Math, asked by themarvel, 2 months ago

Solve it and answer it as soon as possible ​

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Answered by mathdude500
3

\large\underline\purple{\bold{Solution :-  }}

\tt \:   \: Let \: \dfrac{ {x}^{4} }{(x - 1)(x - 2)}  =  {x}^{2}  + 3x + \dfrac{a}{x - 1}  + \dfrac{b}{x - 2}  -  - (1)

☆ On taking LCM, we get

\tt \:   {x}^{4}  =  ({x}^{2}  + 3x)(x - 1)(x - 2) + a(x - 2) + b(x - 1)

☆ On substituting x = 1, we get

\tt \:  ⟼ \: 1 = a(1 - 2)

\tt \:  ⟼ \: a \:  =  \:  - 1

☆ On substituting x = 2, we get

\tt \:  ⟼ \:  {2}^{4}  = b(2 - 1)

\tt \:  ⟼ \: b \:  =  \: 16

☆ On substituting the values of a and b in equation (1), we get

\tt \: \dfrac{ {x}^{4} }{(x - 1)(x - 2)}  =  {x}^{2}  + 3x + \dfrac{( - 1)}{x - 1}  + \dfrac{16}{x - 2}

☆ Now, Differentiate n times w.r.t x, we get

\tt \:   \dfrac{ {d}^{n} y}{ {dx}^{n} }  \: \dfrac{ {x}^{4} }{(x - 1)(x - 2)}  =\dfrac{ {d}^{n} }{ {dx}^{n} }   \bigg({x}^{2}  + 3x + \dfrac{( - 1)}{x - 1}  + \dfrac{16}{x - 2} \bigg)

\tt \:   =  ( - 1)\dfrac{ {( - 1)}^{n}n! }{ {(x  -  1)}^{n + 1} }  + 16\dfrac{ {( - 1)}^{n}n! }{ {(x - 2)}^{n + 1} }

\tt \:  \bf\implies \:  a \:  = 16 {( - 1)}^{n} n!  \:  \:  \:  \: \\ \bf\implies \:b \:  =  {( - 1)}^{n + 1} n! \:

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