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 the value of $p$ is

• A. $1$
• B. $3$
• C. $-2$
• D. $-1$

Answer 1

The correct option is D $-1$

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

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

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

Using $\frac{1}{4}+\frac{p}{2}+q=0$
$\Rightarrow \frac{1}{4}+\frac{p}{2}\pm \frac{1}{4}=0\Rightarrow p=-1,0$
From given options $p=-1$