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Class 12 Continuity And Differentiability · Long Answer3 Markssecond_derivatives

Class 12 Maths · Continuity And Differentiability · Long Answer #148

For y=\(\left(\sin{\left(x \right)} + \cos{\left(x \right)}\right) e^{x}\), prove \(1\)y″ + \(-2\)y′ + \(2\)y = \(0\). Assume all open intervals where the displayed expressions are real and differentiable.

✅ Step-by-Step Solution

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

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