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Q.
If $\frac{d}{d x}\left[(x+1)\left(x^{2}+1\right)\left(x^{4}+1\right)\left(x^{8}+1\right)\right]=\left(15 x^{p}-16 x^{q}+1\right)(x-1)^{-2}$, then $(p, q)$ is equal to
EAMCETEAMCET 2013
Solution:
$\frac{d}{d x}\left[(x+1)\left(x^{2}+1\right)\left(x^{4}+1\right)\left(x^{8}+1\right)\right]$
$=\frac{\left(15 x^{p}-16 x^{q}+1\right)}{(x-1)^{2}}$ ...(i)
$LHS =\frac{d}{d x}\left[\frac{\left(x^{2}-1\right)\left(x^{2}+1\right)\left(x^{4}+1\right)\left(x^{8}+1\right)}{(x-1)}\right]$
$=\frac{d}{d x}\left[\frac{\left(x^{4}-1\right)\left(x^{4}+1\right)\left(x^{8}+1\right)}{(x-1)}\right]$
$=\frac{d}{d x}\left[\frac{\left(x^{8}-1\right)\left(x^{8}+1\right)}{(x-1)}\right]$
$=\frac{d}{d x}\left[\frac{\left(x^{16}-1\right)}{(x-1)}\right]$
$=\frac{(x-1)\left(16 x^{15}\right)-\left(x^{16}-1\right)}{(x-1)^{2}}$
$=\frac{16 x^{16}-16 x^{15}-x^{16}+1}{(x-1)^{2}}$
$=\frac{15 x^{16}-16 x^{15}+1}{(x-1)^{2}}$
On comparing LHS = RHS, we get
$p=16$ and $q=15$
$\therefore (p, q)=(16, 15)$