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(1) \(3\sqrt[3]{2}\)

(4) √25 = ±5

\(\frac{73}{(2^4\times5^3)}\) = \(\frac{73\times5}{(2^4\times5^4)}\)

= \(\frac{365}{(2\times5)^4}\)

= \(\frac{365}{(10)^4}\)

= \(\frac{365}{10000}\)

= 0.0365 

Thus, the decimal expansion of \(\frac{73}{(2^4\times5^3)}\) is 0.0365.

Let the two irrationals be 4 √5 and 3 √5 

(4 √5) × (3 √5) = 60 

Thus, product (i.e., 60) is a rational number.

Answer is (B) p³

(4) π 

π represents a irrational number

(c) π is an irrational number 

π is an irrational number because it is a non-repeating and non-terminating decimal. 

False

Prime numbers are always odd numbers and the sum of odd numbers is even.

Example:

Let two prime numbers be 3 and 5.

Sum of 3 and 5 = 3 + 5 = 8 which is not a prime number.

We can write: 

(2ⁿ × 5ⁿ) = (2 × 5)ⁿ 

= 10ⁿ 

For any value of n, we get 0 in the end. 

Thus, there is no value of n for which (2ⁿ × 5ⁿ ) ends in 5.

Let the two irrationals be 4 - √5 and 4 + √5 

(4 - √5) + (4 + √5) = 8 

Thus, sum (i.e., 8) is a rational number.

Answer is (C) 343/(2³ x 5² x 7³)

We can write: 

(2ⁿ × 5ⁿ ) = (2 × 5) ⁿ

= 10ⁿ 

For any value of n, we get 0 in the end. 

Thus, there is no value of n for which (2ⁿ × 5ⁿ ) ends in 5. 

Correct option is (D) 47

\((7+\sqrt{2})(7-\sqrt{2})\) \(=7^2-(\sqrt2)^2\)

= 49 - 2 = 47

Correct option is  D) 47

True

π is an irrational number

Answer is (D) Euclid

No, it is not possible to have two numbers whose HCF is 25 and LCM is 520. 

Since, HCF must be a factor of LCM, but 25 is not a factor of 520.

No, it is not possible to have two numbers whose HCF is 25 and LCM is 520. Since, HCF must be a factor of LCM, but 25 is not a factor of 520. 

No, it is not possible to have two numbers whose HCF is 18 and LCM is 760. Since, HCF must be a factor of LCM, but 18 is not factor of 760.

Answer is (C) 2m

Correct option is (B) \(7+2\sqrt{10}\)

\((\sqrt{5}+\sqrt{2})(\sqrt{5}+\sqrt{2})\) \(=5+\sqrt{10}+\sqrt{10}+2\)

= \(7+2\sqrt{10}\)

 B) 7 + 2\(\sqrt{10}\)

Correct option is (D) (10)^-3 = 0.001

\(log_{10}\) 0.001 = -3

Its exponential form 0.001 \(=10^{-3}\)

Correct option is D) (10)^-3 = 0.001

Answer is (B) non-terminating repeating decimal

HCF is always a factor is its LCM.

Given: HCF is 25 and LCM is 520

But 25 is not a factor of 520

It is not possible to have two numbers having HCF is 25 and LCM is 520.

Answer is (C) odd number

Let us assume, to the contrary, that √3 is rational.

That is, we can find integers a and b (≠0) such that √3 =a/b 

Suppose a and b have a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime.

So, b√3 =a

Squaring on both sides, and rearranging, we get 3b² = a².

Therefore, a² is divisible by 3, and by Theorem 1.3, it follows that a is also divisible by 3.

So, we can write a = 3c for some integer c.

Substituting for a, we get 3b² = 9c², that is, b² = 3c².

This means that b² is divisible by 3, and so b is also divisible by 3 (using Theorem 1.3 with p = 3).

Therefore, a and b have at least 3 as a common factor.

But this contradicts the fact that a and b are coprime.

This contradiction has arisen because of our incorrect assumption that 3 is rational.

So, we conclude that √3 is irrational.

