Question

If $\lambda$ is real and $(\lambda^2 + \lambda - 2)x^2 + (\lambda + 2)x < 1$ for all real $x$, then number of integral values of $\lambda$ is

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (C)

3

Answer 2

The inequality can be rewritten as $(\lambda^2 + \lambda - 2)x^2 + (\lambda + 2)x - 1 < 0$. For this quadratic inequality to hold for all real $x$, two conditions must be met:

1. The leading coefficient must be negative: $A = \lambda^2 + \lambda - 2 < 0$. Factoring gives $(\lambda+2)(\lambda-1) < 0$, so $-2 < \lambda < 1$.
2. The discriminant must be negative: $D = (\lambda+2)^2 - 4(\lambda^2 + \lambda - 2)(-1) < 0$. $D = (\lambda^2 + 4\lambda + 4) + 4(\lambda^2 + \lambda - 2) = 5\lambda^2 + 8\lambda - 4$. For $D < 0$, we find the roots of $5\lambda^2 + 8\lambda - 4 = 0$: $\lambda = \frac{-8 \pm \sqrt{64 - 4(5)(-4)}}{10} = \frac{-8 \pm \sqrt{144}}{10} = \frac{-8 \pm 12}{10}$. The roots are $\lambda = -2$ and $\lambda = 2/5$. So, $D < 0$ when $-2 < \lambda < 2/5$. Combining the conditions for $A<0$ and $D<0$, we need $-2 < \lambda < 1$ and $-2 < \lambda < 2/5$. The intersection is $-2 < \lambda < 2/5$. The integral values in this range are $-1$ and $0$.

We must also consider the case when the leading coefficient is zero: $A = \lambda^2 + \lambda - 2 = 0$. This occurs when $(\lambda+2)(\lambda-1) = 0$, so $\lambda = -2$ or $\lambda = 1$.

• If $\lambda = -2$: The inequality becomes $0x^2 + (-2+2)x - 1 < 0$, which simplifies to $-1 < 0$. This is true for all $x$, so $\lambda = -2$ is a valid integral value.
• If $\lambda = 1$: The inequality becomes $0x^2 + (1+2)x - 1 < 0$, which simplifies to $3x - 1 < 0$. This is not true for all $x$ (e.g., $x=1$). So $\lambda = 1$ is not valid.

The integral values of $\lambda$ are $-2, -1, 0$. There are 3 such values.