If d d x x 4 + x 2 + 1 x 2 + x + 1 = a x + b , then a - b =

If

\frac{d}{d x} (\frac{x^{4} + x^{2} + 1}{x^{2} + x + 1}) = a x + b,

then

a - b =

$3$
$4$
$1$
$2$

Solution

$\frac{d}{d x} (\frac{x^{4} + x^{2} + 1}{x^{2} + x + 1})$

$= \frac{d}{d x} (\frac{(x^{2} + x + 1) (x^{2} - x + 1)}{x^{2} + x + 1})$

$= \frac{d}{d x} (x^{2} - x + 1) = 2 x - 1$

$a = 2, b = - 1$

$a - b = 3$

Asked in: AP EAMCET 2020 (23 Sep Shift 1)

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