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If Y = Sin (M Sin−1 X), Then (1 − X2) Y2 − Xy1 is Equal to (A) M2y (B) My (C) −M2y (D) None of These - Mathematics

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प्रश्न

If y = sin (m sin−1 x), then (1 − x2) y2 − xy1 is equal to

विकल्प

  • m2y

  • my

  • −m2y

  • none of these

MCQ
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उत्तर

(c)−m2

Here,

\[y = \sin\left( m \sin^{- 1} x \right)\]
\[ \Rightarrow y_1 = \cos\left( m \sin^{- 1} x \right)\frac{m}{\sqrt{1 - x^2}}\]
\[ \Rightarrow y_2 = - \sin\left( m \sin^{- 1} x \right)\frac{m^2}{\left( 1 - x^2 \right)} + \frac{mx\cos\left( m \sin^{- 1} x \right)}{\left( 1 - x^2 \right)^{3/2}}\]
\[ \Rightarrow y_2 = - \sin\left( m \sin^{- 1} x \right)\frac{m^2}{\left( 1 - x^2 \right)} + \frac{xm\cos\left( m \sin^{- 1} x \right)}{\left( 1 - x^2 \right) \times \sqrt{1 - x^2}}\]
\[ \Rightarrow y_2 = - \sin\left( m \sin^{- 1} x \right)\frac{m^2}{\left( 1 - x^2 \right)} + \frac{x y_1}{\left( 1 - x^2 \right)}\]
\[ \Rightarrow \left( 1 - x^2 \right) y_2 = - y m^2 + x y_1 \]
\[ \Rightarrow \left( 1 - x^2 \right) y_2 - x y_1 = - m^2 y\]

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अध्याय 12: Higher Order Derivatives - Exercise 12.3 [पृष्ठ २४]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 12 Higher Order Derivatives
Exercise 12.3 | Q 15 | पृष्ठ २४

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