(A) \( s^{2}+\frac{1}{a} \cdot\left(\frac{1}{a}\right)^{4} \) (1i) \( x^{3}+\left(\frac{1}{1}\right)^{x} \times \frac{1}{x} \) (C) \( \frac{1}{n} \cdot\left(\frac{1}{8}\right)^{0}

Ask by Reid Schofield. in Singapore

Mar 10,2025

**(A)** We have \[ s^2 + \frac{1}{a} \cdot \left(\frac{1}{a}\right)^4. \] Step 1. Evaluate the power: \[ \left(\frac{1}{a}\right)^4 = \frac{1}{a^4}. \] Step 2. Multiply by \(\frac{1}{a}\): \[ \frac{1}{a} \cdot \frac{1}{a^4} = \frac{1}{a^5}. \] Thus, the simplified expression is \[ s^2 + \frac{1}{a^5}. \] --- **(1i)** We have \[ x^3 + \left(\frac{1}{1}\right)^x \times \frac{1}{x}. \] Step 1. Note that \[ \frac{1}{1} = 1 \quad \text{so} \quad 1^x = 1. \] Step 2. The expression becomes \[ x^3 + 1 \times \frac{1}{x} = x^3 + \frac{1}{x}. \] --- **(C)** We have the inequality \[ \frac{1}{n} \cdot \left(\frac{1}{8}\right)^0 < s^\circ. \] Step 1. Recognize that any nonzero number raised to the 0th power equals 1, so \[ \left(\frac{1}{8}\right)^0 = 1. \] Step 2. Multiply by \(\frac{1}{n}\): \[ \frac{1}{n} \cdot 1 = \frac{1}{n}. \] Thus, the inequality reduces to \[ \frac{1}{n} < s^\circ. \] --- **(D)** We have \[ \frac{1}{8} + x^2 + \left(\frac{1}{x}\right)^2. \] Step 1. Evaluate the square: \[ \left(\frac{1}{x}\right)^2 = \frac{1}{x^2}. \] Thus, the expression becomes \[ x^2 + \frac{1}{x^2} + \frac{1}{8}. \] --- **(1)** We have the inequality \[ \left(\frac{1}{n}\right)^0 < n^3 < \frac{1}{n}. \] Step 1. Since \(\left(\frac{1}{n}\right)^0 = 1\) (provided \(n \neq 0\)), the inequality becomes \[ 1 < n^3 < \frac{1}{n}. \] Step 2. Consider \(n>0\): - From \(1 < n^3\) we deduce that \[ n > 1. \] - From \(n^3 < \frac{1}{n}\) multiply both sides by \(n\) (which is positive) to obtain \[ n^4 < 1 \quad \Longrightarrow \quad n < 1. \] These two conditions, \(n > 1\) and \(n < 1\), contradict one another. Step 3. Consider \(n<0\): - For negative \(n\), note that \(n^3\) is negative while \(1\) is positive. Thus, \[ 1 < n^3 \] cannot hold. Thus, there is no real number \(n\) satisfying the inequality. --- **Summary of Answers:** - (A) \(\quad s^2 + \frac{1}{a^5}\) - (1i) \(\quad x^3 + \frac{1}{x}\) - (C) \(\quad \frac{1}{n} < s^\circ\) - (D) \(\quad x^2 + \frac{1}{x^2} + \frac{1}{8}\) - (1) \(\quad\) The inequality reduces to \(\,1 < n^3 < \frac{1}{n},\,\) which has no real solution.