Question

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

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

Answer 1

Answer: (D)

(16, 15)

Multiplying and dividing LHS by (x-1)
L.H.S.$=\frac{d}{dx}\left[\frac{(x-1)(x+1)(x^2+1)(x^4+1)(x^8+1)}{x-1}\right]$

$=\frac{d}{dx}\left[\frac{x^{16}-1}{x-1}\right]$

$=\frac{16x^{15}(x-1)-x^{16}+1}{(x-1{)}^2}$

$=\frac{15x^{16}-16x^{15}+1}{(x-1{)}^2}$
On comparing with RHS of question

$p=16$ and $q=15$