Question

The number of values of the pair (a, b) for which $a(x+1{)}^2+b(x^2-3x-2)+x+1$ = $0\forall xϵR$

• A. 0
• B. $1$
• C. $2$
• D. infinite

Answer 1

The correct option is A 0

We have, $f\left(x\right)=a(x+1{)}^2+b(x^2-3x-2)+x+1$

On expanding $(x+1{)}^2$, we get

$f\left(x\right)=a(x^2+2x+1)+b(x^2-3x-2)+x+1=0$

$f\left(x\right)=x^2(a+b)+x(2a-3b+1)+(a-2b+1)$

If the above quadratic equation $f\left(x\right)=0\forall x=0$, then

$a+b=0\dots \left(1\right)2a-3b+1=0\dots \left(2\right)a-2b+1=0\dots \left(3\right)$

From (1), $a=-b$

Put $a=-b$ in equaton 2 and 3, we get

$b=\frac{1}{5}$ from equation 2

and $b=\frac{1}{3}$ from equation 3, which is not possible.

$\therefore (a,b)\in \varphi$ for which $f\left(x\right)=0\forall x\in R$