Question

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

• A. $2b=-a=\pm (2^{20}-1{)}^{1/20},p=q=\frac{1}{2}$
• B. $b=-2a=\pm (2^{20}-1{)}^{1/20},p=q=\frac{1}{4}$
• C. $2b=-a=\pm (2^{20}-1{)}^{1/20},4q=-p=1$
• D. $Noneofthese$

Answer 1

The correct option is C $2b=-a=\pm (2^{20}-1{)}^{1/20},4q=-p=1$

Equating the coefficients of $x^{20}$,
$2^{20}-a^{20}=1$
$a^{20}=2^{20}-1$
Now, put $x=\frac{1}{2}$
${(\frac{a}{2}+b)}^{20}+{(\frac{1}{4}+\frac{p}{2}+q)}^{10}=0$
$\Rightarrow \frac{a}{2}=-b$ and $\Rightarrow \frac{p}{2}+q=-\frac{1}{4}$
So, the given equation becomes
$2^{20}{(x-\frac{1}{2})}^{20}-a^{20}{(x-\frac{1}{2})}^{20}=(x^2+px+q{)}^{10}$
$\Rightarrow {(x-\frac{1}{2})}^{20}(2^{20}-a^{20})=(x^2+px+q{)}^{10}$
${(x-\frac{1}{2})}^{20}=(x^2+px+q{)}^{10}$
${(x-\frac{1}{2})}^2=x^2+px+q$
$\Rightarrow p=-1,q=\frac{1}{4}$
Hence, $2b=-a=\pm (2^{20}-1{)}^{1/20},4q=-p=1$