (x + 1)²
So, the two zeroes of  x² + 2x + 1 are -1 and -1.


Question 4:
On dividing x³-3x²+x+2   by a polynomial g(x), the quotient and remainder were x – 2

and –2x + 4, respectively. Find g(x).

Answer:
Divided, p(x) = x³ - 3x² + x + 2
Quotient = (x - 2)
Remainder = (-2x + 4)
Let g(x) be the divisor.

According to the division algorithm,

Dividend = Divisor x Quotient + Remainder





Concept insight: When a polynomial is divided by any other non-zero polynomial, then it satisfies the division algorithm which is as below:
Dividend = Divisor x Quotient + Remainder


Divisor x Quotient = Dividend - Remainder

So, from this relation, the divisor can be obtained by dividing the result of (Dividend - Remainder) by the quotient.


Question 5:

Give examples of polynomial p(x), g(x), q(x) and r(x), which satisfy the division algorithm and
(i) deg p(x) = deg q(x)
(ii) deg q(x) = deg r(x)
(iii) deg r(x) = 0
Answer:
According to the division algorithm, if p(x) and g(x) are two polynomials with g(x)  0, then we can find polynomials q(x) and r(x) such that
p(x) = g(x) x q(x) + r(x), where r(x) = 0 or degree of r(x) < degree of g(x).

(i)    Degree of quotient will be equal to degree of dividend when divisor is constant.
Let us consider the division of 18x² + 3x + 9  by 3.
Here, p(x) = 18x² + 3x + 9  and g(x) = 3
q(x) = 6x² + x + 3  and r(x) = 0
Here, degree of p(x) and q(x) is the same which is 2.


Checking:
p(x) = g(x) x q(x) + r(x)

Thus, the division algorithm is satisfied.

(ii)    Let us consider the division of 2x⁴ + 2x by 2x³,
Here, p(x) = 2x⁴ + 2x and g(x) = 2x³
q(x) = x and r(x) = 2x
Clearly, the degree of q(x) and r(x) is the same which is 1.

Checking,
p(x) = g(x) x q(x) + r(x)
2x⁴ + 2x =  (2x³ ) x x  + 2x
2x⁴ + 2x = 2x⁴ + 2x
Thus, the division algorithm is satisfied.

(iii)    Degree of remainder will be 0 when remainder obtained on division is a constant.
Let us consider the division of 10x³ + 3 by 5x².
Here, p(x) = 10x³ + 3 and g(x) = 5x²
q(x) = 2x and r(x) = 3
Clearly, the degree of r(x) is 0.

Checking:
p(x) = g(x) x q(x) + r(x)
10x³ + 3 = (5x² ) x 2x  +  3
10x³ + 3 = 10x³ + 3
Thus, the division algorithm is satisfied.

 Exercise 2.4

Question 1:

Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:

Answer:


On comparing the given polynomial with the polynomial ax³ + bx² + cx + d, we obtain a = 2, b = 1, c = -5, d = 2



Thus, the relationship between the zeroes and the coefficients is verified.



On comparing the given polynomial with the polynomial ax_³ + bx_² + cx + d, we obtain a = 1, b = -4, c = 5, d = -2.



Thus, the relationship between the zeroes and the coefficients is verified.

Question 2:

 Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, − 7, − 14 respectively.
Answer:

Let the polynomial be ax_³  + bx_² + cx + d and its zeroes be   .
According to the given information:



If a = k, then b = -2k, c = -7k, d = 14k

Thus, the required cubic polynomial is k(x_³ - 2x_² - 7x + 14), where k is any real number.
The simplest polynomial will be obtained by taking k = 1.

Question 3:
If the zeroes of the polynomial x³ – 3x² + x + 1  are a – b, a, a + b, find a and b.
Answer:
Let p(x) = x³ - 3x² + x + 1
The zeroes of the polynomial p(x) are given as a - b, a, a + b.
Comparing the given polynomial with dx³ + ex² + fx + g, we can observe that
d = 1, e = -3, f = 1, g = 1



Therefore, the zeroes of p(x) are 1-b, 1, 1+b .



Question 4:

Answer:


Therefore, x^₂ - 4x + 1 is a factor of the given polynomial.

For finding the remaining zeroes of the given polynomial, we will carry out the division of



Hence, 7 and -5 are also zeroes of the given polynomial.

Question 5:

Answer: 

By division algorithm,
Dividend = Divisor x Quotient + Remainder
Divisor x Quotient = Dividend - Remainder





It will be perfectly divisible by x^₂ - 2x + k.




10 - a - 8 x 5 + 25 = 0
10 - a - 40 + 25 = 0
- 5 - a = 0
a = -5

Thus, k = 5 and a = -5.