मराठी

If D D X [ X N − a 1 X N − 1 + a 2 X N − 2 + . . . + ( − 1 ) N a N ] E X = X N E X Then the Value of Ar, 0 < R ≤ N, is Equal to - Mathematics

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प्रश्न

If \[\frac{d}{dx}\left[ x^n - a_1 x^{n - 1} + a_2 x^{n - 2} + . . . + \left( - 1 \right)^n a_n \right] e^x = x^n e^x\] then the value of ar, 0 < r ≤ n, is equal to 

 

पर्याय

  • \[\frac{n!}{r!}\]

  • \[\frac{\left( n - r \right)!}{r!}\]

  • \[\frac{n!}{\left( n - r \right)!}\]

  • none of these

MCQ
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उत्तर

(c) \[\frac{n!}{\left( n - r \right)!}\]

According to the given equation,

\[\frac{d}{dx}\left[ x^n - a_1 x^{n - 1} + a_2 x^{n - 2} + . . . + \left( - 1 \right)^n a_n \right] e^x = x^n e^x \]

\[ \Rightarrow \frac{d}{dx}\left[ x^n - a_1 x^{n - 1} + a_2 x^{n - 2} + . . . + \left( - 1 \right)^n a_n \right] e^x = \frac{d}{dx}\left[ x^n - n x^{n - 1} + n\left( n - 1 \right) x^{n - 2} + . . . + \left( - 1 \right)^n a_n \right] e^x \]

\[\text { Comparing the coefficients of the above equation we get }, \]

\[ a_1 = n\]

\[ a_2 = n\left( n - 1 \right)\]

\[\text { Similarly }, \]

\[ a_r = n\left( n - 1 \right)\left( n - 2 \right)\left( n - 3 \right) . . . \left( n - r + 1 \right)\]

\[ \Rightarrow a_r = \frac{n!}{\left( n - r \right)!}\]

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

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आरडी शर्मा Mathematics [English] Class 12
पाठ 12 Higher Order Derivatives
Exercise 12.3 | Q 23 | पृष्ठ २४

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