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 6n, n being a natural number, end with the digit 5? Give reason.
Answer:
Given number = 6n ; n ∈ N
6n to be end in 5; it should be divisible by 5
6n = (2 × 3)n
The prime factors of 6n are 2 and 3.
It can’t end with the digit 5.

AP SSC 10th Class Maths Solutions Chapter 1 Real Numbers Optional Exercise

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
AP SSC 10th Class Maths Solutions Chapter 1 Real Numbers Optional Exercise 1
AP SSC 10th Class Maths Solutions Chapter 1 Real Numbers Optional Exercise 2
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) = a2 – b2]
= 4 × 3 – 5
= 12 – 5 = 7, a rational number.

AP SSC 10th Class Maths Solutions Chapter 1 Real Numbers Optional Exercise

Question 4.
If x2 + y2 = 6xy, prove that 2 log (x + y) = log x + log y + 3 log 2.
Answer:
Given: x2 + y2 = 6xy
x2 + y2 + 2xy = 6xy + 2xy
(x + y)2 = 8xy
Taking logarithms on both sides log (x + y)2 = log8xy
⇒ 2log(x + y)= log8 + logx + logy [∵ logxm = mlogx]
[∵ logxy = logx + logy]
= log23 + logx + logy
⇒ 2log(x + y) = logx + logy + 3log2

Question 5.
Find the number of digits in 42013, if log102 = 0.3010.
Answer:
Given:
log102 = 0.3010
42013 = (22)2013 = 24026
∴ log10 24026 = 4026 log102
[∵ log xm = m log x]
= 4026 × 0.3010 = 1211.826.
So 1211 + 1 = 1212
∴ 42013 has 1212 digits in its expansion.
(∵ characteristic 1211)