Question

please explain answer for this question ​

Answer 1

Answer:

$\bold{c) \: \: 0}$

Step-by-step explanation:

$\bold{Given} = > \alpha \: and \: \beta \: are \: the \: roots \: of \: equation \\ {x}^{2} - a(x - 1) + b = 0$

$\bold{To \: find} = > value \: of \\ \dfrac{1}{ { \alpha }^{2} - a \alpha } + \dfrac{1}{ { \beta }^{2} - a \beta } + \dfrac{2}{a + b}$

$\bold{ Concept \: used} = > \\ roots \: of \: equation \: always \: satisfy \: its \: own\: equation$

$\bold{Solution} = > \\ {x}^{2} - a(x - 1) + b = 0 \\ roots \: of \: given \: equation \: is \: \alpha \: and \beta$

$satisfying \: given \: equation \: by \: \alpha$

$= > { \alpha }^{2} - a( \alpha - 1) + b = 0$

$= > { \alpha }^{2} - a \alpha + a + b = 0$

$= > { \alpha }^{2} - a \alpha = - (a + b)$

$similarly \: we \: get \\ { \beta }^{2} - a \beta = - (a + b)$

$now$

$\dfrac{1}{ { \alpha }^{2} - a \alpha } + \dfrac{1}{ { \beta }^{2} - a \beta } + \dfrac{2}{a + b}$

$= \dfrac{1}{ - (a + b)} + \dfrac{1}{ - (a + b)} + \dfrac{2}{(a + b)}$

$= - \dfrac{1}{a + b} - \dfrac{1}{a + b} + \dfrac{2}{a + b}$

$= - \dfrac{2}{a + b} + \dfrac{2}{a + b}$

$= 0$