NDA 2015
Exam NDA
Year 2015
Subject Mathematics
Chapter Differentiation NDA 2015 · Mathematics · Differentiation
Given that (d)/(d x)((1+x^2+x^4)/(1+x+x^2))=A x+B What is the value...
Given that \frac{d}{d x}\left(\frac{1+x^2+x^4}{1+x+x^2}\right)=A x+B$
What is the value of B?
Step-by-step solution
Given
d d x ( 1 + x 2 + x 4 1 + x + x 2 ) = d d x [ 1 + x + x 2 + x 4 − x 1 + x + x 2 ] \frac{d}{d x}\left(\frac{1+x^2+x^4}{1+\mathrm{x}+\mathrm{x}^2}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left[\frac{1+\mathrm{x}+\mathrm{x}^2+\mathrm{x}^4-\mathrm{x}}{1+\mathrm{x}+\mathrm{x}^2}\right] d x d ( 1 + x + x 2 1 + x 2 + x 4 ) = dx d [ 1 + x + x 2 1 + x + x 2 + x 4 − x ] = d d x [ 1 + x 4 − x x 2 + x + 1 ] = d d x [ 1 + x ( x 3 − 1 ) x 2 + x + 1 ] = d d x [ 1 + x ( x − 1 ) ] = d d x [ 1 + x 2 − x ] = 2 x − 1... ( i ) \begin{aligned}
& =\frac{d}{d x}\left[1+\frac{x^4-x}{x^2+x+1}\right] \\
& =\frac{d}{d x}\left[1+\frac{x\left(x^3-1\right)}{x^2+x+1}\right] \\
& =\frac{d}{d x}[1+x(x-1)] \\
& =\frac{d}{d x}\left[1+x^2-x\right]=2 x-1...(i)
\end{aligned} = d x d [ 1 + x 2 + x + 1 x 4 − x ] = d x d [ 1 + x 2 + x + 1 x ( x 3 − 1 ) ] = d x d [ 1 + x ( x − 1 )] = d x d [ 1 + x 2 − x ] = 2 x − 1... ( i ) $
N o w c o m p a r i n g e q u a t i o n ( i ) w i t h Now comparing equation (i) with N o w co m p a r in g e q u a t i o n ( i ) w i t h \mathrm{AX}+\mathrm{B}
, w e g e t , we get , w e g e t \mathrm{A}=2
a n d and an d \mathrm{B}=-1$.
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