Question 20 of 42advanced🔧 ApplyShort Answer3 marks

Write the following functions in the simplest form: tan⁻¹[(3a²x - x³)/(a³ - 3ax²)], a > 0; -a/3\sqrt{3} < x < a/3\sqrt{3}

Correct Answer

3tan⁻¹(x/a)

Exercise: EXERCISE 2.2 | Q: 6 | (Page 12)
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Explanation

This question requires using the substitution method similar to Example 3 in the textbook. The expression inside tan⁻¹ resembles the formula for tan(3θ). By substituting x = a·tan θ, we can simplify the compound fraction to tan(3θ), and then apply the inverse tangent function. The domain constraint ensures that 3θ remains within the principal range of tan⁻¹.

Solution Steps

  1. Step 1: Let x = a·tan θ. Then tan θ = x/a, which gives θ = tan⁻¹(x/a).

  2. Step 2: Substitute x = a·tan θ in the numerator: 3a²x - x³ = 3a²(a·tan θ) - a³tan³θ = a³(3tan θ - tan³θ)

  3. Step 3: Substitute x = a·tan θ in the denominator: a³ - 3ax² = a³ - 3a(a²tan²θ) = a³(1 - 3tan²θ)

  4. Step 4: Therefore, (3a²x - x³)/(a³ - 3ax²) = [a³(3tan θ - tan³θ)]/[a³(1 - 3tan²θ)] = (3tan θ - tan³θ)/(1 - 3tan²θ) = tan(3θ)

  5. Step 5: Given -a/3\sqrt{3} < x < a/3\sqrt{3}, we have -1/3\sqrt{3} < tan θ < 1/3\sqrt{3}, so -π/6 < θ < π/6, which means -π/2 < 3θ < π/2.

  6. Step 6: Hence tan⁻¹[(3a²x - x³)/(a³ - 3ax²)] = tan⁻¹(tan 3θ) = 3θ = 3tan⁻¹(x/a)

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