OP Malhotra Class 9 Maths Solutions Chapter 6 Indices/Exponents Chapter Test

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S Chand Class 9 ICSE Maths Solutions Chapter 6 Indices/Exponents Chapter Test

Question 1.
Determine whether each equation is true or false. Change the right side of the equation to make a true equation.
(i) (2a)^-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 ≠ \(\frac { 1 }{ 3 }\), then (3x – 1)⁰ = (1 – 3x)⁰
Why is the condition x ≠ \(\frac { 1 }{ 3 }\) given in part (iv) above?
Solution:

Which is not possible

Simplify:

Question 2.
\(\frac{\left(5 x^3 y^{-3} z\right)^{-2}}{y^4 z^{-2}}\)
Solution:

Question 3.
\(\left(\frac{8 a^3 b^{-4}}{64 a^{-9} b^2}\right)^{\frac{2}{3}}\)
Solution:

Question 4.
– \(\sqrt[4]{16 a^4 b^8}\)
Solution:
– \(\sqrt[4]{16 a^4 b^8}\)
= – \(\left(2^{4 \times \frac{1}{4}} a^{4 \times \frac{1}{4}} \cdot b^{8 \times \frac{1}{4}}\right)\)
= – (2¹a¹b²) = – 2ab²

Question 5.
\(\left(\frac{y^{\frac{2}{3}} \cdot y^{-\frac{5}{6}}}{y^{\frac{1}{9}}}\right)^9\)
Solution:

Question 6.
\(\sqrt[3]{\sqrt{x^6}}\)
Solution:
\(\sqrt[3]{\sqrt{x^6}}=\left[\left(x^6\right)^{\frac{1}{2}}\right]^{\frac{1}{3}}=x^{6 \times \frac{1}{2} \times \frac{1}{3}}\)
= x¹ = x

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
Solution:
(b) \(\frac { 1 }{ 4 }\)
\(16^{-\frac{1}{2}}=\left(4^2\right)^{\frac{-1}{2}}=4^{2 \times\left(\frac{-1}{2}\right)}\)
= 4^-1

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)^{\(\frac { 1 }{ 2 }\)}
Solution:
(d) (-25)^{\(\frac { 1 }{ 2 }\)}
(- 25)^{\(\frac { 1 }{ 2 }\)} is not defined as square root of negative term is not possible.

Question 9.
True or False?
(a) \(\frac{a^{4 n}}{a^n}=a^4\)
(b) \(\frac{1}{a^{m-n}}=d^{n-m}\)
(c) a^-n.aⁿ = 1
(d) \(\frac{a^n}{b^m}=\left(\frac{a}{b}\right)^{n-m}\)
Solution:
(a) \(\frac{a^{4 n}}{a^n}=a^{4 n-n}=a^{3 n} \neq a^4\) False

(b) \(\frac{1}{a^{m-n}}=a^{-(m-n)}=a^{-m+n}\)
= a^n-m = a^n-m True

(c) a^-n.aⁿ = 1
a^-n.aⁿ = a^-n+n = a⁰ = 1 = 1 True

(d) \(\frac{a^n}{b^m} \neq\left(\frac{a}{b}\right)^{n-m}\) 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
Solution:
(d) 0.2
(i) (- 4.8)^k = 1
⇒ (- 4.8)^k = (- 4.8)⁰ (∵ a⁰ = 1)
Comparing, we get
K = 0

(ii) \(\sqrt[3]{\sqrt{0.000064}}\)
= (0.04 x 0.04 x 0.04)^{\(\frac { 1 }{ 2 }\)x\(\frac { 1 }{ 2 }\)}
= (0.2 x 0.2 x 0.2 x 0.2 x 0.2 x 0.2)\(\frac { 1 }{ 6 }\)
= [(0.2)⁶]\(\frac { 1 }{ 6 }\)
= (0.2)¹
= 0.2