Question

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

• A. $\frac{\sqrt[20]{202^{20}+1}}{2}$
• B. $\sqrt[20]{202^{20}+1}$
• C. $\sqrt[20]{202^{20}-1}$
• D. $\frac{\sqrt[20]{202^{20}-1}}{2}$

Answer 1

The correct option is D $\frac{\sqrt[20]{202^{20}-1}}{2}$

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

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 $\frac{1}{4}+\frac{p}{2}+q=0$

Now, $b=-\frac{a}{2}=\pm \frac{\sqrt[20]{202^{20}-1}}{2}$
$\therefore |b|=\frac{\sqrt[20]{202^{20}-1}}{2}$