Correct option is (A) 16

\(\frac{p}{q}=0.\bar7\)

\(\therefore\) \(\frac{p}{q}=0.777.....\)   ________(1)

Multiply equation (1) by 10, we get

\(10\frac{p}{q}=7.777.....\)   ________(2)

Subtract (1) from (2), we get

\(9\frac{p}{q}=7\)

\(\Rightarrow\) \(\frac{p}{q}=\frac79\)

\(\therefore\) p = 7 and q = 9

\(\therefore\) p+q = 7+9 = 16

Correct option is A) 16

Correct option is (C) \(\sqrt\frac{5}{2}+\sqrt\frac{1}{2}\)

Let \(\sqrt{3+\sqrt5} \) \(=\sqrt a+\sqrt b\)

\(\Rightarrow\) \(3+\sqrt5\) \(=(\sqrt a+\sqrt b)^2\)   (By squaring both sides)

\(\Rightarrow\) \(a+b+2\sqrt{ab}\) \(=3+\sqrt5\)

\(\Rightarrow\) \(a+b+\sqrt{4ab}\) \(=3+\sqrt5\)

\(\Rightarrow\) a+b = 3 and 4ab = 5  _______(1) (By comparing rational and irrational parts of both rational numbers)

Now, \((a-b)^2=(a+b)^2-4ab\)

\(=3^2-5\) = 9 - 5 = 4

\(\therefore\) a - b = 2  (Let)             _______(2)

From (1) and (2), we obtain

(a+b) + (a-b) = 3+2

\(\Rightarrow\) 2a = 5

\(\Rightarrow\) \(a=\frac52\)

Then from (1), b = 3 - a

\(=3-\frac52=\frac12\)

\(\therefore\) \(\sqrt{3+\sqrt5} \) \(=\sqrt a+\sqrt b\)

\(=\sqrt{\frac52}+\sqrt{\frac12}\)

Correct option is C) \(\sqrt{\cfrac{7}{2}}\) - \(\sqrt{\cfrac{1}{2}}\)

Correct option is B) (ii), (iv)

Let us take n = 1.

2 ¹ × 5¹ = 10

So, for no value of n, 2ⁿ × 5ⁿ ends in 5.

No value of n

2ⁿ x 5ⁿ can also be written as

2ⁿ x 5ⁿ = (2 x 5)ⁿ = (10)ⁿ

Which always ends in a zero

There is no value of n for which (2ⁿ x 5ⁿ) ends in 5.

Let the prime factorisation of a be as follows :

a = p₁p₂ . . . p_n, where p₁,p₂, . . ., p_n are primes, not necessarily distinct.

Therefore, a² = ( p₁p₂ . . . p_n)( p₁p₂ . . . p_n) = p²₁p₂² . . . p_n².

Now, we are given that p divides a². Therefore, from the Fundamental Theorem of Arithmetic, it follows that p is one of the prime factors of a². However, using the uniqueness part of the Fundamental Theorem of Arithmetic, we realise that the only prime factors of a² are p₁, p₂, . . ., p_n. So p is one of p₁, p₂, . . ., p_n.

Now, since a = p₁ p₂ . . . p_n, p divides a.

We are now ready to give a proof that √2 is irrational.

The proof is based on a technique called ‘proof by contradiction’. (This technique is discussed in some detail in Appendix 1).

Correct option is (C) (x + 3) (x – 2)

\(\because\) \(x^2-4x+4\) = (x-2) (x-2) and

\(x^2+2x-3\) = (x+3) (x-1)

\(\therefore\) HCF of polynomials \((x^2-4x+4)(x+3)\) \(and\,(x^2+2x-3)(x-2)\)

= Highest common factor of polynomials \((x-2)^2(x+3)\) \(and\,(x-2)(x-1)(x+3)\)

= (x – 2) (x + 3)

Correct option is C) (x + 3) (x – 2)

Correct option is (D) None

\(2^x\times5^x=(2\times5)^x=10^x\)

\(\therefore\) \(2^x\times5^x\) is a multiple of 10.

\(\therefore\) \(2^x\times5^x\) is always ends with digit 0.

Thus, \(2^x\times5^x\) never ends with 8 or 5.

i.e., for no value of \(x,2^x\times5^x\) ends with 8 or 5.

Correct option is D) None

Let log₁₀100 = x 

⇒ log₁₀10² = x 

⇒ 2 log₁₀10 = x = 2 

∴ log 100 is rational. 

∴ log 100 = 2 

Hence rational.

Let log₁₀2 = x 

Then 10^x= 2 

But 2 can’t be written as 10^x for any value of x 

∴ log 2 is irrational.

We have :

6 = 2 × 3, 72 = 2³ × 3², 120 = 2³ × 3 × 5

Here, 2¹ and 3¹ are the smallest powers of the common factors 2 and 3 respectively.

