Question

If $\frac{d}{dx}\left[(x+1)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\right]=\left(15x^p-16x^q+1\right)(x-1{)}^{-2}$, then $(p,q)$ is equal to

• A. (12,11)
• B. (15,14)
• C. (16,14)
• D. (16,15)

Answer 1

Answer: (D)

$\frac{d}{dx}\left[(x+1)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\right]$
$=\frac{\left(15x^p-16x^q+1\right)}{(x-1{)}^2}$ ...(i)
$LHS=\frac{d}{dx}\left[\frac{\left(x^2-1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)}{(x-1)}\right]$
$=\frac{d}{dx}\left[\frac{\left(x^4-1\right)\left(x^4+1\right)\left(x^8+1\right)}{(x-1)}\right]$
$=\frac{d}{dx}\left[\frac{\left(x^8-1\right)\left(x^8+1\right)}{(x-1)}\right]$
$=\frac{d}{dx}\left[\frac{\left(x^{16}-1\right)}{(x-1)}\right]$
$=\frac{(x-1)\left(16x^{15}\right)-\left(x^{16}-1\right)}{(x-1{)}^2}$
$=\frac{16x^{16}-16x^{15}-x^{16}+1}{(x-1{)}^2}$
$=\frac{15x^{16}-16x^{15}+1}{(x-1{)}^2}$
On comparing LHS = RHS, we get
$p=16$ and $q=15$
$\therefore (p,q)=(16,15)$