Question

If $(1+ax+b{x}^2{)}^4=a_0+a_{1}x+a_2{x}^2+\dots +a_8{x}^8;a,b,a_0,a_1\dots a_8ϵR$ and are such that $a_0+a_1+a_2\ne 0$ and $\left|\begin{array}{ccc}{a}_0& a_1& a_2\\ a_1& a_2& a_0\\ a_2& a_0& a_1\end{array}\right|=0$, then

• A. $a=\frac{3}{4},b=\frac{5}{8}$
• B. $a=\frac{1}{4},b=\frac{5}{32}$
• C. $a=1,b=\frac{2}{3}$
• D. $a=1,b=\frac{5}{32}$

Answer 1

The correct option is B

$a=\frac{1}{4},b=\frac{5}{32}$

Put $x=0$ in $(1+ax+b{x}^2{)}^4=a_0+a_{1}x+a_2{x}^2+\dots +a_8{x}^8\dots (1)$

$\therefore a_0=1$

Differentiate equation (1) and substitute $x=0,$ we get $a_1=4a.$

Differentiating again and substituting $x=0,$ we get $a_2=6a^2+4b$

Now $\left|\begin{array}{ccc}{a}_0& a_1& a_2\\ a_1& a_2& a_0\\ a_2& a_0& a_1\end{array}\right|=0$

$\Rightarrow a_0^3+a_1^3+a_2^3–3a_0{a}_1{a}_2=0$

$\Rightarrow (a_0+a_1+a_2)\left(\right(a_0-a_1{)}^2+(a_1-a_2{)}^2+(a_2-a_0{)}^2)=0$

But $a_0+a_1+a_2\ne 0\Rightarrow a_0=a_1=a_2=1$

$\therefore 1=4a,6a^2+4b=1$

$\therefore a=\frac{1}{4},b=\frac{5}{32}$