So, HCF (6, 72, 120) = 2¹ × 3¹ = 2 × 3 = 6

2³, 3² and 5¹ are the greatest powers of the prime factors 2, 3 and 5 respectively

involved in the three numbers.

So, LCM (6, 72, 120) = 2³ × 3² × 5¹ = 360

Remark: Notice, 6 × 72 × 120≠ HCF (6, 72, 120) × LCM (6, 72, 120). So, the product of three numbers is not equal to the product of their HCF and LCM.

We have : 

6 = 2¹ × 3¹ and 20 = 2 × 2 × 5 = 2² × 5¹.

You can find HCF(6, 20) = 2 and LCM(6, 20) = 2 × 2 × 3 × 5 = 60, as done in your earlier classes.

Note that HCF(6, 20) = 2¹ = Product of the smallest power of each common prime factor in the numbers.

LCM (6, 20) = 2² × 3¹ × 5¹ = Product of the greatest power of each prime factor, involved in the numbers.

From the example above, you might have noticed that HCF(6, 20) × LCM(6, 20)

= 6 × 20. In fact, we can verify that for any two positive integers a and b, HCF (a, b) × LCM (a, b) = a × b. We can use this result to find the LCM of two positive integers, if we have already found the HCF of the two positive integers.

Correct option is (A) -3

\(7^{x+3}=5^{x+3}\) is true if x+3 = 0

\(\Rightarrow\) x = 0 - 3 = -3

i.e., \(7^{x+3}=5^{x+3}\) is true if x = -3.

Correct option is A) -3

y = a^x here a ≠ 0 

We can determine the value of ‘x’ for a given y. 

for example y = 5, a = 2 

We cannot express y = a^x for y = 5, a = 2 and for y = 7, a = 3, we cannot express seven (7) as a power of 3.

(i) Prime factors of 20570 = 2 × 5 × 11 × 11 × 17

(ii) Prime factors of 58500 = 2 × 2 × 3 x 3 × 5 x 5 × 13

(iii) Prime factors of 45470971 = 7 × 7 × 13 x 13 × 17 × 17 × 19

For any n ∈ N, e and 5ⁿ end with 6 and 5 respectively.

Therefore,

6ⁿ – 5ⁿ always ends with 6 - 5 = 1.

Correct option is (D) 7

\((x-3)^2\) \(=\left(\frac{2}{3+\sqrt7}-3\right)^2\)

\(=\left(\frac{2-9-3\sqrt7}{3+\sqrt7}\right)^2\) \(=\left(\frac{-7-3\sqrt7}{3+\sqrt7}\right)^2\)

\(=\frac{(7+3\sqrt7)^2}{(3+\sqrt7)^2}\) \(=\frac{(\sqrt7(\sqrt7+3))^2}{(3+\sqrt7)^2}\)

\(=\frac{(\sqrt7)^2(3+\sqrt7)^2}{(3+\sqrt7)^2}\) = 7

Correct option is D) 7

From the graph log₂ 2 = log₄ 4 = log₈ 8 = log₁₀ 10 = 1 

We conclude that log_a a = 1 where a is a natural number.

9 ²ⁿ – 4 ²ⁿ is of the form a ²ⁿ — b ²ⁿ.

It is divisible by both a - b and a + b.

So, 9 ²ⁿ – 4 ²ⁿ is divisible by both 9 - 4 = 5 and 9 + 4 = 13.

Every composite number can be expressed (factorized) as a product of primes, and this factorization is unique, apart from the order in which the prime factors occur.

Prime factorization: 

360 = 2³ × 3² × 5

Let x = p/q be a rational number, such that the prime factorization of q is of the form 2ⁿ 5^m, where n, m are non - negative integers.

Then x has a decimal expansion which terminates.

So 1/3 should be multiplied by 3/10 so that it is in the form of 2ⁿ5^m.

\(\sqrt{27}\) = \(\sqrt{3\times 3\times 3}\)

If we multiply √3 we will get a perfect square 9 which is a rational number.

Correct option is (A) 4

The fundamental theorem of arithmetic states that every composite number can be factorized as a product of primes and this factorization is unique, apart from the order in which the prime factors occurs.

It means fundamental theorem of arithmetic is applicable to composite numbers and among given numbers only 4 is a composite number.

Thus, the fundamental theorem of arithmetic is applicable to 4.

Correct option is A) 4

The fundamental theorem of arithmetic, states that every integer greater than 1 either is  prime itself or is the product of prime numbers, and this product is unique.