Students often turn to OP Malhotra Class 9 Solutions Chapter 1 Rational and Irrational Numbers Ex 1(C) to clarify doubts and improve problem-solving skills.
S Chand Class 9 ICSE Maths Solutions Chapter 1 Rational and Irrational Numbers Ex 1(C)
Simplify:
Question 1.
(a) \(\sqrt{\frac{1}{3}}\)
(b) \(\sqrt{\frac{5}{12}}\)
(c) \(\sqrt{1 \frac{46}{75}}\)
Solution:
Question 2.
\(\sqrt{112}-\sqrt{63}+\frac{224}{\sqrt{28}}\)
Solution:
Question 3.
\(\frac{4 \sqrt{18}}{\sqrt{12}}-\frac{8 \sqrt{75}}{\sqrt{32}}+\frac{9 \sqrt{2}}{\sqrt{3}}\)
Solution:
Question 4.
Rationalise the denominators of
(a) \(\frac{1}{4-\sqrt{3}}\)
(b) \(\frac{2}{\sqrt{5}+\sqrt{3}}\)
(c) \(\frac{1}{2 \sqrt{5}-\sqrt{3}}\)
(d) \(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\)
Solution:
Question 5.
Rationalise the denominator and simplify :
(i) \(\frac{4+\sqrt{5}}{4-\sqrt{5}}+\frac{4-\sqrt{5}}{4+\sqrt{5}}\)
(ii) \(\frac{3}{5-\sqrt{3}}+\frac{2}{5+\sqrt{3}}\)
Solution:
Question 6.
Find the values of a and b if
(i) \(\frac{3+\sqrt{2}}{3-\sqrt{2}}=a+b \sqrt{2}\)
(ii) \(\frac{5+2 \sqrt{3}}{7+4 \sqrt{3}}=a+b \sqrt{3}\)
(iii) \(\frac{\sqrt{7}-1}{\sqrt{7}+1}-\frac{\sqrt{7}+1}{\sqrt{7}-1}=a+b \sqrt{7}\)
Solution:
Question 7.
Rationalise the denominator of \(\frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}\).
Solution:
\(\frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}=\frac{1}{(\sqrt{2}+\sqrt{3})+\sqrt{5}}\)
Rationalising denominator
= \(\frac{(\sqrt{2}+\sqrt{3})-(\sqrt{5})}{[(\sqrt{2}+\sqrt{3})+\sqrt{5}][(\sqrt{2}+\sqrt{3})-\sqrt{5}]}\)
= \(\frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{(\sqrt{2}+\sqrt{3})^2-(\sqrt{5})^2}=\frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{2+3+2 \sqrt{6}-5}\)
= \(\frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{2 \sqrt{6}}\)
Again rationalising,
= \(\frac{(\sqrt{2}+\sqrt{3}-\sqrt{5}) \sqrt{6}}{2 \times \sqrt{6} \times \sqrt{6}}=\frac{\sqrt{12}+\sqrt{18}-\sqrt{30}}{2 \times 6}\)
= \(\frac{\sqrt{4 \times 3}+\sqrt{9 \times 2}-\sqrt{30}}{12}\)
= \(\frac{2 \sqrt{3}+3 \sqrt{2}-\sqrt{30}}{12}\)
Question 8.
Taking \(\sqrt{2}\) = 1.414 and \(\sqrt{3}\) = 1.732, find without using tables or long division, the value of
(a) \(\frac{1}{3-\sqrt{2}}\)
(b) \(\frac{2}{\sqrt{3}-\sqrt{2}}\)
Solution:
(a) \(\frac{1}{3-\sqrt{2}}=\frac{1(3+\sqrt{2})}{(3-\sqrt{2})(3+\sqrt{2})}\)
(Rationalising the denominator)
= \(\frac{3+\sqrt{2}}{(3)^2-(\sqrt{2})^2}\)
= \(\frac{3+\sqrt{2}}{9-2}=\frac{3+\sqrt{2}}{7}=\frac{3+1.414}{7}\)
= \(\frac { 4.414 }{ 7 }\) = 0.631
(b) \(\frac{2}{\sqrt{3}-\sqrt{2}}=\frac{2(\sqrt{3}+\sqrt{2})}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}\)
(Rationalising the denominator)
= \(\frac{2(\sqrt{3}+\sqrt{2})}{(\sqrt{3})^2-(\sqrt{2})^2}=\frac{2(\sqrt{3}+\sqrt{2})}{3-2}\)
= 2(\(\sqrt{3}\) + \(\sqrt{2}\))
= 2 (1.732 + 1.414) = 2 x 3.146
= 6.292
Question 9.
Express \(\frac{3-5 \sqrt{5}}{3+2 \sqrt{5}}\) in the form (a\(\sqrt{5}\) – b) where a and b are simple fractions.
Solution:
Question 10.
Prove that \(\frac{1}{\sqrt{2}-1}+\frac{2}{\sqrt{3}+1}=\sqrt{2}+\sqrt{3}\).
Solution:
Question 11.
Simplify:
\(\frac{6 \sqrt{2}}{\sqrt{3}+\sqrt{6}}-\frac{4 \sqrt{3}}{\sqrt{6}+\sqrt{2}}+\frac{2 \sqrt{6}}{\sqrt{2}+\sqrt{3}}\) (NDA)
Solution:
Question 12.
