Question

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

• A. $2p+3q=1$
• B. $a+2b=0$
• C. $a=\sqrt[20]{202^{20}-1}$
• D. $4q+p=0$

Answer 1

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

The correct options are
B $a+2b=0$
C $a=\sqrt[20]{202^{20}-1}$
D $4q+p=0$

$(2x-1{)}^{20}-(ax+b{)}^{20}={(x^2+px+q)}^{10}$
Equating the coefficient of $x^{20},$ we get $2^{20}-a^{20}=1$
$\Rightarrow a=\sqrt[20]{202^{20}-1}$

Put $x=\frac{1}{2},$ we get
$0-{(\frac{a}{2}+b)}^{20}={(\frac{1}{4}+\frac{p}{2}+q)}^{10}$
$\Rightarrow {(\frac{a}{2}+b)}^{20}+{(\frac{1}{4}+\frac{p}{2}+q)}^{10}=0$
$\Rightarrow \frac{a}{2}=-b$
and $\frac{1}{4}+\frac{p}{2}+q=0\cdots \left(1\right)$
$a+2b=0$
$\therefore b=\frac{-\sqrt[20]{202^{20}-1}}{2}$

Put $x=0,$ we get
$1-b^{20}=q^{10}$
$\Rightarrow 1-{\left(\frac{-\sqrt[20]{202^{20}-1}}{2}\right)}^{20}=q^{10}$
$\Rightarrow 1-\frac{(2^{20}-1)}{2^{20}}=q^{10}$
$\Rightarrow \frac{1}{2^{20}}=q^{10}$
$\Rightarrow q=\frac{1}{4}$

From equation $\left(1\right),$
$\frac{1}{4}+\frac{p}{2}+q=0$
$\Rightarrow \frac{1}{4}+\frac{p}{2}+\frac{1}{4}=0$
$\Rightarrow p=-1$