Trigonometry & Inverse Trigonometry Question - None

Question

Statement I: Let $f(x) = \sin^{-1}\left(\frac{2x}{1+x^2}\right)$. Then $f'(2) = -\frac{2}{5}$
Statement II: $\sin^{-1}\left(\frac{2x}{1+x^2}\right) = \pi$

Accepted Answer

Step 1: Analyze Statement II

The range of sin⁻¹ is [-π/2, π/2]. Since π ≈ 3.14 lies outside this range, sin⁻¹(2x/(1+x²)) can never equal π for any real value of x. Therefore, Statement II is FALSE.

Step 2: Verify Statement I by finding f'(x)

Given: f(x) = sin⁻¹(2x/(1+x²))

Using the chain rule: f'(x) = 1/√(1 - (2x/(1+x²))²) · d/dx[2x/(1+x²)]

Step 3: Calculate d/dx[2x/(1+x²)]

Using quotient rule: d/dx[2x/(1+x²)] = [2(1+x²) - 2x(2x)]/(1+x²)² = (2 + 2x² - 4x²)/(1+x²)² = (2 - 2x²)/(1+x²)²

Step 4: Simplify 1 - (2x/(1+x²))²

1 - (2x/(1+x²))² = 1 - 4x²/(1+x²)² = [(1+x²)² - 4x²]/(1+x²)² = (1 + 2x² + x⁴ - 4x²)/(1+x²)² = (1 - 2x² + x⁴)/(1+x²)² = (1-x²)²/(1+x²)²

Therefore: √[1 - (2x/(1+x²))²] = |1-x²|/(1+x²)

Step 5: Calculate f'(x)

f'(x) = [(1+x²)/|1-x²|] · [(2-2x²)/(1+x²)²] = [2(1-x²)]/[|1-x²|(1+x²)]

For x = 2: |1-4| = 3, so f'(2) = [2(1-4)]/[3(1+4)] = [2(-3)]/[3(5)] = -6/15 = -2/5

Therefore, Statement I is TRUE.

Step 6: Determine the correct option

Statement I is True; Statement II is False.

∴ Answer: c

Question:

Step-by-Step Solution

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