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Class 12 Inverse Trigonometric Functions · Short Answer2 Marksinverse_function_graphs

Class 12 Maths · Inverse Trigonometric Functions · Short Answer #56

Draw \(y=\sin^{-1}x\) for \(x\in[-1/\sqrt2,1/\sqrt2]\), and state its range.
y=arcsin x for −1/√2≤x≤1/√2, with range [−π/4,π/4]. Both endpoints are closed.
y=arcsin x for −1/√2≤x≤1/√2, with range [−π/4,π/4]. Both endpoints are closed.

✅ Step-by-Step Solution

Final Answer: [−π/4,π/4]
  1. The endpoint values are arcsin(−1/√2)=−π/4 and arcsin(1/√2)=π/4.
  2. Arcsine is continuous and strictly increasing; draw its curve through the origin between these endpoints, both closed. The attached SVG plots the restricted curve.
  3. The range is [−π/4,π/4].

📋 CBSE Stepwise Marking Scheme

  • [0.5 marks] The endpoint values are arcsin(−1/√2)=−π/4 and arcsin(1/√2)=π/4.
  • [1 mark] Arcsine is continuous and strictly increasing; draw its curve through the origin between these endpoints, both closed. The attached SVG plots the restricted curve.
  • [0.5 marks] The range is [−π/4,π/4].

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