Question

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

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

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
$0-{\left(\frac{a}{2}+b\right)}^{20}={\left(\frac{1}{4}+\frac{p}{2}+q\right)}^{10}$
$\therefore {\left(\frac{a}{2}+b\right)}^{20}+{\left(\frac{1}{4}+\frac{p}{2}+q\right)}^{10}=0$
$\frac{a}{2}=-b\text{ and }\frac{1}{4}+\frac{p}{2}+q=0$
$a+2b=0\Rightarrow \text{ (B) }$
$b=\frac{-\sqrt[20]{2^{20}-1}}{2}.$
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.