If x3-6x2-10x-2-x2-5x-6-fx-A-x-2-B-x-3- then fx- Note - A-B are constants

The correct option is A x 1 Given, x^3 6x^2+10x 2/x^2 5x+6=f(x)+A/x 2+B/x 3 nbsp; nbsp; nbsp;....(1)Consider, nbsp; x^ 3 6x^ 2 +10x 2 ...

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Byju's Answer

Standard X

Mathematics

Rational Expression

If x3-6x2+1...

Question

If $\frac{x^{3} - 6 x^{2} + 10 x - 2}{x^{2} - 5 x + 6} = f (x) + \frac{A}{x - 2} + \frac{B}{x - 3}$ , then $f (x) =$
(Note : $A, B$ are constants)

x - 1

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x + 1

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x

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x + 2

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Solution

The correct option is A $x - 1$
Given, $\frac{x^{3} - 6 x^{2} + 10 x - 2}{x^{2} - 5 x + 6} = f (x) + \frac{A}{x - 2} + \frac{B}{x - 3}$ ....(1)
Consider, $\frac{x^{3} - 6 x^{2} + 10 x - 2}{x^{2} - 5 x + 6}$
Here, degree of numerator > degree of denominator
$\frac{x^{3} - 6 x^{2} + 10 x - 2}{x^{2} - 5 x + 6} = x - 1 + \frac{- x - 2}{x^{2} - 5 x + 6}$ ....(2)
Resolving $\frac{- x - 2}{x^{2} - 5 x + 6}$ into partial fractions
$\frac{- x - 2}{x^{2} - 5 x + 6} = \frac{- x - 2}{(x - 2) (x - 3)} = \frac{A}{x - 2} + \frac{B}{x - 3}$ .....(3)
Put this in (2), we get
$\frac{x^{3} - 6 x^{2} + 10 x - 2}{x^{2} - 5 x + 6} = x - 1 + \frac{A}{x - 2} + \frac{B}{x - 3}$ .....(4)
Comparing with (1), $f (x) = x - 1$
From (3), it follows
$\Rightarrow - x - 2 = A (x - 3) + B (x - 2)$
$\Rightarrow - x - 2 = (A + B) x + (- 3 A - 2 B)$
$A + B = - 1, - 3 A - 2 B = - 2$
Solving these equations, we get
$A = 4, B = - 5$
Hence, option A is correct

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If x3−6x2+10x−2x2−5x+6=f(x)+Ax−2+Bx−3, then f(x)= (Note : A,B are constants)

If $\frac{x^{3} - 6 x^{2} + 10 x - 2}{x^{2} - 5 x + 6} = f (x) + \frac{A}{x - 2} + \frac{B}{x - 3}$ , then $f (x) =$
(Note : $A, B$ are constants)