Question

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 a_r, 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

Answer 1

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

According to the given equation,

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

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

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

\[ a_1 = n\]

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

\[\text { Similarly }, \]

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

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