A Correct Answer Key That Is Wrong on Half the Domain
At a Kendriya Vidyalaya, a PGT Mathematics is moderating the Class XII half-yearly answer key for 38 scripts. Question 22 (4 marks) asks for the derivative of sin⁻¹(2x√(1−x²)). The key gives 2/√(1−x²) for all x in (−1, 1). While checking, the PGT finds that for x > 1/√2 the function equals π − 2 sin⁻¹x, so the derivative is −2/√(1−x²). The key is therefore wrong on part of the domain. No student has written the piecewise answer: 11 wrote the key's expression, 6 left it blank, and one wrote '2' with no working. The head examiner calls the issue 'close enough' because 'the board never tests such subtleties', and a parent has already complained about a different question. Scripts must be submitted in three days, and the school follows the CBSE marking scheme.
1. Mathematically, where exactly does the key fail, and what is the correct piecewise derivative? Show the reasoning, including the domain restriction that makes 2x√(1−x²) a valid sine input.
Read the worked answer
Put x = sin θ with θ in the principal range [−π/2, π/2]. Then √(1−x²) = cos θ (non-negative on this range), so 2x√(1−x²) = 2 sin θ cos θ = sin 2θ. Thus f(x) = sin⁻¹(sin 2θ). The identity sin⁻¹(sin 2θ) = 2θ holds only while 2θ stays inside the principal range [−π/2, π/2], i.e. θ in [−π/4, π/4], i.e. x in [−1/√2, 1/√2].
For θ in (π/4, π/2], 2θ lies in (π/2, π], so sin⁻¹(sin 2θ) = π − 2θ. Therefore for x > 1/√2, f(x) = π − 2 sin⁻¹x and f′(x) = −2/√(1−x²). Similarly, for x < −1/√2, f(x) = −π − 2 sin⁻¹x and f′(x) = −2/√(1−x²). At x = ±1/√2 the left and right derivatives are 2√2 and −2√2 (and their negatives), so the derivative does not exist there; the graph has a corner.
The key is right only on [−1/√2, 1/√2] and wrong outside it. It also fails to mention the non-differentiability at the two join points.
2. How should the PGT handle the marking of the 11 scripts and the head examiner's instruction? Give a decision and the procedural steps, keeping in mind fairness to students and the CBSE scheme.
Read the worked answer
First separate the mathematical defect from the marking decision. The question as printed does not specify the domain, so the key is defective; the 11 students who wrote the key's expression followed the intended method and should not be penalised for the setter's omission. Award them full marks. The six blank scripts get zero. The script that wrote '2' with no working should get no more than 1 mark for recognising a constant derivative on some interval: 2 is not the derivative; the derivative is 2/√(1−x²) on the principal interval.
Do not single out the head examiner publicly. Put the correct piecewise result in the moderation file with a dated note, ask the principal to forward it to the school's examination committee, and apply the same policy to all 38 scripts. If any student had given the piecewise answer, award full marks and record it as evidence that the question was capable of a better answer. The parent's separate complaint must not affect this decision. Three days is enough for a written note and a re-check. The aim is consistency and a corrected record, not a confrontation.
3. Design a 15-minute follow-up for the class that turns this error into learning, without confusing weaker students.
Read the worked answer
Start with a table: x = 0, 0.5, 0.7, 0.8, 0.9; compute f(x) and compare it with 2 sin⁻¹x. Students see agreement until about x = 0.7 and divergence after. Ask them to plot both; the graph shows a corner near x = 0.707. Then substitute x = sin θ and use the unit circle to show that sin⁻¹(sin 2θ) equals 2θ only while 2θ stays within [−π/2, π/2]. Finish with the piecewise derivative and one exam-style item.
Weaker students only need the rule, the interval and the corner: 'the formula changes when 2θ leaves the principal range.' Stronger students derive the second branch. Do not present this as a trick or as a way to catch the examiner out; present it as domain discipline, which is a standard part of inverse trigonometric functions. The key error becomes a lesson on checking the interval before applying an identity.