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Class 12 Continuity And Differentiability · Long Answer3 Markssecond_derivatives

Class 12 Maths · Continuity And Differentiability · Long Answer #98

For y=\(\log{\left(\left(\sqrt{x} + \frac{1}{\sqrt{x}}\right)^{2} \right)}\), prove \(x \left(x + 1\right)^{2}\)y″ + \(\left(x + 1\right)^{2}\)y′ + \(0\)y = \(2\). Assume x>0.

✅ Step-by-Step Solution

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

📋 CBSE Stepwise Marking Scheme

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

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