AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 1 Real Numbers Optional Exercise Textbook Questions and Answers.

## AP State Syllabus SSC 10th Class Maths Solutions 1st Lesson Real Numbers Optional Exercise

### 10th Class Maths 1st Lesson Real Numbers Optional Exercise Textbook Questions and Answers

Question 1.

Can the number 6^{n}, n being a natural number, end with the digit 5? Give reason.

Answer:

Given number = 6^{n} ; n ∈ N

6^{n} to be end in 5; it should be divisible by 5

6^{n} = (2 × 3)^{n}

The prime factors of 6^{n} are 2 and 3.

It can’t end with the digit 5.

Question 2.

Is 7 × 5 × 3 × 2 + 3 a composite number? Justify your answer.

Answer:

Given:

7 × 5 × 3 × 2 + 3

= 3 (7 × 5 × 2 + 1)

= 3 × (70 + 1)

= 3 × 71

∴ The given number has two factors namely 3 and 71.

Hence it is a composite number.

Question 3.

Prove that (2√3 + √5 ) is an irrational number. Also check whether (2√3 + √5) (2√3 – √5) is rational or irrational.

Answer:

To prove:

2√3 + √5 is an irrational number. On contrary, let us suppose that 2√3 + √5 be a rational number.

Then 2√3 + √5 = \(\frac{p}{q}\)

Squaring on both sides, we get

L.H.S = an irrational number.

R.H.S = p, q being integers, \(\frac{p^{2}-17 q^{2}}{4 q^{2}}\) is a rational number.

This is a contradiction to the fact that √l5 is an irrational. This is due to our assumption that 2√3 + √5 is a rational. Hence our assumption is wrong and 2√3 + √5 is an irrational number. Also,

(2√3 + √5) (2√3 – √5)

= (2√3)^{2} – (√5)^{2}

[∵ (a + b) (a – b) = a^{2} – b^{2}]

= 4 × 3 – 5

= 12 – 5 = 7, a rational number.

Question 4.

If x^{2} + y^{2} = 6xy, prove that 2 log (x + y) = log x + log y + 3 log 2.

Answer:

Given: x^{2} + y^{2} = 6xy

x^{2} + y^{2} + 2xy = 6xy + 2xy

(x + y)^{2} = 8xy

Taking logarithms on both sides log (x + y)^{2} = log8xy

⇒ 2log(x + y)= log8 + logx + logy [∵ logx^{m} = mlogx]

[∵ logxy = logx + logy]

= log2^{3} + logx + logy

⇒ 2log(x + y) = logx + logy + 3log2

Question 5.

Find the number of digits in 42013, if log_{10}2 = 0.3010.

Answer:

Given:

log_{10}2 = 0.3010

4^{2013} = (2^{2})^{2013} = 2^{4026}

∴ log_{10} 2^{4026} = 4026 log_{10}2

[∵ log x^{m} = m log x]

= 4026 × 0.3010 = 1211.826.

So 1211 + 1 = 1212

∴ 4^{2013} has 1212 digits in its expansion.

(∵ characteristic 1211)