Question

The exhaustive set of values of a for which inequation $(a-1)x^2-(a+1)x+a-1\geq0$ is true $\forall x\geq2$.

• A. $(-\infty,1)$
• B. $[\frac{7}{3},\infty)$
• C. $[\frac{3}{7},\infty)$
• D. None of these

Answer 1

Answer: (B)

[$7/3, ∞)

Answer 2

Let $f(x) = (a-1)x^2-(a+1)x+a-1$. We need $f(x) \geq 0$ for all $x \geq 2$.

1. If $a=1$: $f(x) = -2x$. For $f(x) \geq 0$, we need $-2x \geq 0 \implies x \leq 0$. This contradicts $x \geq 2$. So $a=1$ is not a solution.

2. If $a<1$: The parabola opens downwards. As $x \to \infty$, $f(x) \to -\infty$. Thus, $f(x) \geq 0$ for all $x \geq 2$ is impossible. No solutions for $a<1$.

3. If $a>1$: The parabola opens upwards.

   • If $D \leq 0$: The discriminant $D = -3a^2+10a-3$. For $D \leq 0$, we have $3a^2-10a+3 \geq 0$, which gives $a \leq 1/3$ or $a \geq 3$. Since $a>1$, we take $a \geq 3$. In this case, $f(x) \geq 0$ for all $x \in \mathbb{R}$, so it's true for $x \geq 2$.
   • If $D > 0$: This means $1/3 < a < 3$. Combined with $a>1$, we have $1 < a < 3$. For $f(x) \geq 0$ for $x \geq 2$, the roots of $f(x)=0$ must satisfy $x_1 \leq x_2 \leq 2$. This requires two conditions:
     • $f(2) \geq 0 \implies 3a-7 \geq 0 \implies a \geq 7/3$.
     • Axis of symmetry $x_v = \frac{a+1}{2(a-1)} \leq 2 \implies a+1 \leq 4a-4 \implies 5 \leq 3a \implies a \geq 5/3$. Combining $1 < a < 3$, $a \geq 7/3$, and $a \geq 5/3$, we get $\frac{7}{3} \leq a < 3$.

   The union of solutions for $a>1$ is $[3, \infty) \cup [\frac{7}{3}, 3)$, which simplifies to $[\frac{7}{3}, \infty)$.

The exhaustive set of values for $a$ is $[\frac{7}{3}, \infty)$.