Question

The real values of $a,b,p,q$ for which
${(2x-1)}^{20}-{(ax+b)}^{20}={(x^2+px+q)}^{10}$

• A. $2b=-a=\pm {(2^{20}-1)}^{\frac{1}{20}},p=q=\frac{1}{2}$
• B. $b=-2a=\pm {(2^{20}-1)}^{\frac{1}{20}},p=q=\frac{1}{4}$
• C. $2b=-a=\pm {(2^{20}-1)}^{\frac{1}{20}},4q=-p=1$
• D. none of these

Answer 1

Answer: (C)

$2b=-a=\pm {(2^{20}-1)}^{\frac{1}{20}},4q=-p=1$

Equating coefficients of $x^{20}$, we get
$2^{20}-a^{20}=1$
$\Rightarrow a=\pm {(2^{20}-1)}^{\frac{1}{20}}$
Next, putting $x=1/2$ we get
${(\frac{a}{2}+b)}^2+{(\frac{1}{4}+\frac{1}{2}p+q)}^{10}=0\Rightarrow b=-\frac{a}{2}$
The given equation can now be written as
$2^{20}{(x-\frac{1}{2})}^{20}-a^{20}{(x-\frac{1}{2})}^{20}={(x^2+px+q)}^{10}$
$or{(x-\frac{1}{2})}^2=x^2+px+q$
$\Rightarrow p=-1,q=\frac{1}{4}$