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Class 12 Continuity And Differentiability · Long Answer3 Markssecond_derivatives

Class 12 Maths · Continuity And Differentiability · Long Answer #92

Here y is principal arccosecant x. For y=\(\operatorname{asin}{\left(\frac{1}{x} \right)}\), prove \(x \left(x^{2} - 1\right)\)y″ + \(2 x^{2} - 1\)y′ + \(0\)y = \(0\). Assume x>1.

✅ Step-by-Step Solution

Final Answer: Identity proved.
  1. Differentiate y=\(\operatorname{asin}{\left(\frac{1}{x} \right)}\) to obtain y′=\(- \frac{1}{x^{2} \sqrt{1 - \frac{1}{x^{2}}}}\).
  2. Differentiate again, using the product and chain rules: y″=\(\frac{2 x^{2} - 1}{x^{3} \sqrt{\frac{x^{2} - 1}{x^{2}}} \left(x^{2} - 1\right)}\).
  3. Substitute these derivatives into the left side \(x \left(x^{2} - 1\right)\)y″ + \(2 x^{2} - 1\)y′ + \(0\)y = \(0\). Collecting like terms gives \(0\).
  4. This equals \(0\), proving the identity on the stated domain.

📋 CBSE Stepwise Marking Scheme

  • [0.5 marks] Differentiate y=\(\operatorname{asin}{\left(\frac{1}{x} \right)}\) to obtain y′=\(- \frac{1}{x^{2} \sqrt{1 - \frac{1}{x^{2}}}}\).
  • [1 mark] Differentiate again, using the product and chain rules: y″=\(\frac{2 x^{2} - 1}{x^{3} \sqrt{\frac{x^{2} - 1}{x^{2}}} \left(x^{2} - 1\right)}\).
  • [1 mark] Substitute these derivatives into the left side \(x \left(x^{2} - 1\right)\)y″ + \(2 x^{2} - 1\)y′ + \(0\)y = \(0\). Collecting like terms gives \(0\).
  • [0.5 marks] This equals \(0\), proving the identity on the stated domain.

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