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STD 12 - 5. continuity and differentiation — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 5. continuity and differentiation1 Mark
MCQ
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
    $-23$
  • ✓
    $-24$
  • C
    $26$
  • D
    $-26$
✓

Answer

Correct option: B.
$-24$
b
${y^{1/5}} + {y^{ - 1/5}} = 2x$

$\left( {\frac{1}{5}{y^{ - 4/5}} - \frac{1}{5}{y^{ - 6/5}}} \right).\frac{{dy}}{{dx}} = 2$

$y'\left( {{y^{1/5}} - {y^{ - 1/5}}} \right) = 10y$

${y^{1/5}} + {y^{ - 1/5}} = 2x$

${y^{1/5}} - {y^{ - 1/5}} = \sqrt {4{x^2} - 4} $

$y'\left( {2\sqrt {{x^2} - 1} } \right) = 10y$

$y''\left( {2\sqrt {{x^2} - 1} } \right) + y'2\frac{{2x}}{{2\sqrt {{x^2} - 1} }} = 10y'$

                   $y''\left( {{x^2} - 1} \right) + xy' = 5\sqrt {{x^2} - 1} \left( {y'} \right)$

$\boxed{y''\left( {{x^2} - 1} \right) + xy' - 25y = 0}$

$\lambda  = 1,k =  - 25$

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Let $E$ and $F$ be two independent events. The probability that exactly one of them occurs is $\frac{11}{25}$ and the probability of none of them occurring is $\frac{2}{25}$. If $P(T)$ denotes the probability of occurrence of the event $T$, then

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