Question 1
Sol:
(i)$(2 a)^{-3}=\frac{2}{a^{3}}$
$(2 a)^{-3}=\frac{1}{(2 a)^{3}}=\frac{1}{8 a^{3}} \neq \frac{2}{a^{3}} \quad$ [it is not equal]
(ii) $\begin{aligned}\left(\left(a^{-1}\right)^{-1}\right)^{-1} &=\frac{1}{a} \\\left(\left(a^{-1}\right)^{-1}\right)^{-1} &=a\left(^{-1)} \times(-1) \times(-1)\right.\\ &=a^{-1}=\frac{1}{a}=\frac{1}{a} \end{aligned}$
[It is made true equation]
(iii) $(2+3)^{-1}=2^{-1}+3^{-1}$
$(2+3)^{-1}=5^{-1}=\frac{1}{5} .$
$=2^{-1}+3^{-1} .$
$=\frac{1}{2}+\frac{1}{3}=\frac{3+2}{6}$
$\begin{aligned} &=\frac{5}{6} \\ \because \frac{1}{5} & \neq \frac{5}{6} \end{aligned}$ it is not equal
[It is not make proper equation]
(iv) $x \neq \frac{1}{3}$
$(3 x-1)^{0}=(1-3 x)^{\circ}$
$(3 x-1)^{\circ}=1$
$(1-3 x)^{0}=1 \quad\left(x^{\circ}=1\right)$
1=1 [It is make true equation]
According to Question if $x=1 / 3$,
Then
$\left(3 x-1^{0}\right)=\left(3 \times \frac{1}{3}-1\right)^{0}$
$=(1-1)^{\circ}=0^{\circ}$
It is not make a proper equation which is not possible
Question 2
Sol:
$\begin{aligned} & \frac{\left(5 x^{3} y^{-3} z\right)^{-2}}{y^{4} z^{-2}} \\=& \frac{\left(s^{-2}\right) x^{-6} y+1 z^{-2}}{y^{4} z^{-2}} \\=& \frac{1}{25} \frac{1}{x^{6}} \times y^{6-4} \cdot z^{-2+2} \\=& \frac{1}{25} \frac{y^{2}}{x^{6}} \times z^{0}=\frac{y^{2}}{25 x^{6}} \end{aligned}$ Ans
Question 3
Sol:
$\begin{aligned} &\left(\frac{8 a^{3} b^{-4}}{64 a^{-9} b^{2}}\right)^{2 / 3} \\=&\left[\frac{2 \times a^{3+9}}{8 \times b^{2+4}}\right]^{2 / 3} \\=&\left(\frac{a^{12}}{8 b^{6}}\right)^{2 / 3} \\=&\left(\frac{a^{12}}{2^{3} \times b^{6}}\right)^{2 / 3} \end{aligned}$
$\frac{a^{12 \times 2 / 3}}{2^{3 \times 2 / 3} \cdot b^{6 \times 2 / 3}}=\frac{a^{24 / 3}}{2^{6 / 3} \cdot b^{12 / 3}}$
$\frac{a^{8}}{2^{2} \cdot b^{4}} \Rightarrow \frac{a^{8}}{4 b^{4}}$
Question 4
Sol:
$\begin{aligned}-\sqrt[4]{16 a^{4} b^{8}} &=-\left(2^{4} a^{4} b^{8}\right)^{1 / 4} \\ &=-\left(2^{a \times \frac{1}{4}} a^{4 \times \frac{1}{4}} b^{8 \times \frac{1}{4}}\right) \\ &=-\left(2^{1} a^{1} b^{2}\right) \\ &=-2 a b^{2} \text { Ans } \end{aligned}$
Question 5
Sol:
=$\left[\frac{y^{2 / 3} \cdot y^{-5 / 6}}{y^{1 / 5}}\right]^{9}$
=$\left(y \frac{12-15-2}{18}\right)^{9}$
$=y^{-\frac{5}{2}}$
$=\frac{1}{y^{\frac{2}{5}}}$
Question 6
Sol:
$\begin{aligned} &\left[\sqrt[3]{\sqrt{x^{6}}}\right.\\=&\left[\left(x^{6}\right)^{1 / 2}\right]^{1 / 3} \\=& x^{6 \times 1 / 2} \times 1 / 3 \\=& x^{1}=x \end{aligned}$
Question 7
Sol:
=$16^{-1\2}$
=$\left(4^{2}\right)^{\frac{-1}{2}}$
=$4^{2 \times(-1 / 2)}$
=$4^{-1}$
$=\frac{1}{4}$ option (b) and (d) both are correct
Question 8
Sol: option (d) $(-25)^{1 / 2}$ is correct
because- Square root of negative number is not defined
Question 9
(i)
Sol: $\begin{aligned} \frac{a^{4 n}}{a^{n}} &=a^{4 n-n} \\ &=a^{3 n} \neq a^{4} \end{aligned}$
It is false
(ii)
$\begin{aligned} \frac{1}{a^{m-n}} &=a^{-(m-n)}=a^{-m+n} \\ &=a^{n-m} \\ &=a^{n-m} \text { it is true } \end{aligned}$
(iii) $a^{-n} \cdot a^{n}=1$
$a^{-n} \cdot a^{n}=a^{-n+n}$
$=a^{\circ}=1$ $=1$ is also true
(iv) $\frac{a^{n}}{b^{m}} \neq\left(\frac{a}{b}\right)^{a-n}$
It is not equal so it is false
Question 10
(i)
$\begin{aligned}&(-4 \cdot 8)^{k}=1 \\&(-4 \cdot 8)^{k}=(-4 \cdot 8)^{0}\end{aligned}$
comparing both sides of power
k=0
(ii) $\sqrt[3]{0.000064}$
$=(0.04 \times 0.04 \times 0.04)^{1 / 2 \times 1 / 3}$
$=(0.2 \times 0.2 \times 0.2 \times 0.2 \times 0.2 \times 0.2)^{1 / 6}$
$=\left[(0.2)^{6}\right]^{1 / 6}=(0.2)^{6 \times \frac{1}{6}}$
$=(0.2)^{2}$
$=0.2$ Ans
option (d) is correct
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