Students can practice TS 10th Class Maths Solutions Chapter 1 Real Numbers Optional Exercise to get the best methods of solving problems.

## TS 10th Class Maths Solutions Chapter 1 Real Numbers Optional Exercise

Question 1.

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

Solution:

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.

Solution:

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\(\sqrt{3}\) + \(\sqrt{5}\) is an irrational number. Also check whether (2\(\sqrt{3}\) + \(\sqrt{5}\))(2\(\sqrt{3}\) – \(\sqrt{5}\)) is rational or irrational.

Solution:

To prove : 2\(\sqrt{3}\) + \(\sqrt{5}\) is an irrational number. On contrary, let us suppose that 2\(\sqrt{3}\) + \(\sqrt{5}\) be a rational number then 2\(\sqrt{3}\) + \(\sqrt{5}\) = \(\frac{\mathrm{p}}{\mathrm{q}}\)

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 \(\sqrt{15}\) is an irrational. This is due to our assumption that 2\(\sqrt{3}\) + \(\sqrt{5}\) is a rational. Hence our assumption is wrong and 2\(\sqrt{3}\) + \(\sqrt{5}\) is an irrational number.

Also,

(2\(\sqrt{3}\) + \(\sqrt{5}\)) (2\(\sqrt{3}\) – \(\sqrt{5}\)) = (2\(\sqrt{3}\))^{2} -(\(\sqrt{5}\))^{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) = logx + logy + 3 log 2

Solution:

Given : x^{2} + y2^{2} = 6xy Adding both sides

⇒ x^{2} + y^{2} + 2xy = 6xy + 2xy (x + y)^{2} = 8xy

Taking logarithms on both sides

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

⇒ 2 log(x + y) = log 8 + log x + log y

[∵ log x^{m} = mlog x]

[∵ log xy = log x + log y]

= log 2^{3} + log x + log y

⇒ 2log (x + y) = log x + log y + 3 log 2

Question 5.

Find the number of digits in 4^{2013}, if log_{10} 2 = 0.3010.

Solution:

Given log_{10}2 = 0.3010

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

∴ log_{10} 2^{4026} = 4026 log 2

[∵ log x^{m} = m log x]

= 4026 × 0.3010

= 1211.826 \(\simeq\) 1212

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