Question

If $(2 x-1)^{20}-(a x+b)^{20}=\left(x^2+p x+q\right)^{10}$ holds true $\forall x \in R$ where $a, b, p$ and $q$ are real numbers, then which of the following is/are true?

• A. $2 p+3 q=1$
• B. $a+2 b=0$
• C. $a =\sqrt[20]{2^{20}-1}$
• D. $4 q+p=0$

Answer 1

Answer: (B), (C), (D)

Equating the coefficient of $x ^{20}$, we get
$2^{20}-a^{20}=1 \Rightarrow a=\sqrt[20]{2^{20}-1} \Rightarrow \text { (C) }$
put, $x =\frac{1}{2}$, we get
$0-\left(\frac{a}{2}+b\right)^{20}=\left(\frac{1}{4}+\frac{p}{2}+q\right)^{10} $
$\therefore \left(\frac{a}{2}+b\right)^{20}+\left(\frac{1}{4}+\frac{p}{2}+q\right)^{10}=0 $
$\frac{a}{2}=-b \text { and } \frac{1}{4}+\frac{p}{2}+q=0 $
$a+2 b=0 \Rightarrow \text { (B) } $
$b=\frac{-\sqrt[20]{2^{20}-1}}{2} .$
put, $ x =0$ we get
$1-b^{20}=q^{10} $
$ 1-\left(\frac{-\sqrt[20]{2^{20}-1}}{2}\right)^{20}=q^{10}$
$1-\frac{\left(2^{20}-1\right)}{2^{20}}= q ^{10} $
$\frac{1}{2^{20}}= q ^{10} \Rightarrow q =\frac{1}{4}$
Using, $\frac{1}{4}+\frac{p}{2}+q=0$
$\frac{1}{4}+\frac{p}{2}+\frac{1}{4}=0 \Rightarrow p=-1$
Hence B, C, D.