Simplify:
(i) \(\frac{6}{2 \sqrt{3}-\sqrt{6}}+\frac{\sqrt{6}}{\sqrt{3}+\sqrt{2}}-\frac{4 \sqrt{3}}{\sqrt{6}-\sqrt{2}}\)
(ii) \(\frac{7 \sqrt{3}}{\sqrt{10}+\sqrt{3}}-\frac{2 \sqrt{5}}{\sqrt{6}+\sqrt{5}}-\frac{3 \sqrt{2}}{\sqrt{15}+3 \sqrt{2}}\)
Solution:
(i) \(\frac{6}{2 \sqrt{3}-\sqrt{6}}+\frac{\sqrt{6}}{\sqrt{3}+\sqrt{2}}-\frac{4 \sqrt{3}}{\sqrt{6}-\sqrt{2}}\)
Now from (i), (ii), (iii)
(2\(\sqrt{3}\) + \(\sqrt{6}\)) + (3\(\sqrt{2}\) – 2\(\sqrt{3}\)) – (3\(\sqrt{2}\) + \(\sqrt{6}\))
= 2\(\sqrt{3}\) + \(\sqrt{6}\) + 3\(\sqrt{2}\) – 2\(\sqrt{2}\) – 3\(\sqrt{2}\) – \(\sqrt{6}\) = 0
(ii) \(\frac{7 \sqrt{3}}{\sqrt{10}+\sqrt{3}}-\frac{2 \sqrt{5}}{\sqrt{6}+\sqrt{5}}-\frac{3 \sqrt{2}}{\sqrt{15}+3 \sqrt{2}}\)
From (i), (ii) and (iii)
(\(\sqrt{30}\) – 3) – (2\(\sqrt{30}\) – 10) – (- \(\sqrt{30}\) + 6)
= \(\sqrt{30}\) – 3 – 27\(\sqrt{30}\) + 10 + \(\sqrt{30}\) – 6
= 10 – 9 = 1
Question 13.
If x = 2 + \(\sqrt{3}\), find the value of x² + \(\frac{1}{x^2}\).
Solution:
x = 2 + \(\sqrt{3}\)
\(\frac { 1 }{ x }\) = \(\frac{1}{2+\sqrt{3}}=\frac{2-\sqrt{3}}{(2+\sqrt{3})(2-\sqrt{3})}\)
= \(\frac{2-\sqrt{3}}{4-3}=\frac{2-\sqrt{3}}{1}=2-\sqrt{3}\)
Now x + \(\frac { 1 }{ x }\) = 2 + \(\sqrt{3}\) + 2 – \(\sqrt{3}\) = 4
Squaring both sides,
⇒ \(\left(x+\frac{1}{x}\right)^2\) = (4)²
⇒ x² + \(\frac { 1 }{ x² }\) + 2 = 16
⇒ x² + \(\frac { 1 }{ x² }\) = 16 – 2 = 14
Question 14.
If x = \(\sqrt{2}\) + 1, find the value of x² + \(\frac{1}{x^2}\).
Solution:
x = \(\sqrt{2}\) + 1
∴ \(\frac { 1 }{ x }\) = \(\frac{1}{\sqrt{2}+1}=\frac{\sqrt{2}-1}{(\sqrt{2}+1)(\sqrt{2}-1)}\)
= \(\frac{\sqrt{2}-1}{2-1}=\sqrt{2}\) – 1
Now x + \(\frac { 1 }{ x }\) = \(\sqrt{2}\) + 1 + \(\sqrt{2}\) – 1
Squaring both sides,
⇒ \(\left(x+\frac{1}{x}\right)^2=(2 \sqrt{2})^2\)
⇒ x² + \(\frac { 1 }{ x² }\) + 2 = 4 x 2 = 8
⇒ x² + \(\frac { 1 }{ x² }\) = 8 – 2 = 6
Question 15.
If x = \(\frac{5-\sqrt{21}}{2}\), find the value of
(i) x + \(\frac { 1 }{ x }\) and (ii) x² + \(\frac { 1 }{ x² }\)
Solution:
x = \(\frac{5-\sqrt{21}}{2}\)
∴ \(\frac { 1 }{ x }\) = \(\frac{2}{5-\sqrt{21}}=\frac{2(5+\sqrt{21})}{(5-\sqrt{21})(5+\sqrt{21})}\)
= \(\frac{2(5+\sqrt{21})}{25-21}=\frac{2(5+\sqrt{21})}{4}=\frac{5+\sqrt{21}}{2}\)
(i) x + \(\frac { 1 }{ x }\) = \(\frac{5-\sqrt{21}}{2}+\frac{5+\sqrt{21}}{2}\)
= \(\frac{5-\sqrt{21}+5+\sqrt{21}}{2}\)
= \(\frac { 10 }{ 2 }\) = 5
and 1 + \(\frac { 1 }{ x }\) = 5
Squaring both sides,
(x + \(\frac { 1 }{ x }\))² = (5)²
⇒ x² + \(\frac { 1 }{ x² }\) + 2 = 25
⇒ x² + \(\frac { 1 }{ x² }\) = 25 – 2 = 23
Hence x + \(\frac { 1 }{ x }\) = 5 and x² + \(\frac { 1 }{ x² }\) = 23