Hint
Use the stated units and precision. Divide the coefficients and add or subtract fractional powers using a common denominator.
Worked solution
(a)
The coefficient is \(\displaystyle \frac{3}{2}\). Combining exponents gives \(\displaystyle \frac{1}{2}\) − \(\displaystyle \frac{5}{3}\) + \(\displaystyle \frac{1}{2}\) = \(\displaystyle - \frac{2}{3}\).
Answer: \(\displaystyle \frac{3}{2 x^{\frac{2}{3}}}\), x>0.
(b)
Since \(\displaystyle 64=2^{6}\), the variable part becomes \(2^{-4}\). Multiply by \(\displaystyle \frac{3}{2}\).
Answer: \(\displaystyle \frac{3}{32}\)
(c)
Square the coefficient and double the complete exponent.
Answer: \(\displaystyle \frac{9}{4 x^{\frac{4}{3}}}\), x>0.
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official mark scheme. Credit complete valid alternative methods and consequential reasoning. An unexplained result does not earn a method mark.
(a) · 3 marks
- 1 mark: Use a common denominator to combine every exponent with its correct sign.
- 1 mark: Simplify the coefficient and preserve the positive-domain restriction.
- 1 mark: Give the complete requested result, using the required units, precision and boundary inclusion.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: Keep intermediate values exact until the final step.
(b) · 2 marks
- 1 mark: Express the positive input as a convenient power before applying the rational exponent.
- 1 mark: Give the complete requested result, using the required units, precision and boundary inclusion.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: Check the answer against the original conditions.
(c) · 2 marks
- 1 mark: Apply the outer power to both the coefficient and the variable power.
- 1 mark: Give the complete requested result, using the required units, precision and boundary inclusion.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: Keep intermediate values exact until the final step.
Total: 7 marks.