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Question

If $$\dfrac { d }{ dx } \left( \dfrac { 1+{ x }^{ 2 }+{ x }^{ 4 } }{ 1+x+{ x }^{ 2 } }  \right) = ax+b,\quad then\quad (a, b)=$$


A
(1,2)
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B
(2,1)
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C
(2,1)
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D
(1,2)
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Solution

The correct option is A $$(2, -1)$$
Given,

$$\dfrac{d}{dx}\left(\dfrac{1+x^2+x^4}{1+x+x^2}\right)$$ 

$$=\dfrac{\frac{d}{dx}\left(1+x^2+x^4\right)\left(1+x+x^2\right)-\frac{d}{dx}\left(1+x+x^2\right)\left(1+x^2+x^4\right)}{\left(1+x+x^2\right)^2}$$

$$=\dfrac{\left(4x^3+2x\right)\left(1+x+x^2\right)-\left(2x+1\right)\left(1+x^2+x^4\right)}{\left(1+x+x^2\right)^2}$$

$$=\dfrac{2x^5+3x^4+4x^3+x^2-1}{\left(1+x+x^2\right)^2}$$

$$=\dfrac{2x^5+3x^4+4x^3+x^2-1}{x^4+2x^3+3x^2+2x+1}$$

$$=2x+\dfrac{-x^4-2x^3-3x^2-2x-1}{x^4+2x^3+3x^2+2x+1}$$

$$=2x-1$$

$$\Rightarrow 2x-1=ax+b$$

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

Mathematics

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