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JEE Mains · Maths · STD 12 - 5. continuity and differentiation
If \(2x = {y^{\frac{1}{5}}} + {y^{ - \frac{1}{5}}}\) and \((x^2 -1) \frac{{{d^2}y}}{{d{x^2}}} + \lambda x\frac{{dy}}{{dx}} + ky = 0\) , then \( \lambda + k\) is equal to
- A \(-23\)
- B \(-24\)
- C \(26\)
- D \(-26\)
Answer & Solution
Correct Answer
(B) \(-24\)
Step-by-step Solution
Detailed explanation
\({y^{1/5}} + {y^{ - 1/5}} = 2x\) \(\left( {\frac{1}{5}{y^{ - 4/5}} - \frac{1}{5}{y^{ - 6/5}}} \right).\frac{{dy}}{{dx}} = 2\) \(y'\left( {{y^{1/5}} - {y^{ - 1/5}}} \right) = 10y\) \({y^{1/5}} + {y^{ - 1/5}} = 2x\) \({y^{1/5}} - {y^{ - 1/5}} = \sqrt {4{x^2} - 4} \)…
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