SChand CLASS 9 Chapter 6 Indices/Exponent TEST

 TEST

 Question 1

Determine whether each equation is true or false. Change the right side of the equation to make a true equation.

(i) $(2 a)^{-3}=\frac{2}{a^{3}}$

(ii) $\left(\left(a^{-1}\right)^{-1}\right)^{-1} =\frac{1}{a}$

(iii) $(2+3)^{-1}=2^{-1}+3^{-1}$

(iv) If $x \neq \frac{1}{3}$ , then $(3x – 1)^0 = (1 – 3x)^0$

Why is the condition $x \neq \frac{1}{3}$ given in part (iv) above?

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 

Simplify:

Question 2

$\frac{\left(5 x^{3} y^{-3} z\right)^{-2}}{y^{4} z^{-2}}$

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

$\left(\frac{8 a^{3} b^{-4}}{64 a^{-9} b^{2}}\right)^{2 / 3}$

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

$-\sqrt[4]{16 a^{4} b^{8}}$

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

$\left[\frac{y^{2 / 3} \cdot y^{-5 / 6}}{y^{1 / 5}}\right]^{9}$

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

$\left[\sqrt[3]{\sqrt{x^{6}}}\right.$

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

Which of the following is (are) equivalent to $16\frac{–1}{2}$?

(a) – 8

(b) $\frac{1}{4}$

(c) – 4

(d) $4^{-1}$

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

Which of the following is undefined?

(a) $– 25\frac{1}{2}$

(b) $25\frac{1}{2}$

(c) $– 25\frac{–1}{2}$

(d) $(-25)\frac{1}{2}$

Sol: 

option (d) $(-25)^{1 / 2}$ is correct

because- Square root of negative number is not defined

 Question 9

True or False?

(a) $\frac{a^{4n}}{a^n}=a^4$

(b) $\frac{1}{am−n}=d^{n−m}$

(c) $a^{-n}.a^n = 1$

(d) $\frac{a^n}{b^m}=\left(\frac{a}{b}\right)^{n−m}$

Sol :

(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) Solve : $(- 4.8)^k = 1$

(ii) $\sqrt[3]{\sqrt{0.000064}}$ is equal to

(a) 0.0002

(b) 0.002

(c) 0.02

(d) 0.2

Sol :

(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