Test
Question 1
A number is an irrational number if and only if its decimal
representation is
(a) non-terminating
(b) non-terminating and repeating
(c) non-terminating and non-repeating
(d) terminating
Sol :
(c) non-terminating and non-repeating
Question 2
Which of the following is an irrational number ?
(a) √29
(b) √441
(c) 0.5948
(d) $5.\sqrt{318}$
Sol :
(a) √29 is irrational number as 29 is not a perfect square.
Question 3
$(-2-\sqrt{3})(-2+\sqrt{3})$ when simplified is
(a) positive and irrational
(b) positive and rational
(c) negative and irrational
(d) negative and rational
Sol :
(b) positive and rational
$(-2-\sqrt{3})(-2+\sqrt{3})=(-2)^2-(\sqrt{3})^2$
=4-3=1
Which is positive and rational.
Question 4
If $\sqrt{6} \times \sqrt{15}=x\sqrt{10}$ , then the value of x is
(a) 3
(b) ± 3
(c) √3
(d) √6
Sol :
⇒√6×√15=x√10
⇒$\sqrt{6 \times 15}$=x√10
⇒√90=x√10
⇒$\sqrt{9 \times 10}=x\sqrt{10}$
⇒3$\sqrt{10}=x\sqrt{10}$
Comparing, we get
∴ x = 3
Question 5
Two rational numbers between $\frac{2}{7}$ and $\frac{2}{14}$ are
(a) $\frac{1}{14}\text{ and }\frac{2}{14}$
(b) $\frac{1}{2}\text{ and }\frac{3}{2}$
(c) $\frac{3}{14}\text{ and }\frac{3}{7}$
(d) $\frac{5}{14}\text{ and }\frac{8}{14}$
Sol :
(d) $\frac{5}{14}\text{ and }\frac{8}{14}$
Two rational numbers between $\frac{2}{7} \text{ and } \frac{2}{14}$ are $\frac{5}{14}\text{ and }\frac{8}{14}$
∵$\frac{2}{7} \text{ and } \frac{5}{7}=\frac{4}{14} \text{ and } \frac{10}{14}$ and 5 and 8 lie between 4 and 10
Question 6
An irrational number between $\frac{5}{7} \text{ and } \frac{7}{9}$ is
(a) 0.75
(b) √6
(c) 0.7507500075000…
(d) 0.7512
Sol :
An irrational number between $\frac{5}{7} \text{ and } \frac{7}{9}$ either √6 or 0.7507500075000....
∵ 0.75 and 0.7512 are rational number
Which does not line between $\frac{5}{7} \text{ and } \frac{7}{9}$
∴ 0.7507500075000… is irrational number between $\frac{5}{7} \text{ and } \frac{7}{9}$
Question 7
If √2=1.4142 , then the value of $\frac{7}{3+\sqrt{2}}$ correct to two decimal places is
(a) 1.59
(b) 1.60
(c) 2.58
(d) 2.57
Sol :
(a) 1.59
√2=1.4142
$\frac{7}{3+\sqrt{2}}=\frac{7\times (3-\sqrt{2})}{(3-\sqrt{2})(3-\sqrt{2})}$
Rationalising denominator
$=\frac{7(3-\sqrt{2})}{(9-2)}=\frac{7(3-\sqrt{2})}{7}$
=3-√2
= 3 – 1.4142 = 1.5858 = 1.59
Question 8
Taking √3 as i.732 and √2=1.414 , the value of $\frac{1}{\sqrt{3}+\sqrt{2}}$ is
(a) 0.064
(b) 0.308
(c) 0.318
(d) 2.146
Sol :
(c) 0.318
√3 = 1.732, √2= 1.414
$\frac{1}{\sqrt{3}+\sqrt{2}}=\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}$
(Rationalising denominator)
$\frac{\sqrt{3}-\sqrt{2}}{3-2}=\frac{\sqrt{3}-\sqrt{2}}{1}=\sqrt{3}-\sqrt{2}$
=1.732-1.414=0.318
Question 9
If $x=\sqrt{3}+\sqrt{2}$ , then the value of $\left(x+\frac{1}{x}\right)$ is
(a) 2
(b) 3
(c) $2\sqrt{2}$
Sol :
(d) $2\sqrt{3}$
$x=\sqrt{3}+\sqrt{2}$
$\frac{1}{x}=\frac{1}{\sqrt{3}+\sqrt{2}}=\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}$
(Rationalising denominator)
$=\frac{\sqrt{3}-\sqrt{2}}{3-2}=\frac{\sqrt{3}-\sqrt{2}}{1}=\sqrt{3}-\sqrt{2}$
$x+\frac{1}{x}=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}=2\sqrt{3}$
Question 10
If $x=2+\sqrt{3}$ , then the value of $\sqrt{x}+\frac{1}{\sqrt{x}}$ is
(a) $3+\sqrt{3}$
(b) √6
(c) 2√6
(d) 6
Sol :
x=2+√3
$\frac{1}{x}=\frac{1}{2+\sqrt{3}}=\frac{1(2-\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})}$
(Rationalising denominator)
$\frac{2-\sqrt{3}}{(2)^2-(\sqrt{3})^2}=\frac{2-\sqrt{3}}{4-3}=2-\sqrt{3}$
∴$x+\frac{1}{x}=2+\sqrt{3}+2-\sqrt{3}=4$ and $\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^2=x+\frac{1}{x}+2$
=4+2=6
$\sqrt{x}+\frac{1}{\sqrt{x}}=\pm \sqrt{6}$
=√6
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