Question

Solve it and answer it as soon as possible ​

Answer 1

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

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

☆ 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

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

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

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