Which of the following expressions has the greatest value?

Aptitude Surds and Indices Difficulty: Medium
Choose an option
  • A
    $[(2^{-1})^0]^2$
  • B
    $[(4^0)^{-\frac{1}{2}}]^2$
  • C
    $[(2^{-2})^{-1}]^2$
  • D
    $[(2^{-1})^2]^2$

Answer

Correct Answer: $[(2^{-2})^{-1}]^2$

Explanation

### Concept & Formula This problem tests your mastery of the power of a power rule and the zero exponent rule. When an exponent is raised to another exponent, you multiply the powers. Any non-zero number raised to the power of zero equals one. $$ (x^a)^b = x^{ab} $$ $$ x^0 = 1 $$ ### Step-by-Step Solution * **Given:** Four separate exponential expressions to evaluate and compare. * **Calculation:** 1. Evaluate option (a): $[(2^{-1})^0]^2$ Multiply the exponents: $-1 \times 0 \times 2 = 0$. Result: $2^0 = 1$. 2. Evaluate option (b): $[(4^0)^{-\frac{1}{2}}]^2$ Multiply the exponents: $0 \times -\frac{1}{2} \times 2 = 0$. Result: $4^0 = 1$. 3. Evaluate option (c): $[(2^{-2})^{-1}]^2$ Multiply the exponents: $-2 \times -1 \times 2 = 4$. Result: $2^4 = 16$. 4. Evaluate option (d): $[(2^{-1})^2]^2$ Multiply the exponents: $-1 \times 2 \times 2 = -4$. Result: $2^{-4} = \frac{1}{16}$. 5. Compare the results: $1$, $1$, $16$, and $\frac{1}{16}$. The greatest value is $16$. ### Exam Strategy & Shortcut Scan the options for the number $0$ in the exponent chain. In options (a) and (b), there is a $0$ in the multiplication sequence of the powers. Because anything multiplied by $0$ is $0$, you instantly know those expressions evaluate to a base raised to $0$, which is $1$. You can immediately eliminate them and only calculate (c) and (d). Recognizing the double negatives in (c) guarantees a positive power, while (d) remains a negative power (a fraction), making (c) the obvious largest choice. ### Common Pitfall A common mistake is trying to evaluate the inside of the brackets first instead of just multiplying all the exponents together simultaneously. This wastes time and increases the chance of arithmetic errors, especially with fractions and negative signs. ### Final Answer **Therefore, the correct answer is $[(2^{-2})^{-1}]^2$.**
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