Question

Let $f\left(x\right)=({\lambda }^2+\lambda -2)x^2+(\lambda +2)x$ be a quadratic polynomial. The sum of all integral values of $\lambda$ for which $f\left(x\right)<1\forall x\in R$, is

• A. $-1$
• B. $-3$
• C. $0$
• D. $-2$

Answer 1

The correct option is A $-1$

$({\lambda }^2+\lambda -2)x^2+(\lambda +2)x<1({\lambda }^2+\lambda -2)x^2+(\lambda +2)x-1<0$
Let $g\left(x\right)=({\lambda }^2+\lambda -2)x^2+(\lambda +2)x-1$

As $g\left(x\right)<0$, so the graph of $g\left(x\right)$ always lies below the $x-$ axis, therefore the required conditions are
$\left(i\right)\text{leading coefficient}<0$ and $\left(ii\right)D<0$

$\left(i\right)\text{leading coefficient}<0\Rightarrow {\lambda }^2+\lambda -2<0\Rightarrow (\lambda +2)(\lambda -1)<0$
$\Rightarrow \lambda \in (-2,1)\cdots \left(1\right)$

$\left(ii\right)D<0\Rightarrow (\lambda +2{)}^2-4({\lambda }^2+\lambda -2\left)\right(-1)<0\Rightarrow (\lambda +2{)}^2+4(\lambda +2)(\lambda -1)<0\Rightarrow (\lambda +2)(\lambda +2+4\lambda -4)<0\Rightarrow (5\lambda -2)(\lambda +2)<0\Rightarrow \lambda \in (-2,\frac{2}{5})\cdots \left(2\right)$

From equation $\left(1\right)$ and $\left(2\right)$,
$\Rightarrow \lambda \in (-2,\frac{2}{5})$
Integral values in the interval is
$-1,0$
Sum of integer is $-1.$