DAV Class 8 Maths Chapter 3 HOTS
Exponents and Radicals HOTS
1. Evaluate: \( (6^{-1} - 8^{-1})^{-1} + (2^{-1} - 3^{-})^{-1} \)
Solution
\[ \begin{align*} & = (6^{-1} - 8^{-1})^{-1} + (2^{-1} - 3^{-})^{-1} \\[8pt] & = \left(\frac{1}{6} - \frac{1}{8}\right)^{-1} + \left(\frac{1}{2} - \frac{1}{3}\right)^{-1} \\[8pt] & = \left(\frac{4-3}{24}\right)^{-1} + \left(\frac{3-2}{6}\right)^{-1} \\[8pt] & = \left(\frac{1}{24}\right)^{-1} + \left(\frac{1}{6}\right)^{-1} \\[8pt] & = 24+6 \\[8pt] & = 30 \\[8pt] \end{align*} \]
Answer \( \color{red} 30 \)
2. If \( 3^{x-1} = 9 \) and \( 4^{y+2} = 64 \), find the value of \( \dfrac{y}{x} - \dfrac{x}{y} \)
Solution
\[ \begin{align*} 3^{x-1} &= 9 \\ 3^{x-1} &= 3^2 \\ x-1 &= 2 \\ x &= 2 + 1 \\ \implies \color{magenta} x &= \color{magenta} 3 \\ \\ 4^{y+2} &= 64 \\ 4^{y+2} &= 4^3 \\ y+2 &= 3 \\ y &= 3-2 \\ \implies \color{magenta} y &= \color{magenta} 1 \\ \\ & = \frac{y}{x} - \frac{x}{y} \\[8pt] & = \frac{1}{3} - \frac{3}{1} \\[8pt] & = \frac{1 - 9}{3} \\[8pt] & = \frac{-8}{3} \\[8pt] \end{align*} \]
Answer \( \dfrac{y}{x} - \dfrac{x}{y} = \color{red} \dfrac{-8}{3} \)