If $\frac{d}{d x} (\frac{1 + x^{2} + x^{4}}{1 + x + x^{2}}) = a x + b,$ then the value of 'a' and 'b' are respectively

2 and 1

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-2 and 1

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2 and -1

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none of these

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Solution

The correct option is C 2 and -1
$y = \frac{x^{4} + x^{2} + 1}{x^{2} + x + 1}$

$∴ y = x^{2} - x + 1$

Now $\frac{d y}{d x} = 2 x - 1 = a x + b$

$\Rightarrow a = 2, b = - 1$

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