Exponents and Powers — Statement I-II Type Questions

Statement I: The value of $\Big(\dfrac{-1}{2}\Big)^{-4}$ is 16.

Statement II: $a^{-m} = \dfrac{1}{a^m}$, where a is any non-zero integer and m is a positive integer.

1. Statement I is true but Statement II is false.

2. Statement I is false but Statement II is true.

3. Both Statement I and Statement II are true.

4. Both Statement I and Statement II are false.

Answer

Solving,

$\Rightarrow \Big(\dfrac{-1}{2}\Big)^{-4}\\[1em] \Rightarrow \Big(\dfrac{2}{-1}\Big)^{4}\\[1em] \Rightarrow (-2)^4\\[1em] \Rightarrow (-2) \times (-2) \times (-2) \times (-2)\\[1em] \Rightarrow 16$

So, the value is 16 and Statement I is true.

$a^{-m} = \dfrac{1}{a^m}$, where a is any non-zero integer and m is a positive integer, is a standard law of exponents.

So, Statement II is true.

Hence, option 3 is the correct option.

Statement I: The multiplicative inverse of 0 is 0.

Statement II: $\Big(\dfrac{p}{q}\Big)^{-m} = \dfrac{q^m}{p^m}$, where p and q are non-zero integers and m is a positive integer.

1. Statement I is true but Statement II is false.

2. Statement I is false but Statement II is true.

3. Both Statement I and Statement II are true.

4. Both Statement I and Statement II are false.

Answer

For a number to have a multiplicative inverse, its product with the inverse must equal 1. Since 0 multiplied by any number gives 0 (never 1), the multiplicative inverse of 0 does not exist.

So, Statement I is false.

$\Big(\dfrac{p}{q}\Big)^{-m} = \dfrac{1}{\Big(\dfrac{p}{q}\Big)^m} = \Big(\dfrac{q}{p}\Big)^m = \dfrac{q^m}{p^m}$, where p and q are non-zero integers and m is a positive integer, is a standard law of exponents.

So, Statement II is true.

Hence, option 2 is the correct option.

Statement I: $(a + b)^{-1}(a + b)(a^{-1} + b^{-1}) = \dfrac{1}{a} + \dfrac{1}{b}$, where a and b are non-zero rational numbers.

Statement II: $a^m - a^n = a^{m-n}$, where a is a non-zero rational number and m, n are positive integers.

1. Statement I is true but Statement II is false.

2. Statement I is false but Statement II is true.

3. Both Statement I and Statement II are true.

4. Both Statement I and Statement II are false.

Answer

Solving L.H.S.,

$\Rightarrow (a + b)^{-1}(a + b)(a^{-1} + b^{-1})\\[1em] \Rightarrow \Big[(a + b)^{-1}(a + b)\Big](a^{-1} + b^{-1})\\[1em] \Rightarrow 1 \times (a^{-1} + b^{-1})\\[1em] \Rightarrow \dfrac{1}{a} + \dfrac{1}{b}$

Since, L.H.S. = R.H.S., Statement I is true.

$a^m - a^n = a^{m-n}$ is not a valid law of exponents. Subtraction of powers does not simplify by subtracting exponents.

So, Statement II is false.

Hence, option 1 is the correct option.