If ddx 1-x2-x41-x-x2-ax-b- then the value of a-b is

1+x^2+x^4 =1+2x^2+x^4 x^2 =(1+x^2)^2 x^2 =(1+x^2+x)(1+x^2 x) ∴( 1+x^2+x^41+x+x^2 nbsp; nbsp;)=1+x^2 x So, nbsp;ddx( 1+x^2+x^41+x+x^2 nbsp; nbsp;)= d ...

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Standard XII

Mathematics

Sum of n Terms

If ddx 1+x2+x...

Question

If $\frac{d}{d x} (\frac{1 + x^{2} + x^{4}}{1 + x + x^{2}}) = a x + b,$ then the value of $a - b$ is

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Solution

$1 + x^{2} + x^{4}$
$= 1 + 2 x^{2} + x^{4} - x^{2}$
$= (1 + x^{2})^{2} - x^{2}$
$= (1 + x^{2} + x) (1 + x^{2} - x)$
$∴ (\frac{1 + x^{2} + x^{4}}{1 + x + x^{2}}) = 1 + x^{2} - x$
So, $\frac{d}{d x} (\frac{1 + x^{2} + x^{4}}{1 + x + x^{2}}) = \frac{d}{d x} (1 + x^{2} - x) = 2 x - 1$
$\Rightarrow 2 x - 1 = a x + b$
$\Rightarrow a = 2, b = - 1$
$∴ a - b = 3$

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Sum of n Terms

Standard XII Mathematics

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If ddx(1+x2+x41+x+x2)=ax+b, then the value of a−b is

1+x2+x4 =1+2x2+x4−x2 =(1+x2)2−x2 =(1+x2+x)(1+x2−x) ∴(1+x2+x41+x+x2)=1+x2−x So, ddx(1+x2+x41+x+x2)=ddx(1+x2−x)=2x−1 ⇒2x−1=ax+b ⇒a=2,b=−1 ∴a−b=3

If $\frac{d}{d x} (\frac{1 + x^{2} + x^{4}}{1 + x + x^{2}}) = a x + b,$ then the value of $a - b$ is